MathsCBSE Class 10 · Chapter 7

Coordinate Geometry Class 10: Concepts, Formulas and Board Questions

Updated 13 October 2026 · Fact-checked

Coordinate Geometry in Class 10 uses the coordinates of points to find lengths and divide line segments. You learn the distance formula and the section formula for internal division, including midpoints and points of trisection. Board questions ask you to find distances, test figures and collinearity, find unknown coordinates, and find ratios of division.

What Coordinate Geometry is about

Coordinate Geometry joins algebra and geometry. You place points on the Cartesian plane as ordered pairs $(x, y)$ . Then you answer geometric questions (how long is this side, is this a square, where does this line cross the axis) using only arithmetic and algebra. This chapter has two main tools. The distance formula finds the length between two points. The section formula finds the point that divides a segment internally in a given ratio. The midpoint and points of trisection are special cases of it.

The chapter links closely to others. The distance formula comes straight from Pythagoras theorem, and its converse lets you prove a triangle is right-angled. The section formula is proved using similar triangles (Triangles chapter). Finding an unknown coordinate usually ends in a linear or quadratic equation, so your algebra from Pair of Linear Equations and Quadratic Equations is used here too.

Board questions take a few clear shapes:

short MCQs on distances, midpoints and points on the axes
2- and 3-mark questions asking you to find an unknown coordinate, a ratio of division or points of trisection
longer questions asking you to prove that given points form a particular triangle or quadrilateral
case studies that place a park, a classroom seating plan or a map on a grid and ask distances and dividing points

Weightage: This chapter is Unit III, Coordinate Geometry, which carries 6 of the 80 theory marks.Source: CBSE Mathematics syllabus 2026-27, Unit III Coordinate Geometry

In real board papers: Standard, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 6 marks per paper; Basic, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 6 marks per paper. See the questions

Key concepts

The Cartesian plane: coordinates, quadrants and points on the axes

NCERT section 7.1 Introduction

You can locate every point in a plane by an ordered pair $(x, y)$ .

The x-coordinate (abscissa) is the signed distance from the y-axis.
The y-coordinate (ordinate) is the signed distance from the x-axis.

This is why the distance of $(x, y)$ from the x-axis is $∣ y ∣$ , not $∣ x ∣$ . Students often swap these.

The signs tell you the quadrant: $(+, +)$ in I, $(-, +)$ in II, $(-, -)$ in III, $(+, -)$ in IV. A point on the x-axis has $y = 0$ , so it is $(x, 0)$ . A point on the y-axis is $(0, y)$ . The origin is $(0, 0)$ .

The mirror image of $(x, y)$ in the x-axis is $(x, - y)$ . In the y-axis it is $(- x, y)$ .

You use these facts all the time: to write an unknown point on an axis, to check a sign, and to spot horizontal or vertical segments. The length of such a segment is simply a difference of coordinates.

Worked example

A point P lies in the second quadrant. It is 4 units from the x-axis and 3 units from the y-axis. Q is the mirror image of P in the x-axis, and O is the origin. Find the coordinates of P and Q, the length PQ and the length OP.

Try it first, then show the solutionSolution

Distance from the x-axis is $∣ y ∣$ , so $∣ y ∣ = 4$ . Distance from the y-axis is $∣ x ∣$ , so $∣ x ∣ = 3$ .
In the second quadrant $x < 0$ and $y > 0$ , so $P (- 3, 4)$ .
The image in the x-axis keeps $x$ and changes the sign of $y$ , so $Q (- 3, - 4)$ .
P and Q have the same x-coordinate, so PQ is vertical: $P Q = ∣4 - (- 4) ∣ = 8$ units.
$O P = (- 3)^{2} + 4^{2} = 9 + 16 = 25 = 5$ units.

Answer: $P (- 3, 4)$ , $Q (- 3, - 4)$ , $P Q = 8$ units, $O P = 5$ units.

The distance formula and why it works

NCERT section 7.2 Distance Formula

Take two points $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ . Draw a horizontal line through P and a vertical line through Q. They meet at $R (x_{2}, y_{1})$ , which makes a right triangle PRQ with the right angle at R.

The sides are $P R = ∣ x_{2} - x_{1} ∣$ and $QR = ∣ y_{2} - y_{1} ∣$ . By Pythagoras theorem, $P Q^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ , so $P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ .

Each difference is squared. So the order of the points does not matter, and a negative difference becomes positive. The subtraction itself must still be done carefully, for example $3 - (- 5) = 8$ .

Special cases save time:

From the origin, $O P = x^{2} + y^{2}$ .
If two points share the same y-coordinate, the distance is just $∣ x_{2} - x_{1} ∣$ .

Leave answers in simplest surd form, like $35$ units, and always write units.

Worked example

Find the perimeter of the triangle with vertices $A (0, 4)$ , $B (3, 0)$ and $C (3, 8)$ . State, with reason, what type of triangle it is, and check whether it is right-angled.

Try it first, then show the solutionSolution

$A B = (3 - 0)^{2} + (0 - 4)^{2} = 9 + 16 = 5$ units.
B and C have the same x-coordinate, so $B C = ∣8 - 0∣ = 8$ units.
$C A = (0 - 3)^{2} + (4 - 8)^{2} = 9 + 16 = 5$ units.
Perimeter $= 5 + 8 + 5 = 18$ units.
$A B = C A$ , so the triangle is isosceles.
Check for a right angle: $A B^{2} + C A^{2} = 25 + 25 = 50 \neq = 64 = B C^{2}$ , so it is not right-angled.

Answer: Perimeter $= 18$ units. The triangle is isosceles ( $A B = C A = 5$ units) and not right-angled.

Using distances to identify triangles and quadrilaterals

NCERT section 7.2 Distance Formula

Once you can find lengths, you can prove what shape a set of points makes. The method is always the same: find every side (and the diagonals if needed), then compare.

Triangle: three equal sides means equilateral, two equal sides means isosceles. If the squares of two sides add to the square of the third, it is right-angled (converse of Pythagoras theorem).
Quadrilateral ABCD (vertices taken in order):
four equal sides and equal diagonals: a square
four equal sides but unequal diagonals: a rhombus
opposite sides equal and diagonals equal: a rectangle
opposite sides equal only: a parallelogram

The common trap is stopping after the sides. Four equal sides only prove a rhombus. To get a square, you must also show the diagonals are equal.

Work with squared lengths such as $A B^{2}$ . This avoids surds until the end.

Worked example

Show that the points $A (2, 1)$ , $B (5, 3)$ , $C (3, 6)$ and $D (0, 4)$ , taken in order, are the vertices of a square.

Try it first, then show the solutionSolution

$A B^{2} = (5 - 2)^{2} + (3 - 1)^{2} = 9 + 4 = 13$
$B C^{2} = (3 - 5)^{2} + (6 - 3)^{2} = 4 + 9 = 13$
$C D^{2} = (0 - 3)^{2} + (4 - 6)^{2} = 9 + 4 = 13$
$D A^{2} = (2 - 0)^{2} + (1 - 4)^{2} = 4 + 9 = 13$
So $A B = B C = C D = D A = 13$ units: all four sides are equal.
Diagonals: $A C^{2} = (3 - 2)^{2} + (6 - 1)^{2} = 1 + 25 = 26$ and $B D^{2} = (0 - 5)^{2} + (4 - 3)^{2} = 25 + 1 = 26$ .
So $A C = B D = 26$ units: the diagonals are equal.
A quadrilateral with all sides equal and equal diagonals is a square.

Answer: All sides are $13$ units and both diagonals are $26$ units, so ABCD is a square.

Collinearity using the distance formula

NCERT section 7.2 Distance Formula

Three points A, B, C are collinear (on one straight line) exactly when the two shorter distances add up to the longest one. If B lies between A and C, then $A B + B C = A C$ .

Why does this work? If the points formed a real triangle, the sum of any two sides would be strictly greater than the third. Equality can only happen when the triangle has flattened into a line.

Steps:

Find $A B$ , $B C$ and $A C$ .
Identify the longest.
Check whether the other two add to it.

Simplify surds before adding. For example, $22 + 32 = 52$ , but you cannot add $8$ and $18$ to get $26$ .

There is a bonus. When $A B + B C = A C$ , the middle point B divides AC internally in the ratio $A B : B C$ . This links the distance formula to the section formula.

Worked example

Show that the points $A (- 1, - 1)$ , $B (2, 3)$ and $C (8, 11)$ are collinear. Also find the ratio in which B divides AC.

Try it first, then show the solutionSolution

$A B = (2 + 1)^{2} + (3 + 1)^{2} = 9 + 16 = 5$ units.
$B C = (8 - 2)^{2} + (11 - 3)^{2} = 36 + 64 = 10$ units.
$A C = (8 + 1)^{2} + (11 + 1)^{2} = 81 + 144 = 15$ units.
$A B + B C = 5 + 10 = 15 = A C$ , so A, B, C are collinear with B between A and C.
B divides AC in the ratio $A B : B C = 5 : 10 = 1 : 2$ .
Check with the section formula: $(\frac{1 ( 8 ) + 2 ( - 1 )}{3}, \frac{1 ( 11 ) + 2 ( - 1 )}{3}) = (2, 3)$ , which is B.

Answer: The points are collinear because $A B + B C = A C$ ( $5 + 10 = 15$ ). B divides AC in the ratio $1 : 2$ .

Finding unknown coordinates from distance conditions

NCERT section 7.2 Distance Formula

Many board questions give you a condition about lengths and ask for an unknown coordinate. Turn each condition into an equation.

Equidistant from A and B: write $P A = P B$ , then square both sides to get $P A^{2} = P B^{2}$ . Squaring removes the roots safely because lengths are never negative.
Point on an axis: a point on the x-axis is $(x, 0)$ and a point on the y-axis is $(0, y)$ . Writing this first removes one unknown.
Given distance: for a condition such as $P Q = 10$ , square both sides to get $P Q^{2} = 100$ .

In equidistance questions, the $x^{2}$ (or $y^{2}$ ) terms cancel when you expand, leaving a linear equation.

In a given-distance question, you usually get a square equal to a number, which gives two values. Keep both unless the question rules one out, for example by naming a quadrant.

Finally, substitute back to check.

Worked example

Find the point P on the y-axis which is equidistant from $A (5, - 2)$ and $B (- 3, 2)$ . Also find the length PA.

Try it first, then show the solutionSolution

P lies on the y-axis, so let $P (0, y)$ .
$P A = P B$ , so $P A^{2} = P B^{2}$ .
$(5 - 0)^{2} + (- 2 - y)^{2} = (- 3 - 0)^{2} + (2 - y)^{2}$
$25 + 4 + 4 y + y^{2} = 9 + 4 - 4 y + y^{2}$
$29 + 4 y = 13 - 4 y$ , so $8 y = - 16$ and $y = - 2$ .
So $P (0, - 2)$ , and $P A = (5 - 0)^{2} + (- 2 + 2)^{2} = 5$ units.
Check: $P B = (- 3)^{2} + (2 + 2)^{2} = 9 + 16 = 5$ units, equal to PA.

Answer: $P (0, - 2)$ and $P A = P B = 5$ units.

Section formula for internal division

NCERT section 7.3 Section Formula

Suppose P lies on segment AB with $A P : P B = m_{1} : m_{2}$ . Then P divides AB internally in that ratio. With $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ , the coordinates of P are $(\frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}, \frac{m _{1} y _{2} + m _{2} y _{1}}{m _{1} + m _{2}})$ .

Why? Draw perpendiculars from A, P, B to the x-axis, and horizontal lines through A and P. You get two similar triangles (AA similarity), so the horizontal and vertical steps are in the ratio $m_{1} : m_{2}$ . Solving $\frac{x - x _{1}}{x _{2} - x} = \frac{m _{1}}{m _{2}}$ gives the x-coordinate above. The y-coordinate works the same way.

Notice the cross pairing: $m_{1}$ , the part next to A, multiplies the coordinate of B. Think of it as a weighted average: the point is pulled towards the end with the bigger weight.

For internal division, P always lies between A and B. So its coordinates sit between theirs. That is a quick check on your answer.

Worked example

Find the coordinates of the point P which divides the line segment joining $A (- 2, 5)$ and $B (8, - 5)$ internally in the ratio $2 : 3$ . Verify your answer using the distance formula.

Try it first, then show the solutionSolution

Here $m_{1} = 2$ , $m_{2} = 3$ , $(x_{1}, y_{1}) = (- 2, 5)$ , $(x_{2}, y_{2}) = (8, - 5)$ .
$x = \frac{2 ( 8 ) + 3 ( - 2 )}{2 + 3} = \frac{16 - 6}{5} = 2$
$y = \frac{2 ( - 5 ) + 3 ( 5 )}{2 + 3} = \frac{- 10 + 15}{5} = 1$
So $P (2, 1)$ .
Verify: $A P = (2 + 2)^{2} + (1 - 5)^{2} = 32 = 42$ units.
$P B = (8 - 2)^{2} + (- 5 - 1)^{2} = 72 = 62$ units.
$A P : P B = 42 : 62 = 2 : 3$ , as required.

Answer: $P (2, 1)$

Midpoint formula and points of trisection

NCERT section 7.3 Section Formula

The midpoint is the special case $m_{1} : m_{2} = 1 : 1$ . The formula becomes $(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$ , which is simply the average of the two ends.

Points of trisection divide AB into three equal parts. There are two of them:

P with $A P : P B = 1 : 2$
Q with $A Q : QB = 2 : 1$

Use the section formula twice.

There is a quicker check. Find one third of the total step from A to B, which is $(\frac{x _{2} - x _{1}}{3}, \frac{y _{2} - y _{1}}{3})$ . Add it to A once to get P, and twice to get Q. Also, Q must be the midpoint of PB, which gives a third check.

In the same way, the points that divide a segment into four equal parts are at ratios $1 : 3$ , $1 : 1$ and $3 : 1$ .

Always state both trisection points. Giving only one loses marks.

Worked example

Find the coordinates of the points of trisection of the line segment joining $A (- 3, 4)$ and $B (6, - 2)$ . Show that one of them lies on the y-axis and the other on the x-axis.

Try it first, then show the solutionSolution

Let P and Q be the points of trisection, with $A P : P B = 1 : 2$ and $A Q : QB = 2 : 1$ .
For P: $x = \frac{1 ( 6 ) + 2 ( - 3 )}{3} = 0$ and $y = \frac{1 ( - 2 ) + 2 ( 4 )}{3} = \frac{6}{3} = 2$ , so $P (0, 2)$ .
For Q: $x = \frac{2 ( 6 ) + 1 ( - 3 )}{3} = \frac{9}{3} = 3$ and $y = \frac{2 ( - 2 ) + 1 ( 4 )}{3} = 0$ , so $Q (3, 0)$ .
P has x-coordinate 0, so it lies on the y-axis. Q has y-coordinate 0, so it lies on the x-axis.
Check: one third of the step from A to B is $(\frac{9}{3}, \frac{- 6}{3}) = (3, - 2)$ . Then $A + (3, - 2) = (0, 2)$ and $(0, 2) + (3, - 2) = (3, 0)$ .

Answer: The points of trisection are $P (0, 2)$ on the y-axis and $Q (3, 0)$ on the x-axis.

Finding the ratio of division: the k : 1 method

NCERT section 7.3 Section Formula

When the ratio is unknown, take it as $k : 1$ . This uses one unknown instead of two. The dividing point is then $(\frac{k x _{2} + x _{1}}{k + 1}, \frac{k y _{2} + y _{1}}{k + 1})$ .

Now use whatever you know about the point:

It lies on the x-axis: set the y-coordinate equal to 0.
It lies on the y-axis: set the x-coordinate equal to 0.
One coordinate is given, like $P (x, 5)$ : set that coordinate equal to 5.

Solve for k, then write the ratio in whole numbers. For example, $k = \frac{3}{2}$ means the ratio is $3 : 2$ . A positive k confirms internal division.

Then substitute k into the other coordinate to find the missing value. When both coordinates of the point are given, use the second one as a check: both must give the same k.

Worked example

Find the ratio in which the x-axis divides the line segment joining $A (4, - 2)$ and $B (- 1, 6)$ . Also find the point of division.

Try it first, then show the solutionSolution

Let the x-axis divide AB at P in the ratio $k : 1$ .
Then $P = (\frac{k ( - 1 ) + 4}{k + 1}, \frac{k ( 6 ) + ( - 2 )}{k + 1})$ .
P lies on the x-axis, so its y-coordinate is 0: $\frac{6 k - 2}{k + 1} = 0$ .
$6 k - 2 = 0$ , so $k = \frac{1}{3}$ . The ratio is $\frac{1}{3} : 1 = 1 : 3$ .
x-coordinate: $\frac{1 ( - 1 ) + 3 ( 4 )}{1 + 3} = \frac{11}{4}$ , so $P (\frac{11}{4}, 0)$ .
Check: y rises by 8 from A to B, and reaches 0 after rising 2, which is $\frac{1}{4}$ of the way. At that point $x = 4 - \frac{1}{4} (5) = \frac{11}{4}$ .

Answer: Ratio $1 : 3$ . The point of division is $(\frac{11}{4}, 0)$ .

Finding unknown vertices: the parallelogram and midpoint conditions

NCERT section 7.3 Section Formula

The section formula also works backwards. If you know the dividing point and the ratio, equate coordinates to find a missing end point or a missing value.

The most common use is the parallelogram. Its diagonals bisect each other, so in parallelogram ABCD the midpoint of AC equals the midpoint of BD. This gives two easy equations: $x_{A} + x_{C} = x_{B} + x_{D}$ and $y_{A} + y_{C} = y_{B} + y_{D}$ .

The vertices must be in order: A is opposite C, and B is opposite D.

The same idea finds a missing end. If M is the midpoint of AB and you know A and M, then $B = (2 x_{M} - x_{A}, 2 y_{M} - y_{A})$ .

A rhombus, rectangle and square are all parallelograms, so the midpoint condition works for them too.

Always verify with a side check, such as AB and DC having equal steps in x and in y.

Worked example

ABCD is a parallelogram with vertices $A (1, 2)$ , $B (4, y)$ , $C (x, 6)$ and $D (3, 5)$ , taken in order. Find x and y, and the length of side AB.

Try it first, then show the solutionSolution

The diagonals of a parallelogram bisect each other, so the midpoint of AC = the midpoint of BD.
Midpoint of AC $= (\frac{1 + x}{2}, \frac{2 + 6}{2}) = (\frac{1 + x}{2}, 4)$
Midpoint of BD $= (\frac{4 + 3}{2}, \frac{y + 5}{2}) = (\frac{7}{2}, \frac{y + 5}{2})$
Equate x-coordinates: $\frac{1 + x}{2} = \frac{7}{2}$ , so $x = 6$ .
Equate y-coordinates: $4 = \frac{y + 5}{2}$ , so $y = 3$ .
$A B = (4 - 1)^{2} + (3 - 2)^{2} = 9 + 1 = 10$ units.
Check: $D C = (6 - 3)^{2} + (6 - 5)^{2} = 10$ units, equal to AB as expected.

Answer: $x = 6$ , $y = 3$ , and $A B = 10$ units.

Maths in real life: Coordinate Geometry

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

The distance formula and why it works

Imagine a coach marks your school cricket ground on a grid in metres, with the striker's stumps at the origin $O (0, 0)$ . Fielder A stands at $(24, 18)$ and fielder B stands at $(- 20, 21)$ . A third fielder C stands on the straight line from the stumps to A, at the point that divides OA so that $O C : C A = 1 : 5$ . The coach wants to know who is closer to the stumps, and who is closer to C for a quick relay throw.

Find the distance of each of A and B from the stumps, and say which is closer and by how many metres. Then find the position of C, and decide which of A and B is nearer to C, and by how many metres.

Try it first, then show the solutionSolution

Distance from the origin: $O P = x^{2} + y^{2}$ .
$O A = 2 4^{2} + 1 8^{2} = 576 + 324 = 900 = 30$ m.
$O B = (- 20)^{2} + 2 1^{2} = 400 + 441 = 841 = 29$ m.
$30 - 29 = 1$ , so B is closer to the stumps by 1 m.
C divides OA internally in the ratio $1 : 5$ , so $x = \frac{1 ( 24 ) + 5 ( 0 )}{6} = 4$ and $y = \frac{1 ( 18 ) + 5 ( 0 )}{6} = 3$ . So C is $(4, 3)$ .
$C A = (24 - 4)^{2} + (18 - 3)^{2} = 400 + 225 = 625 = 25$ m.
$C B = (- 20 - 4)^{2} + (21 - 3)^{2} = 576 + 324 = 900 = 30$ m.
Check: $O C = 16 + 9 = 5$ m and $O C + C A = 5 + 25 = 30 = O A$ , as expected.
$30 - 25 = 5$ , so A is nearer to C by 5 m.

Answer: $O A = 30$ m and $O B = 29$ m, so B is closer to the stumps by 1 m. C is at $(4, 3)$ , with $C A = 25$ m and $C B = 30$ m, so A is nearer to C by 5 m.

Memory hook: Closer to the stumps is not closer to everyone: square the across, square the up, add, then take the root for each pair.

Using distances to identify triangles and quadrilaterals

Suppose your family draws a Diwali rangoli on a courtyard marked in a grid where 1 unit is 10 cm. The four corners of the outer border are at $A (1, 2)$ , $B (5, 5)$ , $C (8, 1)$ and $D (4, - 2)$ , taken in order. Your cousin says it looks like a square but is not sure.

Use the distance formula to prove whether ABCD is a square. If it is, find the area of the rangoli in square centimetres.

Try it first, then show the solutionSolution

$A B^{2} = (5 - 1)^{2} + (5 - 2)^{2} = 16 + 9 = 25$ , so $A B = 5$ units.
$B C^{2} = (8 - 5)^{2} + (1 - 5)^{2} = 9 + 16 = 25$ , so $B C = 5$ units.
$C D^{2} = (4 - 8)^{2} + (- 2 - 1)^{2} = 16 + 9 = 25$ , so $C D = 5$ units.
$D A^{2} = (1 - 4)^{2} + (2 - (- 2))^{2} = 9 + 16 = 25$ , so $D A = 5$ units.
All four sides are equal, so ABCD is at least a rhombus.
Diagonal $A C^{2} = (8 - 1)^{2} + (1 - 2)^{2} = 49 + 1 = 50$ .
Diagonal $B D^{2} = (4 - 5)^{2} + (- 2 - 5)^{2} = 1 + 49 = 50$ , so $A C = B D$ .
Equal sides and equal diagonals, so ABCD is a square.
Side $= 5 \times 10 = 50$ cm, so area $= 50 \times 50 = 2500$ cm².

Answer: ABCD is a square with side 50 cm, and the rangoli covers 2500 cm².

Memory hook: A rangoli square needs two checks: equal sides, then equal diagonals.

Collinearity using the distance formula

Imagine your eco-club plans to plant three saplings in one straight row along the school garden. On the garden plan, marked in metres, the first two saplings are at $A (1, 1)$ and $B (4, 5)$ , and the third must go at $C (10, q)$ so that all three are in line. The pit-digging team will start at the point S that divides AB so that $A S : S B = 1 : 2$ .

Find q so that A, B and C are collinear, and prove the collinearity using the distance formula. Then find the starting point S and its distance from C.

Try it first, then show the solutionSolution

Since $1 < 4 < 10$ , B lies between A and C. Let B divide AC in the ratio $k : 1$ .
x-coordinate: $\frac{10 k + 1}{k + 1} = 4$ , so $10 k + 1 = 4 k + 4$ and $k = \frac{1}{2}$ . The ratio is $1 : 2$ .
y-coordinate: $\frac{1 ( q ) + 2 ( 1 )}{1 + 2} = 5$ , so $q + 2 = 15$ and $q = 13$ . So C is $(10, 13)$ .
$A B = (4 - 1)^{2} + (5 - 1)^{2} = 9 + 16 = 5$ m.
$B C = (10 - 4)^{2} + (13 - 5)^{2} = 36 + 64 = 10$ m.
$A C = (10 - 1)^{2} + (13 - 1)^{2} = 81 + 144 = 15$ m.
$A B + B C = 5 + 10 = 15 = A C$ , so A, B, C are collinear.
S divides AB in the ratio $1 : 2$ : $x = \frac{1 ( 4 ) + 2 ( 1 )}{3} = 2$ and $y = \frac{1 ( 5 ) + 2 ( 1 )}{3} = \frac{7}{3}$ . So S is $(2, \frac{7}{3})$ .
$S C = (10 - 2)^{2} + (13 - \frac{7}{3})^{2} = 64 + \frac{1024}{9} = \frac{1600}{9} = \frac{40}{3}$ m.
Check: $A S = \frac{1}{3} A B = \frac{5}{3}$ m, and $A C - A S = 15 - \frac{5}{3} = \frac{40}{3}$ m, which matches SC.

Answer: $q = 13$ , so C is $(10, 13)$ . The saplings are collinear because $A B + B C = 5 + 10 = 15 = A C$ (in metres). The team starts at $S (2, \frac{7}{3})$ , which is $\frac{40}{3}$ m, that is $13 \frac{1}{3}$ m, from C.

Memory hook: Saplings in a row: the two short gaps add up to the full row, and the middle sapling fixes the ratio.

Finding unknown coordinates from distance conditions

Suppose a groundsman places a sprinkler exactly midway between two water taps at $P (1, 7)$ and $Q (5, 1)$ on a sports field marked in metres, where the straight side line of the field is the x-axis. The sprinkler throws water exactly 5 m in every direction. He wants marker cones on the side line at the points where the spray just reaches, and ropes from the sprinkler to each cone mark out a triangular patch.

Find the position of the sprinkler and the positions of the two cones. Then, using distances only, show that the triangle formed by the sprinkler and the two cones is isosceles and find its area.

Try it first, then show the solutionSolution

Sprinkler S is the midpoint of PQ: $S = (\frac{1 + 5}{2}, \frac{7 + 1}{2}) = (3, 4)$ .
A cone on the x-axis has coordinates $(x, 0)$ , and its distance from S is 5 m.
$(x - 3)^{2} + (0 - 4)^{2} = 25$ , so $(x - 3)^{2} = 9$ .
$x - 3 = \pm 3$ , so $x = 6$ or $x = 0$ . The cones are at $A (0, 0)$ and $B (6, 0)$ .
$S A = 9 + 16 = 5$ m and $S B = 9 + 16 = 5$ m, so $S A = S B$ and triangle SAB is isosceles.
Base AB lies on the x-axis: $A B = ∣6 - 0∣ = 6$ m.
The height of the triangle is the distance of S from the x-axis, which is $∣ y ∣ = 4$ m. Check: the foot of this height is $(3, 0)$ , the midpoint of AB, as it must be in an isosceles triangle.
Area $= \frac{1}{2} \times 6 \times 4 = 12$ m².

Answer: The sprinkler is at $(3, 4)$ and the cones are at $(0, 0)$ and $(6, 0)$ . The triangle is isosceles with $S A = S B = 5$ m, and its area is 12 m².

Memory hook: Sprinkler between taps: midpoint first, then put $y = 0$ on the side line and keep both answers.

Section formula for internal division

Imagine a family drives on a straight highway from town A to town B. On a map grid where 1 unit is 10 km, A is at $(- 2, 3)$ and B is at $(8, - 2)$ . Two toll plazas stand near the highway at $T_{1} (5, 5)$ and $T_{2} (6, - 2)$ , and the family wants a fuel stop P on the highway that is equally far from both plazas.

Find the ratio in which P divides AB, the coordinates of P, and the distance of P from each toll plaza and from town A, in kilometres.

Try it first, then show the solutionSolution

Let P divide AB in the ratio $k : 1$ . Then $P = (\frac{8 k - 2}{k + 1}, \frac{- 2 k + 3}{k + 1})$ .
$P T_{1} = P T_{2}$ , so $P T_{1}^{2} = P T_{2}^{2}$ : $(x - 5)^{2} + (y - 5)^{2} = (x - 6)^{2} + (y + 2)^{2}$ .
Expand and cancel $x^{2}$ and $y^{2}$ : $- 10 x - 10 y + 50 = - 12 x + 4 y + 40$ , so $2 x - 14 y + 10 = 0$ , that is $x - 7 y + 5 = 0$ .
Substitute P and multiply by $k + 1$ : $(8 k - 2) - 7 (- 2 k + 3) + 5 (k + 1) = 0$ , so $27 k - 18 = 0$ and $k = \frac{2}{3}$ .
So $A P : P B = 2 : 3$ , and $x = \frac{2 ( 8 ) + 3 ( - 2 )}{5} = 2$ , $y = \frac{2 ( - 2 ) + 3 ( 3 )}{5} = 1$ . P is $(2, 1)$ .
$P T_{1} = (5 - 2)^{2} + (5 - 1)^{2} = 5$ units and $P T_{2} = (6 - 2)^{2} + (- 2 - 1)^{2} = 5$ units, that is 50 km each.
$A P = (2 + 2)^{2} + (1 - 3)^{2} = 20 = 25$ units, that is $205$ km.
Check: $P B = (8 - 2)^{2} + (- 2 - 1)^{2} = 45 = 35$ units, so $A P : P B = 2 : 3$ .

Answer: P divides AB in the ratio $2 : 3$ , so the fuel stop is at $(2, 1)$ . It is 50 km from each toll plaza and $205$ km from town A.

Memory hook: Fuel stop between two plazas: equal distances give a line, and $k : 1$ finds where the highway meets it.

Finding the ratio of division: the k : 1 method

Suppose a straight footpath runs across a city park from gate P at $(- 3, 6)$ to gate Q at $(5, - 2)$ , on a grid where 1 unit is 10 m. A main road runs along the x-axis. The council wants a zebra crossing exactly where the footpath meets the road, and a bench on the footpath at the point B with $P B : B Q = 1 : 3$ .

Find the ratio in which the road divides the footpath PQ and the position of the zebra crossing. Then find the position of the bench and the walking distance along the footpath from the bench to the crossing, in metres.

Try it first, then show the solutionSolution

Let the road divide PQ in the ratio $k : 1$ . The point is $(\frac{5 k - 3}{k + 1}, \frac{- 2 k + 6}{k + 1})$ .
On the x-axis the y-coordinate is 0: $- 2 k + 6 = 0$ , so $k = 3$ , which is positive, so the division is internal. The ratio is $3 : 1$ .
x-coordinate: $\frac{5 ( 3 ) - 3}{3 + 1} = \frac{12}{4} = 3$ , so the zebra crossing is at $Z (3, 0)$ .
Bench B divides PQ in the ratio $1 : 3$ : $x = \frac{1 ( 5 ) + 3 ( - 3 )}{4} = - 1$ and $y = \frac{1 ( - 2 ) + 3 ( 6 )}{4} = 4$ . So B is $(- 1, 4)$ .
$B Z = (3 + 1)^{2} + (0 - 4)^{2} = 32 = 42$ units, that is $402$ m.
Check: $P Q = 64 + 64 = 82$ units. The bench is $\frac{1}{4}$ of the way from P and the crossing is $\frac{3}{4}$ of the way, so $B Z = \frac{1}{2} P Q = 42$ units.

Answer: The road divides the footpath in the ratio $3 : 1$ at the zebra crossing $(3, 0)$ . The bench is at $(- 1, 4)$ , which is $402$ m from the crossing.

Memory hook: Footpath meets the road on the x-axis: take $k : 1$ and set $y = 0$ . The bench and the crossing are then just two section points on one line.

Topics in this chapter

Coordinate Geometry formula sheet

Distance formula: $A B = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$; Length between $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ . Use it for sides, diagonals, equidistance and collinearity.
Distance from the origin: $O P = x^{2} + y^{2}$; Distance of $P (x, y)$ from $O (0, 0)$ . A quick special case of the distance formula.
Horizontal and vertical segments: $A B = ∣ x_{2} - x_{1} ∣ (same y), A B = ∣ y_{2} - y_{1} ∣ (same x)$; Use this when two points share a coordinate. It saves working.
Distance of a point from the axes: $from x-axis = ∣ y ∣, from y-axis = ∣ x ∣$; For a point $(x, y)$ . Do not swap them.
Points on the axes: $(x, 0) on the x-axis, (0, y) on the y-axis$; Write the unknown point in this form first in equidistance and ratio questions.
Quadrant signs: $I (+, +), II (-, +), III (-, -), IV (+, -)$; Use this to choose the right value when a question names the quadrant.
Mirror images in the axes: $(x, y) \to (x, - y) in x-axis, (x, y) \to (- x, y) in y-axis$; Useful in case studies and quick MCQs.
Section formula (internal division): $P = (\frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}, \frac{m _{1} y _{2} + m _{2} y _{1}}{m _{1} + m _{2}})$; P divides AB internally with $A P : P B = m_{1} : m_{2}$ . Remember the cross pairing: $m_{1}$ goes with B.
Section formula in k : 1 form: $P = (\frac{k x _{2} + x _{1}}{k + 1}, \frac{k y _{2} + y _{1}}{k + 1})$; Use this when the ratio is unknown. Solve for k, then write the ratio as $k : 1$ in whole numbers.
Midpoint formula: $M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$; The ratio is $1 : 1$ . Use it for bisected diagonals and missing ends.
Missing end from the midpoint: $B = (2 x_{M} - x_{A}, 2 y_{M} - y_{A})$; Use this when M is the midpoint of AB and A is known.
Points of trisection: $P = (\frac{x _{2} + 2 x _{1}}{3}, \frac{y _{2} + 2 y _{1}}{3}), Q = (\frac{2 x _{2} + x _{1}}{3}, \frac{2 y _{2} + y _{1}}{3})$; Ratios $1 : 2$ and $2 : 1$ from A. Always give both points.
Collinearity condition (distance method): $A B + B C = A C$; Here AC is the longest of the three distances. If the condition holds, A, B, C lie on one line with B between A and C.
Right-angled triangle (converse of Pythagoras theorem): $A B^{2} + B C^{2} = A C^{2} \Rightarrow ∠ B = 9 0^{\circ}$; Use squared lengths to test for a right angle.
Square: $A B = B C = C D = D A and A C = B D$; Four equal sides AND equal diagonals. Four equal sides alone only prove a rhombus.
Rhombus: $A B = B C = C D = D A and A C \neq = B D$; Four equal sides with unequal diagonals.
Rectangle: $A B = C D, B C = D A and A C = B D$; Opposite sides equal with equal diagonals.
Parallelogram (diagonals bisect each other): $x_{A} + x_{C} = x_{B} + x_{D}, y_{A} + y_{C} = y_{B} + y_{D}$; This means the midpoint of AC = the midpoint of BD, with vertices in order. Use it to find a missing vertex. Opposite sides equal also proves a parallelogram.

How the board asks Coordinate Geometry

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Find an unknown coordinate from a given distance	For example: if the distance between $(x, - 1)$ and $(3, 2)$ is 5 units, find x. Sometimes the question names a quadrant so that only one value is valid.	Square both sides of the distance equation to remove the root. Simplify to a square equal to a number, or to a quadratic. Keep both roots, then reject one only if a condition rules it out. Substitute back to check.
Point on an axis equidistant from two points	Find the point on the x-axis (or y-axis) equidistant from A and B. Sometimes it is followed by a request for that common distance.	Write the point as $(x, 0)$ or $(0, y)$ . Set $P A^{2} = P B^{2}$ , expand, and cancel the squared unknown to get a linear equation. Find the point, then the distance.
Prove the type of triangle or quadrilateral	Show that given points are the vertices of a square, rhombus, rectangle, parallelogram, isosceles, equilateral or right-angled triangle. Sometimes you must name the figure yourself.	Find all sides as squared lengths, then the diagonals for quadrilaterals. Compare them against the exact conditions. State the conclusion with the reason, for example: all sides equal and diagonals equal, so it is a square.
Check collinearity	Determine whether three points are collinear. Sometimes the question also asks which point lies between the others, or in what ratio the middle point divides the segment.	Find the three distances in simplest surd form. Add the two smaller ones and compare with the largest. If they are equal, the points are collinear, and the middle point divides the segment in the ratio of the two smaller distances.
Point dividing a segment in a given ratio, midpoint, trisection	Find the point dividing AB in the ratio $m_{1} : m_{2}$ . Or find the points of trisection, or the points dividing AB into four equal parts.	Write $m_{1}$ , $m_{2}$ , $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ clearly first. Apply the section formula with $m_{1}$ paired with B. For trisection, use both $1 : 2$ and $2 : 1$ . Check that the answer lies between A and B.
Find the ratio in which a point or an axis divides a segment	In what ratio does the x-axis (or y-axis, or a given point $P (x, 5)$ ) divide the segment AB? Also find the missing coordinate.	Take the ratio $k : 1$ and write P using the section formula. Set the known coordinate equal to its value, solve for k, and convert it to a whole-number ratio. Substitute k into the other coordinate to complete P.
Missing vertex or parameter using the midpoint condition	Three vertices of a parallelogram are given (or vertices with unknowns), and you must find the fourth vertex or the unknowns. Or: the midpoint of AB is given, find B.	Equate the midpoint of AC with the midpoint of BD, taking the vertices in order. This gives one equation for x and one for y. Verify with a side comparison.
Case study on a grid	A map, park, classroom seating plan or sports field is placed on coordinate axes. Sub-parts ask distances between objects, the midpoint where something is placed, a point dividing a path, or the shape formed by positions.	First write the coordinates of every named object from the description. Then treat each sub-part as a standard distance or section formula question. Give units such as metres in every answer.

Tricks to remember Coordinate Geometry

Short memory aids. Say each one aloud once, then use it on a question the same day.

Section formula: cross the pairs
In $m_{1} : m_{2}$ , each ratio number multiplies the far end's coordinate: $m_{1}$ with $x_{2}$ , $m_{2}$ with $x_{1}$ . Say it as: the part near A goes with B.
Use it when: Stops the most common section formula error when you write the numerator.
Square = rhombus + equal diagonals
Sides tell you rhombus. Diagonals upgrade it to a square. Opposite sides tell you parallelogram. Diagonals upgrade it to a rectangle.
Use it when: Reminds you to calculate the diagonals before concluding a square or rectangle.
Unknown ratio? Take k : 1
One letter, one equation. Make the known coordinate equal to its value, solve for k, then clear fractions: $k = \frac{2}{5}$ gives $2 : 5$ .
Use it when: Every 'find the ratio' question, including those where an axis cuts the segment.
Internal means in between
For internal division, each coordinate of the answer must lie between the matching coordinates of A and B. A midpoint is just the average.
Use it when: A 10-second check that catches sign and pairing slips before you move on.
Axis point: the axis name keeps its letter
On the x-axis only x survives: $(x, 0)$ . On the y-axis only y survives: $(0, y)$ . Distance from the x-axis is the other coordinate, $∣ y ∣$ .
Use it when: Writing unknown points on axes and answering distance-from-axis MCQs.

Coordinate Geometry: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Unknowns from a midpoint (MCQ) · 1 mark · MCQBoard level

M(1, 2a + 1) is the midpoint of the line segment joining $A (2 a, 4)$ and $B (- 2, 3 b)$ . The values of a and b are:
(a) $a = 2, b = 2$
(b) $a = 2, b = 1$
(c) $a = 1, b = 2$
(d) $a = 3, b = 2$

Try it first, then show the solutionSolution

x-coordinate of the midpoint: $\frac{2 a + ( - 2 )}{2} = 1$ , so $2 a - 2 = 2$ and $a = 2$ .
So the y-coordinate of M is $2 a + 1 = 5$ .
y-coordinate of the midpoint: $\frac{4 + 3 b}{2} = 5$ , so $3 b = 6$ and $b = 2$ .
Check: with $A (4, 4)$ and $B (- 2, 6)$ , the midpoint is $(1, 5)$ , and $2 a + 1 = 5$ . It matches.

Answer: (a) $a = 2, b = 2$

Example 2: Ratio of division by the y-axis (assertion-reason) · 1 mark · Assertion-reasonBoard level

Assertion (A): The y-axis divides the line segment joining $(4, - 3)$ and $(- 6, 5)$ internally in the ratio $2 : 3$ .
Reason (R): Every point on the y-axis has x-coordinate 0.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

R is true: a point on the y-axis has the form $(0, y)$ .
Let the y-axis divide the segment in the ratio $k : 1$ . Then $\frac{- 6 k + 4}{k + 1} = 0$ , because of R.
$- 6 k + 4 = 0$ , so $k = \frac{2}{3}$ . The ratio is $2 : 3$ , so A is true.
We used R directly to find the ratio, so R correctly explains A.

Answer: (a) Both A and R are true and R is the correct explanation of A.

Example 3: Point from a given fraction of the segment · 3 marks · Short answerBoard level

A and B are the points $(3, - 4)$ and $(- 5, 8)$ . P is a point on AB such that $A P = \frac{1}{4} A B$ . Find the coordinates of P and verify your answer using the distance formula.

Try it first, then show the solutionSolution

$A P = \frac{1}{4} A B$ means $A P : A B = 1 : 4$ , so $A P : P B = 1 : 3$ . (1 mark)
Take $m_{1} = 1$ and $m_{2} = 3$ . Then $x = \frac{1 ( - 5 ) + 3 ( 3 )}{1 + 3} = \frac{4}{4} = 1$ . (½ mark)
$y = \frac{1 ( 8 ) + 3 ( - 4 )}{4} = \frac{- 4}{4} = - 1$ . So P is $(1, - 1)$ . (½ mark)
Verify: $A P = (1 - 3)^{2} + (- 1 + 4)^{2} = 4 + 9 = 13$ . (½ mark)
$A B = (- 5 - 3)^{2} + (8 + 4)^{2} = 64 + 144 = 208 = 413$ . So $A P = \frac{1}{4} A B$ , as required. (½ mark)

Answer: $P (1, - 1)$ . This is checked by $A P = 13$ and $A B = 413$ .

Example 4: Equidistant points and a point on the y-axis · 3 marks · Short answerBoard level

Find a relation between x and y such that the point $P (x, y)$ is equidistant from $A (2, - 1)$ and $B (- 4, 3)$ . Hence find the point on the y-axis which is equidistant from A and B.

Try it first, then show the solutionSolution

$P A = P B$ gives $P A^{2} = P B^{2}$ : $(x - 2)^{2} + (y + 1)^{2} = (x + 4)^{2} + (y - 3)^{2}$ . (1 mark)
Expand. The $x^{2}$ and $y^{2}$ terms cancel: $- 4 x + 4 + 2 y + 1 = 8 x + 16 - 6 y + 9$ .
$- 4 x + 2 y + 5 = 8 x - 6 y + 25$ , so $12 x - 8 y + 20 = 0$ , which is $3 x - 2 y + 5 = 0$ . (1 mark)
On the y-axis $x = 0$ , so $- 2 y + 5 = 0$ and $y = \frac{5}{2}$ . The point is $(0, \frac{5}{2})$ . (½ mark)
Check: distance squared to A is $4 + (\frac{7}{2})^{2} = \frac{65}{4}$ , and to B is $16 + (\frac{1}{2})^{2} = \frac{65}{4}$ . They are equal. (½ mark)

Answer: The relation is $3 x - 2 y + 5 = 0$ . The point on the y-axis is $(0, \frac{5}{2})$ .

Example 5: Rhombus, not a square, and its area · 5 marks · Long answerBoard level

Show that $A (- 3, 2)$ , $B (1, 5)$ , $C (5, 2)$ and $D (1, - 1)$ , taken in order, are the vertices of a rhombus that is not a square. Find its area, and show that its diagonals bisect each other.

Try it first, then show the solutionSolution

$A B = (1 + 3)^{2} + (5 - 2)^{2} = 16 + 9 = 5$ and $B C = (5 - 1)^{2} + (2 - 5)^{2} = 5$ . (1 mark)
$C D = (1 - 5)^{2} + (- 1 - 2)^{2} = 5$ and $D A = (- 3 - 1)^{2} + (2 + 1)^{2} = 5$ . All four sides are equal. (1 mark)
Diagonals: $A C = (5 + 3)^{2} + 0^{2} = 8$ and $B D = 0^{2} + (- 1 - 5)^{2} = 6$ . (1 mark)
All sides are equal but the diagonals are unequal, so ABCD is a rhombus and not a square. (½ mark)
Area of a rhombus $= \frac{1}{2} \times d_{1} \times d_{2} = \frac{1}{2} \times 8 \times 6 = 24$ square units. (1 mark)
Midpoint of AC $= (\frac{- 3 + 5}{2}, \frac{2 + 2}{2}) = (1, 2)$ . Midpoint of BD $= (\frac{1 + 1}{2}, \frac{5 - 1}{2}) = (1, 2)$ . They are the same point, so the diagonals bisect each other. (½ mark)

Answer: ABCD is a rhombus with side 5 units and diagonals 8 and 6 units, so it is not a square. Its area is 24 square units, and both diagonals have midpoint $(1, 2)$ .

Example 6: Triangle from the midpoints of its sides · 5 marks · Long answerHarder board level

The midpoints of the sides BC, CA and AB of $△ A B C$ are $D (3, 4)$ , $E (- 1, 2)$ and $F (1, - 2)$ respectively. Find the vertices A, B and C. Find the length of the median AD, and show that $△ A B C$ is a right-angled isosceles triangle.

Try it first, then show the solutionSolution

Let $A (x_{1}, y_{1})$ , $B (x_{2}, y_{2})$ and $C (x_{3}, y_{3})$ . From the midpoint formula: $x_{2} + x_{3} = 6$ , $y_{2} + y_{3} = 8$ ; $x_{3} + x_{1} = - 2$ , $y_{3} + y_{1} = 4$ ; $x_{1} + x_{2} = 2$ , $y_{1} + y_{2} = - 4$ . (1 mark)
Add the three x-equations: $2 (x_{1} + x_{2} + x_{3}) = 6$ , so $x_{1} + x_{2} + x_{3} = 3$ . Similarly $y_{1} + y_{2} + y_{3} = 4$ . (½ mark)
$x_{1} = 3 - 6 = - 3$ and $y_{1} = 4 - 8 = - 4$ , so $A (- 3, - 4)$ . (½ mark)
$x_{2} = 3 - (- 2) = 5$ and $y_{2} = 4 - 4 = 0$ , so $B (5, 0)$ . (½ mark)
$x_{3} = 3 - 2 = 1$ and $y_{3} = 4 - (- 4) = 8$ , so $C (1, 8)$ . Check: the midpoint of BC is $(3, 4) = D$ . (½ mark)
Median $A D = (3 + 3)^{2} + (4 + 4)^{2} = 36 + 64 = 10$ units. (1 mark)
$A B = 8^{2} + 4^{2} = 80$ , $B C = (- 4)^{2} + 8^{2} = 80$ and $C A = (- 4)^{2} + (- 12)^{2} = 160$ . Since $A B = B C$ and $A B^{2} + B C^{2} = 160 = C A^{2}$ , the triangle is isosceles and right-angled at B. (1 mark)

Answer: $A (- 3, - 4)$ , $B (5, 0)$ and $C (1, 8)$ . The median AD is 10 units. $A B = B C = 45$ and $A B^{2} + B C^{2} = C A^{2}$ , so $△ A B C$ is right-angled and isosceles with $∠ B = 9 0^{\circ}$ .

Example 7: Find the error: ratio cut by the y-axis · 3 marks · Short answerBoard level

A student was asked for the ratio in which the y-axis divides the segment joining $A (- 2, 5)$ and $B (4, - 1)$ . The student took the ratio as $k : 1$ and wrote: $\frac{k ( - 2 ) + 4}{k + 1} = 0$ , so $k = 2$ and the ratio is $2 : 1$ . Find the student's mistake, then give the correct ratio and the point of intersection.

Try it first, then show the solutionSolution

Mistake: in $\frac{k x _{2} + x _{1}}{k + 1}$ , k must multiply the x-coordinate of B (the second point), which is 4. The student multiplied A's coordinate, $- 2$ , by k instead. (1 mark)
Correct equation: $\frac{4 k + ( - 2 )}{k + 1} = 0$ , so $4 k = 2$ and $k = \frac{1}{2}$ . The ratio is $1 : 2$ . (1 mark)
y-coordinate: $\frac{1 ( - 1 ) + 2 ( 5 )}{3} = \frac{9}{3} = 3$ . The point is $(0, 3)$ . Check: AP is one-third of AB, and $x = - 2 + \frac{1}{3} (6) = 0$ . (1 mark)

Answer: The student swapped the roles of A and B. The y-axis divides AB in the ratio $1 : 2$ at the point $(0, 3)$ .

Example 8: Placing a mobile tower equidistant from three villages · 5 marks · Long answerBoard level

On a district map, 1 unit = 1 km. Three villages are at $A (6, 6)$ , $B (- 1, 5)$ and $C (3, - 3)$ . A mobile tower T is to be built at the same distance from all three villages. Find the position of T and its distance from each village.

Try it first, then show the solutionSolution

Let $T (x, y)$ . From $T A^{2} = T B^{2}$ : $(x - 6)^{2} + (y - 6)^{2} = (x + 1)^{2} + (y - 5)^{2}$ . (½ mark)
Expand and cancel the squares: $- 12 x - 12 y + 72 = 2 x - 10 y + 26$ , so $14 x + 2 y = 46$ , which is $7 x + y = 23$ . (1 mark)
From $T B^{2} = T C^{2}$ : $(x + 1)^{2} + (y - 5)^{2} = (x - 3)^{2} + (y + 3)^{2}$ . (½ mark)
$2 x - 10 y + 26 = - 6 x + 6 y + 18$ , so $8 x - 16 y + 8 = 0$ , which is $x = 2 y - 1$ . (1 mark)
Substitute: $7 (2 y - 1) + y = 23$ , so $15 y = 30$ . This gives $y = 2$ and $x = 3$ . So T is $(3, 2)$ . (1 mark)
$T A = 9 + 16 = 5$ , $T B = 16 + 9 = 5$ and $T C = 0 + 25 = 5$ . All three are equal, which confirms the answer. (1 mark)

Answer: The tower should be at $T (3, 2)$ , 5 km from each village.

Example 9: Unknown vertex of a right-angled triangle, then a square · 5 marks · Long answerHarder board level

The triangle with vertices $A (k, 2)$ , $B (6, - 2)$ and $C (2, - 5)$ is right-angled at B. Find k. Show that the triangle is also isosceles, and find the point D such that ABCD, taken in order, is a square.

Try it first, then show the solutionSolution

$A B^{2} = (6 - k)^{2} + (- 2 - 2)^{2} = (6 - k)^{2} + 16$ . (½ mark)
$B C^{2} = (2 - 6)^{2} + (- 5 + 2)^{2} = 16 + 9 = 25$ . (½ mark)
$A C^{2} = (2 - k)^{2} + (- 5 - 2)^{2} = (2 - k)^{2} + 49$ . (½ mark)
The right angle is at B, so AC is the hypotenuse: $A B^{2} + B C^{2} = A C^{2}$ , which gives $(6 - k)^{2} + 41 = (2 - k)^{2} + 49$ . (½ mark)
$36 - 12 k + k^{2} + 41 = 4 - 4 k + k^{2} + 49$ , so $77 - 12 k = 53 - 4 k$ , giving $8 k = 24$ and $k = 3$ . (1 mark)
With $k = 3$ : $A B^{2} = 9 + 16 = 25$ , so $A B = B C = 5$ units, and $A C^{2} = 1 + 49 = 50$ . Check: $25 + 25 = 50$ . So the triangle is right-angled and isosceles. (½ mark)
The diagonals of a square bisect each other, so the midpoint of AC = the midpoint of BD. Then $D = (3 + 2 - 6, 2 + (- 5) - (- 2)) = (- 1, - 1)$ . (1 mark)
Check: $A D^{2} = (- 1 - 3)^{2} + (- 1 - 2)^{2} = 25$ , $C D^{2} = (- 1 - 2)^{2} + (- 1 + 5)^{2} = 25$ and $B D^{2} = (- 1 - 6)^{2} + (- 1 + 2)^{2} = 50 = A C^{2}$ . All sides are 5 units and the diagonals are equal, so ABCD is a square. (½ mark)

Answer: $k = 3$ . $A B = B C = 5$ units with $A B^{2} + B C^{2} = A C^{2}$ , so the triangle is right-angled and isosceles. $D (- 1, - 1)$ makes ABCD a square.

Example 10: Parallelogram with an unknown vertex, and a diagonal cut by the x-axis · 5 marks · Long answerHarder board level

ABCD is a parallelogram with vertices $A (- 4, - 2)$ , $B (2, - 1)$ , $C (4, y)$ and $D (x, 2)$ , taken in order. (i) Find x and y. (ii) Find the ratio in which the x-axis divides the diagonal AC, and the point of division. (iii) Find the lengths of both diagonals and state, with reason, whether ABCD is a rectangle.

Try it first, then show the solutionSolution

The diagonals of a parallelogram bisect each other, so the midpoint of AC = the midpoint of BD. (½ mark)
Midpoint of AC $= (\frac{- 4 + 4}{2}, \frac{- 2 + y}{2}) = (0, \frac{y - 2}{2})$ . Midpoint of BD $= (\frac{2 + x}{2}, \frac{- 1 + 2}{2}) = (\frac{2 + x}{2}, \frac{1}{2})$ . (½ mark)
$\frac{2 + x}{2} = 0$ gives $x = - 2$ , and $\frac{y - 2}{2} = \frac{1}{2}$ gives $y = 3$ . So $C (4, 3)$ and $D (- 2, 2)$ . (1 mark)
Let the x-axis divide AC in the ratio $k : 1$ . The y-coordinate of the point is $\frac{3 k + ( - 2 )}{k + 1} = 0$ , so $k = \frac{2}{3}$ and the ratio is $2 : 3$ . (1 mark)
x-coordinate $= \frac{2 ( 4 ) + 3 ( - 4 )}{2 + 3} = - \frac{4}{5}$ , so the point is $(- \frac{4}{5}, 0)$ . (½ mark)
$A C = (4 + 4)^{2} + (3 + 2)^{2} = 64 + 25 = 89$ units and $B D = (- 2 - 2)^{2} + (2 + 1)^{2} = 16 + 9 = 5$ units. (1 mark)
A parallelogram is a rectangle only if its diagonals are equal. Here $89 \neq = 5$ , so ABCD is not a rectangle. (½ mark)

Answer: $x = - 2$ and $y = 3$ . The x-axis divides AC in the ratio $2 : 3$ at $(- \frac{4}{5}, 0)$ . The diagonals are $89$ units and 5 units; they are unequal, so ABCD is not a rectangle.

Example 11: Unknown in both the ratio and the coordinate · 3 marks · Short answerHarder board level

The point $P (k, 2 k - 2)$ divides the line segment joining $A (6, - 4)$ and $B (0, 5)$ internally in the ratio $k : 1$ . Find k and the coordinates of P.

Try it first, then show the solutionSolution

By the section formula with ratio $k : 1$ , the x-coordinate of P is $\frac{k ( 0 ) + 6}{k + 1} = \frac{6}{k + 1}$ . (½ mark)
This equals k: $6 = k^{2} + k$ , so $k^{2} + k - 6 = 0$ . (½ mark)
$(k + 3) (k - 2) = 0$ , so $k = 2$ or $k = - 3$ . A ratio of internal division must be positive, so $k = 2$ . (1 mark)
y-coordinate $= \frac{2 ( 5 ) + 1 ( - 4 )}{3} = 2$ , and $2 k - 2 = 2$ , so it agrees. P is $(2, 2)$ and the ratio is $2 : 1$ . (½ mark)
Note: the y-condition alone, $\frac{5 k - 4}{k + 1} = 2 k - 2$ , gives $2 k^{2} - 5 k + 2 = 0$ , so $k = 2$ or $\frac{1}{2}$ . Only $k = 2$ satisfies both coordinates. (½ mark)

Answer: $k = 2$ , so P divides AB in the ratio $2 : 1$ and P is $(2, 2)$ .

Example 12: Milk booth on a road, equally far from two housing societies · 5 marks · Long answerHOTS / competency

On a town map, 1 unit = 100 m. A straight road runs from junction $A (- 2, 5)$ to junction $B (6, 1)$ . Two housing societies are at $H_{1} (1, - 2)$ and $H_{2} (9, 2)$ . A milk booth P is to be set up on the road AB so that it is equally far from both societies. Find the ratio in which P divides AB, the position of P, and its distance from each society.

Try it first, then show the solutionSolution

Let P divide AB in the ratio $k : 1$ . Then $P = (\frac{6 k - 2}{k + 1}, \frac{k + 5}{k + 1})$ . (1 mark)
P is equally far from $H_{1} (1, - 2)$ and $H_{2} (9, 2)$ , so $P H_{1}^{2} = P H_{2}^{2}$ : $(x - 1)^{2} + (y + 2)^{2} = (x - 9)^{2} + (y - 2)^{2}$ . (½ mark)
Expand and cancel $x^{2}$ and $y^{2}$ : $- 2 x + 1 + 4 y + 4 = - 18 x + 81 - 4 y + 4$ , so $16 x + 8 y = 80$ , that is $2 x + y = 10$ . (1 mark)
Substitute P: $\frac{2 ( 6 k - 2 ) + ( k + 5 )}{k + 1} = 10$ , so $13 k + 1 = 10 k + 10$ and $k = 3$ . (1 mark)
So the ratio is $3 : 1$ and $P = (\frac{18 - 2}{4}, \frac{3 + 5}{4}) = (4, 2)$ . (½ mark)
$P H_{1} = (4 - 1)^{2} + (2 + 2)^{2} = 9 + 16 = 5$ units and $P H_{2} = (9 - 4)^{2} + 0^{2} = 5$ units, which is $5 \times 100 = 500$ m. (½ mark)
Check: $A P = 36 + 9 = 35$ , $P B = 4 + 1 = 5$ and $A B = 64 + 16 = 45$ . So $A P + P B = A B$ and $A P : P B = 3 : 1$ . (½ mark)

Answer: P divides AB in the ratio $3 : 1$ , so the booth is at $P (4, 2)$ , 500 m from each society.

Example 13: Right angle on the x-axis: a quadratic condition · 5 marks · Long answerHarder board level

The points $A (7, 7)$ and $B (1, - 1)$ are given. Find all points C on the x-axis such that $△ A B C$ is right-angled at C. For each such point, state whether the triangle is isosceles, and find the perimeter of the isosceles one.

Try it first, then show the solutionSolution

Let $C (x, 0)$ . The right angle is at C, so AB is the hypotenuse: $C A^{2} + C B^{2} = A B^{2}$ (converse of Pythagoras theorem). (½ mark)
$C A^{2} = (x - 7)^{2} + 49$ , $C B^{2} = (x - 1)^{2} + 1$ and $A B^{2} = (1 - 7)^{2} + (- 1 - 7)^{2} = 36 + 64 = 100$ . (1 mark)
$x^{2} - 14 x + 98 + x^{2} - 2 x + 2 = 100$ , so $2 x^{2} - 16 x = 0$ , that is $2 x (x - 8) = 0$ . (1 mark)
$x = 0$ or $x = 8$ , so $C (0, 0)$ or $C (8, 0)$ . Neither coincides with A or B, so both give triangles. (½ mark)
For $C (0, 0)$ : $C A^{2} = 49 + 49 = 98$ and $C B^{2} = 1 + 1 = 2$ , with $98 + 2 = 100$ . $C A = 72 \neq = C B = 2$ , so it is not isosceles. (½ mark)
For $C (8, 0)$ : $C A^{2} = 1 + 49 = 50$ and $C B^{2} = 49 + 1 = 50$ , with $50 + 50 = 100$ . So $C A = C B = 52$ units and the triangle is isosceles. (½ mark)
Perimeter $= 52 + 52 + 10 = 10 + 102$ units. (1 mark)

Answer: $C (0, 0)$ or $C (8, 0)$ . Only $C (8, 0)$ gives an isosceles triangle ( $C A = C B = 52$ units), and its perimeter is $10 + 102$ units.

Example 14: Recovering an end point from the points of trisection · 3 marks · Short answerHarder board level

$P (1, b)$ and $Q (c, 5)$ are the points of trisection of the line segment joining $A (- 2, 1)$ and $B (d, e)$ , with P nearer to A. Find the coordinates of B and the values of b and c. Verify that $A P = P Q = QB$ .

Try it first, then show the solutionSolution

P divides AB in the ratio $1 : 2$ , so its x-coordinate gives $1 = \frac{1 ( d ) + 2 ( - 2 )}{3}$ , so $d - 4 = 3$ and $d = 7$ . (½ mark)
Q divides AB in the ratio $2 : 1$ , so its y-coordinate gives $5 = \frac{2 ( e ) + 1 ( 1 )}{3}$ , so $2 e + 1 = 15$ and $e = 7$ . So B is $(7, 7)$ . (½ mark)
$b = \frac{1 ( 7 ) + 2 ( 1 )}{3} = 3$ and $c = \frac{2 ( 7 ) + 1 ( - 2 )}{3} = 4$ . So $P (1, 3)$ and $Q (4, 5)$ . (1 mark)
$A P = (1 + 2)^{2} + (3 - 1)^{2} = 13$ , $P Q = (4 - 1)^{2} + (5 - 3)^{2} = 13$ and $QB = (7 - 4)^{2} + (7 - 5)^{2} = 13$ , so $A P = P Q = QB$ . (1 mark)

Answer: $B (7, 7)$ , $b = 3$ and $c = 4$ . Each of AP, PQ and QB is $13$ units.

Example 15: Rhombus with a vertex on the x-axis turns out to be a square · 5 marks · Long answerHarder board level

$A (- 1, 2)$ and $C (5, 4)$ are opposite vertices of a rhombus ABCD, and the vertex B lies on the x-axis. Find B and D. Show that ABCD is in fact a square, and find its area.

Try it first, then show the solutionSolution

All sides of a rhombus are equal, so $B A = B C$ . Let $B (x, 0)$ , since B is on the x-axis. (½ mark)
$B A^{2} = B C^{2}$ : $(x + 1)^{2} + 4 = (x - 5)^{2} + 16$ , so $2 x + 5 = - 10 x + 41$ , giving $12 x = 36$ and $x = 3$ . So B is $(3, 0)$ . (1 mark)
The diagonals of a rhombus bisect each other. The midpoint of AC is $(\frac{- 1 + 5}{2}, \frac{2 + 4}{2}) = (2, 3)$ , so $D = (2 \times 2 - 3, 2 \times 3 - 0) = (1, 6)$ . (1 mark)
$A B^{2} = 16 + 4 = 20$ , $B C^{2} = 4 + 16 = 20$ , $C D^{2} = (1 - 5)^{2} + (6 - 4)^{2} = 20$ and $D A^{2} = (1 + 1)^{2} + (6 - 2)^{2} = 20$ . All sides are $25$ units. (1 mark)
Diagonals: $A C^{2} = 36 + 4 = 40$ and $B D^{2} = (1 - 3)^{2} + (6 - 0)^{2} = 40$ . Equal sides and equal diagonals, so ABCD is a square. (1 mark)
Area $= (side)^{2} = A B^{2} = 20$ square units. (½ mark)

Answer: $B (3, 0)$ and $D (1, 6)$ . All sides are $25$ units and both diagonals are $210$ units, so ABCD is a square of area 20 square units.

Example 16: Case study: a water stall on the school boundary · 4 marks · Case-basedHOTS / competency

Suppose a school campus map uses 1 unit = 10 m, with the main gate at the origin and the straight boundary wall along the x-axis. The library is at $L (4, 8)$ and the canteen is at $K (10, 2)$ .
(i) A notice board is placed exactly midway between the library and the canteen. Find its position. (1 mark)
(ii) Find the distance between the library and the canteen in metres. (1 mark)
(iii) A water stall S is to be built against the boundary wall, exactly 100 m from the library. Find both possible positions, choose the one nearer to the canteen, and give its distance from the canteen in metres. (2 marks)

Try it first, then show the solutionSolution

(i) Notice board $= (\frac{4 + 10}{2}, \frac{8 + 2}{2}) = (7, 5)$ . (1 mark)
(ii) $L K = (10 - 4)^{2} + (2 - 8)^{2} = 36 + 36 = 62$ units, that is $602$ m. (1 mark)
(iii) S lies on the x-axis, so let $S (x, 0)$ . 100 m is 10 units, so $(x - 4)^{2} + (0 - 8)^{2} = 100$ , giving $(x - 4)^{2} = 36$ , $x - 4 = \pm 6$ , so $x = 10$ or $x = - 2$ . (1 mark)
$S (10, 0)$ and $K (10, 2)$ share the same x-coordinate, so $S K = 2$ units $= 20$ m. For $S (- 2, 0)$ , $S K = 144 + 4 = 237$ units, which is longer. So the stall goes at $(10, 0)$ . (1 mark)

Answer: (i) $(7, 5)$ . (ii) $602$ m. (iii) The possible positions are $(10, 0)$ and $(- 2, 0)$ ; the stall goes at $(10, 0)$ , which is 20 m from the canteen.

Example 17: Three possible parallelograms from three vertices · 3 marks · Short answerHOTS / competency

$A (1, 2)$ , $B (4, 3)$ and $C (6, 6)$ are three vertices of a parallelogram, but the order of the vertices is not given. Find all possible positions of the fourth vertex D. Which of these parallelograms has the smallest perimeter? Justify your answer.

Try it first, then show the solutionSolution

The diagonals of a parallelogram bisect each other. D is opposite one of A, B, C, so one of AC, AB, BC is a diagonal, and D = (sum of that diagonal's ends) minus the third point. (½ mark)
AC a diagonal: $D = (1 + 6 - 4, 2 + 6 - 3) = (3, 5)$ . AB a diagonal: $D = (1 + 4 - 6, 2 + 3 - 6) = (- 1, - 1)$ . BC a diagonal: $D = (4 + 6 - 1, 3 + 6 - 2) = (9, 7)$ . (1½ marks)
$A B = 9 + 1 = 10$ , $B C = 4 + 9 = 13$ and $C A = 25 + 16 = 41$ . (½ mark)
The sides of each parallelogram are the two segments that are not its diagonal, so the perimeters are $2 (10 + 13)$ , $2 (13 + 41)$ and $2 (10 + 41)$ . The smallest avoids the longest segment $41$ , so it is $2 (10 + 13)$ with $D (3, 5)$ . (½ mark)

Answer: D can be $(3, 5)$ , $(- 1, - 1)$ or $(9, 7)$ . The parallelogram with $D (3, 5)$ (diagonal AC) has the smallest perimeter, $2 (10 + 13)$ units.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Water pipeline in a village

A gram panchayat in Rajasthan is laying a straight underground water pipeline from a well W to a storage tank T. On the planning map, the village chowk is the origin, 1 unit = 10 m, and the well and tank are at $W (- 4, 6)$ and $T (8, - 3)$ . A straight cart road runs along the x-axis and a canal runs along the y-axis. The engineers want to know the pipeline's length, where a valve should go, and where the pipeline crosses the road so that a protective culvert can be built there.

(i) · 1 mark
Find the length of the pipeline in metres.
Answer
Answer: 150 m
$W T = (8 + 4)^{2} + (- 3 - 6)^{2} = 144 + 81 = 225 = 15$ units. Since 1 unit = 10 m, the length is $15 \times 10 = 150$ m.
(ii) · 1 mark
A valve is fitted exactly halfway along the pipeline. Find its coordinates.
Answer
Answer: $(2, \frac{3}{2})$
Midpoint $= (\frac{- 4 + 8}{2}, \frac{6 + ( - 3 )}{2}) = (2, \frac{3}{2})$ .
(iii) · 2 marks
Find the ratio in which the cart road (x-axis) divides the pipeline WT, and the point where the culvert must be built.
Answer
Answer: Ratio $2 : 1$ ; culvert at $(4, 0)$
Let the ratio be $k : 1$ . On the x-axis, $y = 0$ : $\frac{- 3 k + 6}{k + 1} = 0$ , so $k = 2$ and the ratio is $2 : 1$ (1 mark). The x-coordinate is $\frac{2 ( 8 ) + 1 ( - 4 )}{3} = \frac{12}{3} = 4$ , so the culvert is at $(4, 0)$ , which is one of the points of trisection of WT (1 mark).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the ratio in which the y-axis divides the line segment joining $A (- 3, 7)$ and $B (6, - 2)$ . Also find the point of intersection. (2 marks)

Step	Marks
Let the ratio be $k : 1$ . The x-coordinate of the dividing point is $\frac{6 k - 3}{k + 1}$ .	0.5
The point lies on the y-axis, so $6 k - 3 = 0$ , giving $k = \frac{1}{2}$ .	0.5
The required ratio is $1 : 2$ .	0.5
y-coordinate $= \frac{1 ( - 2 ) + 2 ( 7 )}{3} = 4$ , so the point is $(0, 4)$ .	0.5

Find the point P on the x-axis which is equidistant from $A (3, 5)$ and $B (7, 1)$ . Also find the length PA. (3 marks)

Step	Marks
P lies on the x-axis, so let $P (x, 0)$ .	0.5
$P A^{2} = P B^{2}$ : $(x - 3)^{2} + 25 = (x - 7)^{2} + 1$	1
$x^{2} - 6 x + 34 = x^{2} - 14 x + 50$ , so $8 x = 16$ and $x = 2$ .	1
$P (2, 0)$ and $P A = (2 - 3)^{2} + 25 = 26$ units.	0.5

(i) Show that $A (1, 1)$ , $B (5, 4)$ and $C (2, 8)$ are the vertices of a right-angled isosceles triangle. (ii) If M is the midpoint of the hypotenuse AC, show that M is equidistant from A, B and C. (5 marks)

Step	Marks
$A B = (5 - 1)^{2} + (4 - 1)^{2} = 16 + 9 = 5$ units.	1
$B C = (2 - 5)^{2} + (8 - 4)^{2} = 9 + 16 = 5$ units.	0.5
$A C = (2 - 1)^{2} + (8 - 1)^{2} = 1 + 49 = 50 = 52$ units.	0.5
$A B = B C$ , and $A B^{2} + B C^{2} = 25 + 25 = 50 = A C^{2}$ . By the converse of Pythagoras theorem, $∠ B = 9 0^{\circ}$ , so the triangle is right-angled and isosceles.	1
$M = (\frac{1 + 2}{2}, \frac{1 + 8}{2}) = (\frac{3}{2}, \frac{9}{2})$ , and $M A = M C = \frac{A C}{2} = \frac{5 2}{2}$ units.	1
$M B = (5 - \frac{3}{2})^{2} + (4 - \frac{9}{2})^{2} = \frac{49}{4} + \frac{1}{4} = \frac{50}{4} = \frac{5 2}{2}$ units, so $M A = M B = M C$ .	1

Where students lose marks in Coordinate Geometry

The slips examiners see most often, and what to write instead.

Mistake: Pairing the ratio with the wrong end in the section formula
Students multiply $m_{1}$ by $x_{1}$ because both have the subscript 1, so it looks natural.
Wrong working
$A (- 2, 5)$ , $B (8, - 5)$ , ratio $2 : 3$
$x = \frac{2 ( - 2 ) + 3 ( 8 )}{5} = \frac{20}{5} = 4$
Right working
$x = \frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}} = \frac{2 ( 8 ) + 3 ( - 2 )}{5} = \frac{10}{5} = 2$
Do this: Write the formula in full, with $m_{1} x_{2}$ first, before substituting. Then check that the answer lies between the end coordinates and is closer to A when $m_{1} < m_{2}$ .
Usually costs: Usually 1 to 2 marks, plus any later step that uses the wrong point.
Mistake: Sign slip when one coordinate is negative
Students subtract a negative carelessly. In the slip below, $6 - (- 2)$ was worked as if $(- 2)$ were $+ 2$ , so it became $6 - 2$ .
Wrong working
$A (4, - 2)$ , $B (- 2, 6)$
$A B^{2} = (- 2 - 4)^{2} + (6 - 2)^{2} = 36 + 16 = 52$
Right working
$A B^{2} = (- 2 - 4)^{2} + (6 - (- 2))^{2} = 36 + 64 = 100$ , so $A B = 10$ units.
Do this: Put every negative coordinate in brackets when substituting, then simplify each bracket separately.
Usually costs: ½ to 1 mark for the wrong value, and more if the length is used again.
Mistake: Calling a figure a square after checking only the sides
Four equal sides feel like enough, but a rhombus also has four equal sides.
Wrong working
$A B = B C = C D = D A = 13$
Hence ABCD is a square.
Right working
$A B = B C = C D = D A = 13$ and $A C = B D = 26$
All sides are equal and the diagonals are equal, hence ABCD is a square.
Do this: For a square or rectangle, always find both diagonals and state the full reason.
Usually costs: 1 to 2 marks in a 3- or 5-mark proof, because the conclusion is unsupported.
Mistake: Dropping one root when solving a squared equation
Students take only the positive square root out of habit.
Wrong working
$(x - 3)^{2} = 16$
$x - 3 = 4$
$x = 7$
Right working
$(x - 3)^{2} = 16$
$x - 3 = \pm 4$
$x = 7$ or $x = - 1$
Do this: Write $\pm$ every time you take a square root. Reject a value only if the question gives a reason, such as a quadrant.
Usually costs: ½ to 1 mark for the missing value.
Mistake: Writing a point on the x-axis as $(0, y)$
Students confuse 'lies on the x-axis' with 'its x-coordinate is zero'.
Wrong working
Point on the x-axis: let $P (0, y)$ .
Right working
Point on the x-axis: let $P (x, 0)$ , since every point on the x-axis has y-coordinate 0.
Do this: Remember: on the x-axis, the y-coordinate is zero. Picture the point sitting on the horizontal line.
Usually costs: Usually the whole question, since everything after it is wrong.
Mistake: Adding unlike surds in a collinearity check
Students add the numbers under the roots instead of simplifying first.
Wrong working
$A (0, 0)$ , $B (2, 2)$ , $C (5, 5)$
$A B + B C = 8 + 18 = 26 \neq = 50$ , so not collinear.
Right working
$A B + B C = 22 + 32 = 52$ and $A C = 50 = 52$
$A B + B C = A C$ , so A, B, C are collinear.
Do this: Simplify each surd to the form $a b$ with the smallest b before adding or comparing.
Usually costs: 1 to 2 marks, because the conclusion is reversed.

NCERT exercise guide: Coordinate Geometry

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 7.1

What it tests: The distance formula and its uses: the distance between two points, a point's distance from the origin, finding an unknown coordinate from a given distance, and finding a point on an axis equidistant from two points. It also covers checking collinearity and proving that given points form an isosceles, equilateral or right-angled triangle, or a square, rhombus, rectangle or parallelogram.

How to approach it: Write the two points as $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ and substitute carefully, keeping brackets around negative numbers. When a distance is given, square both sides first so the root disappears, then solve. For shape questions, find all four sides and both diagonals before you decide. A square needs equal sides and equal diagonals. A rhombus has equal sides but unequal diagonals.

Like the questions in Exercise 7.1 that give a distance and ask for an unknown coordinate

Find the values of x for which the distance between $A (x, 3)$ and $B (- 2, - 5)$ is 10 units. For each value, state the quadrant in which A lies.

Show the solutionSolution

By the distance formula, $A B^{2} = (x - (- 2))^{2} + (3 - (- 5))^{2} = (x + 2)^{2} + 64$ .
Since $A B = 10$ , $A B^{2} = 100$ . So $(x + 2)^{2} + 64 = 100$ .
$(x + 2)^{2} = 36$ , so $x + 2 = 6$ or $x + 2 = - 6$ .
$x = 4$ or $x = - 8$ .
Check $x = 4$ : $A B = 36 + 64 = 100 = 10$ . Check $x = - 8$ : $A B = 36 + 64 = 10$ . Both values work.
For $x = 4$ , $A (4, 3)$ has both coordinates positive, so it is in quadrant I.
For $x = - 8$ , $A (- 8, 3)$ has a negative x and a positive y, so it is in quadrant II.

Answer: $x = 4$ or $x = - 8$ . $A (4, 3)$ lies in quadrant I and $A (- 8, 3)$ lies in quadrant II.

Exercise 7.2

What it tests: The section formula for internal division. It includes the point dividing a segment in a given ratio, the midpoint and points of trisection, the ratio in which a given point or an axis divides a segment, and missing vertices of a parallelogram using the fact that its diagonals bisect each other. It also covers dividing a segment into equal parts and finding a point when a fraction such as AP : AB is given.

How to approach it: Label A as $(x_{1}, y_{1})$ with weight $m_{2}$ and B as $(x_{2}, y_{2})$ with weight $m_{1}$ . Remember the cross pattern: $m_{1}$ multiplies the coordinate of B. When the ratio is unknown, take it as $k : 1$ , use the coordinate you know to find k, then use k to find the other coordinate. If you are given AP : AB, convert it to AP : PB first.

Like the questions in Exercise 7.2 that ask for the ratio in which a point divides a segment

Find the ratio in which the point $P (m, 6)$ divides the line segment joining $A (- 4, 3)$ and $B (2, 8)$ . Hence find m.

Show the solutionSolution

Let P divide AB in the ratio $k : 1$ .
By the section formula, the y-coordinate of P is $\frac{8 k + 3}{k + 1}$ .
This must equal 6: $8 k + 3 = 6 k + 6$ , so $2 k = 3$ and $k = \frac{3}{2}$ .
So the ratio is $\frac{3}{2} : 1 = 3 : 2$ .
The x-coordinate is $m = \frac{3 ( 2 ) + 2 ( - 4 )}{3 + 2} = \frac{6 - 8}{5} = - \frac{2}{5}$ .
Check: going $\frac{3}{5}$ of the way from A to B gives $x = - 4 + \frac{3}{5} (6) = - \frac{2}{5}$ and $y = 3 + \frac{3}{5} (5) = 6$ . This matches P.

Answer: P divides AB internally in the ratio $3 : 2$ , and $m = - \frac{2}{5}$ .

Coordinate Geometry: previous year questions

Standard: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 6 marks per paper; 79 questions in all, counting OR alternatives. In 4 of 4 sample papers.
Basic: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 6 marks per paper; 74 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Coordinate Geometry

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026 Basic, set 430/1/1, Q5 · 1 mark · MCQ

For a point $(3, - 5)$ , the value of (abscissa $-$ ordinate) is :

(A) $- 8$
(B) $- 2$
(C) $2$
(D) $8$

Try it first, then show our solutionSolution

For the point $(3, - 5)$ , abscissa $= 3$ and ordinate $= - 5$ .
Abscissa − ordinate $= 3 - (- 5) = 8$ . The answer is option (D).
1 mark

Answer: 8, option (D)

Where the marks are

Computes $3 - (- 5) = 8$ (1 mark)

See it in the full 2026 board paper, set 430/1/1

2026 Basic, set 430/1/1, Q38 · 4 marks · Case-basedCompetency-based

Case Study – 3

In a society, there is a circular park having two gates. The gates are placed at points A(10, 20) and B(50, 50), as shown in the figure below. Two fountains are installed at points P and Q on AB such that $A P = P Q = QB$ .

Based on the above information, answer the following questions :

(i)
Find the coordinates of the centre C.
1 mark
(ii)
Find the radius of the circular park.
1 mark
(iii)
(a) Find the coordinates of the point P.
2 marks
OR
(iii)
(b) Find the distance of the fountain at Q from gate A.
2 marks

Figure in the paper: Circle representing the park with a diameter AB; A is at lower left on the circle and B at upper right on the circle. C is the centre on AB. P and Q lie on AB between A and C and between C and B respectively, with AP = PQ = QB. See it in the official paper.

Try it first, then show our solutionSolution

(i) AB is a diameter, so C is its midpoint: $C = (\frac{10 + 50}{2}, \frac{20 + 50}{2}) = (30, 35)$ .
1 mark
(ii) $A B = (50 - 10)^{2} + (50 - 20)^{2} = 1600 + 900 = 50$ units, so radius $= \frac{50}{2} = 25$ units.
1 mark
(iii)(a) Since $A P = P Q = QB$ , P divides AB in the ratio $1 : 2$ . By the section formula, $x = \frac{1 ( 50 ) + 2 ( 10 )}{3}$ and $y = \frac{1 ( 50 ) + 2 ( 20 )}{3}$ .
1 mark
(iii)(a) $P = (\frac{70}{3}, 30) \approx (23.33, 30)$ .
1 mark
(iii)(b) Q divides AB in the ratio $2 : 1$ , so $Q = (\frac{2 ( 50 ) + 10}{3}, \frac{2 ( 50 ) + 20}{3}) = (\frac{110}{3}, 40)$ .
1 mark
(iii)(b) $A Q = (\frac{110}{3} - 10)^{2} + (40 - 20)^{2} = \frac{6400}{9} + \frac{3600}{9} = \frac{100}{3} \approx 33.33$ units. (Check: $A Q = \frac{2}{3} A B = \frac{2}{3} \times 50$ .)
1 mark

Answer: (i) C(30, 35); (ii) 25 units; (iii)(a) P(70/3, 30); (iii)(b) AQ = 100/3 ≈ 33.33 units

Where the marks are

(i) Midpoint formula gives C(30, 35) (1 mark)
(ii) AB = 50 by distance formula, so radius = 25 units (1 mark)
(iii)(a) Section formula with ratio 1:2 (1 mark)
(iii)(a) $P = (70/3, 30)$ (1 mark)
(iii)(b) Finds $Q = (110/3, 40)$ (1 mark)
(iii)(b) $A Q = 100/3$ units (1 mark)

See it in the full 2026 board paper, set 430/1/1

2026 Basic, set 430/4/1, Q23 · 2 marks · Very short answer

A $(-2, 3)$ and B $(4, 1)$ are end points of the diameter of a semi-circle. The semi-circle intersects y-axis at point P. Find the co-ordinates of the point P.

Try it first, then show our solutionSolution

Centre O is the mid-point of diameter AB: $O = (\frac{- 2 + 4}{2}, \frac{3 + 1}{2}) = (1, 2)$ .
½ mark
Let $P = (0, y)$ (it lies on the y-axis). Since P is on the circle, $O P = O B$ (radii): $O P^{2} = O B^{2} \Rightarrow (0 - 1)^{2} + (y - 2)^{2} = (4 - 1)^{2} + (1 - 2)^{2}$
$1 + (y - 2)^{2} = 10 \Rightarrow (y - 2)^{2} = 9 \Rightarrow y - 2 = \pm 3 \Rightarrow y = 5$ or $y = - 1$
1 mark
Hence P is $(0, 5)$ or $(0, - 1)$ (depending on which side of AB the semi-circle lies; the figure shows it above AB, giving $(0, 5)$ ).
½ mark

Answer: P(0, 5) or P(0, −1)

Where the marks are

Finds centre $(1, 2)$ using mid-point formula (½ mark)
Sets $O P^{2} = O B^{2}$ and solves to get $y = 5$ or $- 1$ (1 mark)
States coordinates of P (½ mark)

See it in the full 2026 board paper, set 430/4/1

Every Coordinate Geometry question, by year

Standard (93 questions, sample papers included)

For 2027

Sample paper 2026-27: Q7 (1 mark, solved), Q14 (1 mark, solved), Q19 (1 mark, solved), Q30 (3 marks, solved)

2026

Set 30/1/1: Q8 (1 mark, solved), Q23 (2 marks, solved), Q27 (3 marks, solved)

Set 30/2/1: Q9 (1 mark, solved), Q23 (2 marks, solved), Q27 (3 marks, solved)

Set 30/3/1: Q17 (1 mark, solved), Q23(a) (2 marks, solved), Q23(b) (2 marks, solved), Q30 (3 marks, solved)

Set 30/4/1: Q7 (1 mark, solved), Q15 (1 mark, solved), Q38 (4 marks, solved)

Set 30/5/1: Q2 (1 mark, solved), Q22(a) (2 marks, solved), Q22(b) (2 marks, solved), Q29(a) (3 marks, solved), Q29(b) (3 marks, solved)

Sample paper 2025-26: Q2 (1 mark, solved), Q16 (1 mark, solved), Q37 (4 marks, solved)

2025

Set 30/1/1: Q3 (1 mark, solved), Q23 (2 marks, solved), Q28 (3 marks, solved)

Set 30/2/1: Q5 (1 mark, solved), Q6 (1 mark, solved), Q17 (1 mark, solved), Q28 (3 marks, solved)

Set 30/3/1: Q6 (1 mark, solved), Q7 (1 mark, solved), Q16 (1 mark, solved), Q26(a) (3 marks, solved), Q26(b) (3 marks, solved)

Set 30/4/1: Q4 (1 mark, solved), Q23 (2 marks, solved), Q28 (3 marks, solved)

Set 30/5/1: Q6 (1 mark, solved), Q22 (2 marks, solved), Q29 (3 marks, solved)

Set 30/6/1: Q5 (1 mark, solved), Q23 (2 marks, solved), Q29 (3 marks, solved)

Sample paper 2024-25: Q15 (1 mark, solved), Q17 (1 mark, solved), Q24 (2 marks, solved), Q25 (2 marks, solved)

2024

Set 30/1/1: Q6 (1 mark, solved), Q11 (1 mark, solved), Q17 (1 mark, solved), Q26(A) (3 marks, solved), Q26(B) (3 marks, solved)

Set 30/2/1: Q2 (1 mark, solved), Q15 (1 mark, solved), Q22 (2 marks, solved), Q25(a) (2 marks, solved), Q25(b) (2 marks, solved)

Set 30/3/1: Q19 (1 mark, solved), Q23(a) (2 marks, solved), Q23(b) (2 marks, solved), Q29(a) (3 marks, solved), Q29(b) (3 marks, solved)

Set 30/4/1: Q16 (1 mark, solved), Q19 (1 mark, solved), Q36 (4 marks, solved)

Set 30/5/1: Q24(a) (2 marks, solved), Q24(b) (2 marks, solved), Q37 (4 marks, solved)

Sample paper 2023-24: Q6 (1 mark, solved), Q7 (1 mark, solved), Q37 (4 marks, solved)

2023

Set 30/1/1: Q4 (1 mark, solved), Q13 (1 mark, solved), Q19 (1 mark, solved), Q28 (3 marks, solved)

Set 30/2/1: Q5 (1 mark, solved), Q17 (1 mark, solved), Q23 (2 marks, solved), Q28 (3 marks, solved)

Set 30/4/1: Q4 (1 mark, solved), Q13 (1 mark, solved), Q37 (4 marks, solved)

Set 30/5/1: Q7 (1 mark, solved), Q19 (1 mark, solved), Q21(a) (2 marks, solved), Q21(b) (2 marks, solved), Q24 (2 marks, solved)

Set 30/6/1: Q2 (1 mark, solved), Q22 (2 marks, solved), Q26 (3 marks, solved)

Basic (89 questions, sample papers included)

For 2027

Sample paper 2026-27: Q8 (1 mark, solved), Q25(A) (2 marks, solved), Q25(B) (2 marks, solved), Q28 (3 marks, solved)

2026

Set 430/1/1: Q5 (1 mark, solved), Q6 (1 mark, solved), Q38 (4 marks, solved)

Set 430/2/1: Q6 (1 mark, solved), Q7 (1 mark, solved), Q38 (4 marks, solved)

Set 430/3/1: Q6 (1 mark, solved), Q7 (1 mark, solved), Q38 (4 marks, solved)

Set 430/4/1: Q10 (1 mark, solved), Q23 (2 marks, solved), Q31(a) (3 marks, solved), Q31(b) (3 marks, solved)

Set 430/5/1: Q3 (1 mark, solved), Q23 (2 marks, solved), Q27 (3 marks, solved)

Sample paper 2025-26: Q7 (1 mark, solved), Q8 (1 mark, solved), Q36 (4 marks, solved)

2025

Set 430/1/1: Q5 (1 mark, solved), Q6 (1 mark, solved), Q38 (4 marks, solved)

Set 430/2/1: Q6 (1 mark, solved), Q7 (1 mark, solved), Q38 (4 marks, solved)

Set 430/3/1: Q6 (1 mark, solved), Q7 (1 mark, solved), Q38 (4 marks, solved)

Set 430/4/1: Q2 (1 mark, solved), Q22 (2 marks, solved), Q29(a) (3 marks, solved), Q29(b) (3 marks, solved)

Set 430/5/1: Q3 (1 mark, solved), Q8 (1 mark, solved), Q23 (2 marks, solved), Q25 (2 marks, solved)

Set 430/6/1: Q3 (1 mark, solved), Q7 (1 mark, solved), Q38 (4 marks, solved)

Sample paper 2024-25: Q4 (1 mark, solved), Q21(A) (2 marks, solved), Q21(B) (2 marks, solved), Q27(A) (3 marks, solved), Q27(B) (3 marks, solved)

2024

Set 430/1/1: Q3 (1 mark, solved), Q8 (1 mark, solved), Q16 (1 mark, solved), Q28 (3 marks, solved)

Set 430/2/1: Q5 (1 mark, solved), Q20 (1 mark, solved), Q38 (4 marks, solved)

Set 430/3/1: Q3 (1 mark, solved), Q19 (1 mark, solved), Q21(a) (2 marks, solved), Q21(b) (2 marks, solved), Q24 (2 marks, solved)

Set 430/4/1: Q1 (1 mark, solved), Q2 (1 mark, solved), Q37 (4 marks, solved)

Set 430/5/1: Q15 (1 mark, solved), Q22(a) (2 marks, solved), Q22(b) (2 marks, solved), Q31 (3 marks, solved)

Sample paper 2023-24: Q6 (1 mark, solved), Q19 (1 mark, solved), Q37 (4 marks, solved)

2023

Set 430/1/1: Q5 (1 mark, solved), Q21(a) (2 marks, solved), Q21(b) (2 marks, solved), Q27 (3 marks, solved)

Set 430/2/1: Q2 (1 mark, solved), Q23(a) (2 marks, solved), Q23(b) (2 marks, solved), Q31 (3 marks, solved)

Set 430/4/1: Q22(A) (2 marks, solved), Q22(B) (2 marks, solved), Q36 (4 marks, solved)

Set 430/5/1: Q4 (1 mark, solved), Q24(a) (2 marks, solved), Q24(b) (2 marks, solved), Q29(a) (3 marks, solved), Q29(b) (3 marks, solved)

Set 430/6/1: Q2 (1 mark, solved), Q12 (1 mark, solved), Q29 (3 marks, solved)

Practice questions on Coordinate Geometry

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The point which divides the line segment joining $(2, - 5)$ and $(- 4, 7)$ internally in the ratio $1 : 2$ is:
(a) $(0, - 1)$
(b) $(- 2, 3)$
(c) $(0, 1)$
(d) $(- 1, 0)$
Hint
$m_{1} = 1$ multiplies the coordinates of the second point, $(- 4, 7)$ .
Answer
(a) $(0, - 1)$ . Working: $x = \frac{1 ( - 4 ) + 2 ( 2 )}{3} = 0$ and $y = \frac{1 ( 7 ) + 2 ( - 5 )}{3} = - 1$ . Option (b) is the result if you use the ratio $2 : 1$ by mistake.
Q2 · 2 marks · Very short answerBoard level
The point $A (2, - 4)$ is equidistant from $P (3, 8)$ and $Q (- 10, y)$ . Find the values of y.
Hint
Set $A P^{2} = A Q^{2}$ so you never have to deal with square roots.
Answer
$y = - 3$ or $y = - 5$ . Working: $A P^{2} = 1 + 144 = 145$ and $A Q^{2} = 144 + (y + 4)^{2}$ . So $(y + 4)^{2} = 1$ , which gives $y + 4 = \pm 1$ .
Q3 · 1 mark · Assertion-reasonBoard level
Assertion (A): The midpoint of the line segment joining $(4, - 2)$ and $(- 6, 8)$ lies on the y-axis.
Reason (R): Every point on the y-axis has x-coordinate 0.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Work out the midpoint first and look at its x-coordinate.
Answer
(d). The midpoint is $(\frac{4 - 6}{2}, \frac{- 2 + 8}{2}) = (- 1, 3)$ . Its x-coordinate is not 0, so it does not lie on the y-axis and A is false. R is a true statement.
Q4 · 3 marks · Short answerBoard level
In what ratio does the point $P (x, 2)$ divide the line segment joining $A (12, 5)$ and $B (4, - 3)$ ? Also find x.
Hint
Take the ratio as $k : 1$ and use the y-coordinate, which you know, to find k.
Answer
Ratio $3 : 5$ and $x = 9$ . Working: $\frac{- 3 k + 5}{k + 1} = 2$ gives $5 k = 3$ , so $k = \frac{3}{5}$ . Then $x = \frac{3 ( 4 ) + 5 ( 12 )}{8} = \frac{72}{8} = 9$ .
Q5 · 3 marks · Short answerHarder board level
The centre of a circle is $C (a, 9 - a)$ and the circle passes through $P (1, 1)$ . If the diameter of the circle is 10 units, find the possible values of a and the corresponding centres.
Hint
The radius is 5, so $C P^{2} = 25$ . This gives a quadratic in a.
Answer
$a = 4$ , centre $(4, 5)$ ; or $a = 5$ , centre $(5, 4)$ . Working: $(a - 1)^{2} + (8 - a)^{2} = 25$ gives $2 a^{2} - 18 a + 40 = 0$ , so $a^{2} - 9 a + 20 = 0$ and $(a - 4) (a - 5) = 0$ . Check: from $(4, 5)$ , $C P = 9 + 16 = 5$ .
Q6 · 5 marks · Long answerHarder board level
Show that $A (- 1, 2)$ , $B (3, - 1)$ , $C (9, 7)$ and $D (5, 10)$ , taken in order, are the vertices of a rectangle. Find the length of each diagonal and the point where the diagonals meet.
Hint
Show that the opposite sides are equal and the diagonals are equal. Then use the midpoint formula on both diagonals.
Answer
$A B = C D = 5$ and $B C = D A = 10$ , so the opposite sides are equal. $A C = 100 + 25 = 55$ and $B D = 4 + 121 = 55$ , so the diagonals are equal and ABCD is a rectangle. Each diagonal is $55$ units. Both diagonals have midpoint $(4, \frac{9}{2})$ , so they meet at $(4, \frac{9}{2})$ .
Q7 · 3 marks · Short answerBoard level
The point $P (3, b)$ divides the line segment joining $A (a, - 1)$ and $B (5, 8)$ internally in the ratio $2 : 1$ . Find a and b, and verify that $A P = 2 P B$ .
Hint
Write the section formula for each coordinate and equate it to the given coordinate of P.
Answer
$a = - 1$ and $b = 5$ . Working: $\frac{2 ( 5 ) + a}{3} = 3$ gives $a = - 1$ , and $b = \frac{2 ( 8 ) + ( - 1 )}{3} = 5$ . Check: $A P = 16 + 36 = 213$ and $P B = 4 + 9 = 13$ .
Q8 · 5 marks · Long answerHarder board level
Find the value of p for which the points $A (- 3, 9)$ , $B (2, p)$ and $C (4, - 5)$ are collinear, and find the ratio in which B divides AC. Verify that the points are collinear using the distance formula.
Hint
If B lies on AC, it divides AC in some ratio $k : 1$ . Find k from the x-coordinates.
Answer
$p = - 1$ and the ratio is $5 : 2$ . Working: $\frac{4 k - 3}{k + 1} = 2$ gives $k = \frac{5}{2}$ , and then $p = \frac{5 ( - 5 ) + 2 ( 9 )}{7} = - 1$ . Check: $A B = 55$ , $B C = 25$ and $A C = 75$ , so $A B + B C = A C$ and the points are collinear.
Q9 · 3 marks · Short answerBoard level
Arjun says that the quadrilateral with vertices $P (1, 1)$ , $Q (4, 2)$ , $R (5, 5)$ and $S (2, 4)$ , taken in order, is a square because all four sides are equal. Is he right? Justify your answer.
Hint
Equal sides are not enough for a square. Compare the diagonals too.
Answer
No. $P Q = QR = R S = S P = 10$ , but the diagonals are $P R = 16 + 16 = 42$ and $QS = 4 + 4 = 22$ . The diagonals are unequal, so PQRS is a rhombus, not a square.
Q10 · 4 marks · Case-basedHOTS / competency
A medicine-delivery drone flies in a straight line from a district hospital at $H (- 2, - 3)$ to a village health centre at $V (10, b)$ . On the map, 1 unit = 1 km, the origin O is the town square, a river road runs along the x-axis and a highway runs along the y-axis. When the drone has covered exactly one-third of its journey, it is directly above the river road, at point P. (i) Find b and the coordinates of P. (1 mark) (ii) Find the length of the flight. (1 mark) (iii) The drone crosses the highway at Q. Find the ratio in which the highway divides the flight path, the coordinates of Q, and the perimeter of $△ O P Q$ . (2 marks)
Hint
One-third of the journey means $H P : P V = 1 : 2$ , and P has y-coordinate 0. For Q, take the ratio $k : 1$ and set $x = 0$ .
Answer
(i) P divides HV in the ratio $1 : 2$ , so its y-coordinate is $\frac{1 ( b ) + 2 ( - 3 )}{3} = 0$ , giving $b = 6$ . Its x-coordinate is $\frac{1 ( 10 ) + 2 ( - 2 )}{3} = 2$ , so P is $(2, 0)$ . (ii) $H V = (10 + 2)^{2} + (6 + 3)^{2} = 144 + 81 = 15$ km. (iii) $\frac{10 k - 2}{k + 1} = 0$ gives $k = \frac{1}{5}$ , so the highway divides HV in the ratio $1 : 5$ . Then $y = \frac{1 ( 6 ) + 5 ( - 3 )}{6} = - \frac{3}{2}$ , so $Q (0, - \frac{3}{2})$ . $O P = 2$ km, $O Q = \frac{3}{2}$ km and $P Q = 4 + \frac{9}{4} = \frac{5}{2}$ km, so the perimeter of $△ O P Q$ is $2 + \frac{3}{2} + \frac{5}{2} = 6$ km.
Q11 · 3 marks · Short answerHarder board level
Find the ratio in which the point P, whose abscissa is equal to its ordinate, divides the line segment joining $A (- 4, 6)$ and $B (5, - 3)$ . Also find the coordinates of P.
Hint
Take the ratio as $k : 1$ , write both coordinates of P in terms of k, and set them equal.
Answer
Ratio $5 : 4$ and $P (1, 1)$ . Working: $P = (\frac{5 k - 4}{k + 1}, \frac{- 3 k + 6}{k + 1})$ . Equal coordinates give $5 k - 4 = - 3 k + 6$ , so $8 k = 10$ and $k = \frac{5}{4}$ . Then $x = \frac{5 ( 5 ) + 4 ( - 4 )}{9} = 1$ and $y = \frac{5 ( - 3 ) + 4 ( 6 )}{9} = 1$ .
Q12 · 5 marks · Long answerHarder board level
AB is a diameter of a circle with centre $O (2, - 3)$ , where $A (- 1, a)$ and $B (b, - 7)$ . (i) Find a and b. (ii) Find the radius of the circle. (iii) Show that the point $P (5, 1)$ lies on the circle, and that $∠ A P B = 9 0^{\circ}$ .
Hint
O is the midpoint of AB. A point lies on the circle when its distance from O equals the radius. For the right angle, use the converse of Pythagoras theorem.
Answer
$a = 1$ , $b = 5$ and the radius is 5 units. Working: $\frac{- 1 + b}{2} = 2$ gives $b = 5$ , and $\frac{a - 7}{2} = - 3$ gives $a = 1$ . $O A = (- 1 - 2)^{2} + (1 + 3)^{2} = 9 + 16 = 5$ units. $O P = (5 - 2)^{2} + (1 + 3)^{2} = 5$ units, so P lies on the circle. $P A^{2} = 36 + 0 = 36$ , $P B^{2} = 0 + 64 = 64$ and $A B^{2} = 36 + 64 = 100$ , so $P A^{2} + P B^{2} = A B^{2}$ and $∠ A P B = 9 0^{\circ}$ .
Q13 · 3 marks · Short answerHarder board level
Find the point P on the x-axis which is equidistant from $A (4, 3)$ and $B (0, 1)$ . Hence show that $△ P A B$ is a right-angled isosceles triangle.
Hint
Let $P (x, 0)$ and set $P A^{2} = P B^{2}$ . Then compare $P A^{2} + P B^{2}$ with $A B^{2}$ .
Answer
$P (3, 0)$ . Working: $(x - 4)^{2} + 9 = x^{2} + 1$ gives $- 8 x + 25 = 1$ , so $x = 3$ . Then $P A^{2} = 1 + 9 = 10$ , $P B^{2} = 9 + 1 = 10$ and $A B^{2} = 16 + 4 = 20$ . So $P A = P B$ and $P A^{2} + P B^{2} = A B^{2}$ , and the triangle is isosceles and right-angled at P.
Q14 · 4 marks · Case-basedHOTS / competency
A rectangular garden has corners $O (0, 0)$ , $P (12, 0)$ , $Q (12, 9)$ and $R (0, 9)$ , where 1 unit = 1 m. A water pipe is laid along the diagonal OQ, and two sprinklers are fixed at the points of trisection of OQ. Each sprinkler waters every point within 5 m of it. (i) Find the positions of the two sprinklers. (1 mark) (ii) Show that the corners O and Q just get watered. (1 mark) (iii) Do the corners P and R get watered? Find the least watering range each sprinkler needs so that all four corners get water. (2 marks)
Hint
Points of trisection use the ratios $1 : 2$ and $2 : 1$ . A corner is watered only if its distance from the nearer sprinkler is at most the range.
Answer
(i) $S_{1} = (\frac{1 ( 12 ) + 2 ( 0 )}{3}, \frac{1 ( 9 ) + 2 ( 0 )}{3}) = (4, 3)$ and $S_{2} = (\frac{2 ( 12 ) + 1 ( 0 )}{3}, \frac{2 ( 9 ) + 1 ( 0 )}{3}) = (8, 6)$ . (ii) $O S_{1} = 16 + 9 = 5$ m and $Q S_{2} = 16 + 9 = 5$ m, exactly the range. (iii) $P S_{2} = 16 + 36 = 52$ m and $P S_{1} = 64 + 9 = 73$ m. Similarly $R S_{1} = 52$ m and $R S_{2} = 73$ m. Since $52 > 25$ , neither P nor R is watered. The nearest sprinkler to P or R is $52 = 213$ m away, so the least range needed is $213$ m.
Q15 · 3 marks · Short answerHOTS / competency
(i) Using the $k : 1$ method, show that the x-axis cannot divide the line segment joining $A (2, 3)$ and $B (5, 7)$ internally, and explain the reason geometrically. (ii) The x-axis divides the line segment joining $A (2, 3)$ and $C (5, b)$ internally in the ratio $1 : 2$ . Find b and the point of division.
Hint
For internal division, k must be positive. Look at the signs of the y-coordinates of the two ends.
Answer
(i) Setting the y-coordinate $\frac{7 k + 3}{k + 1} = 0$ gives $k = - \frac{3}{7}$ , which is negative, so there is no internal division. Geometrically, both A and B have positive y-coordinates, so the whole segment lies above the x-axis and never meets it. (ii) $\frac{1 ( b ) + 2 ( 3 )}{3} = 0$ gives $b = - 6$ . The x-coordinate is $\frac{1 ( 5 ) + 2 ( 2 )}{3} = 3$ , so the point of division is $(3, 0)$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the distance of the point $(- 5, 7)$ from the x-axis?Show answer
7 units, because the distance from the x-axis is $∣ y ∣$ .
2.What is the general form of a point on the y-axis?Show answer
$(0, y)$ : its x-coordinate is 0.
3.State the distance formula.Show answer
$A B = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
4.What is the distance of $(6, - 8)$ from the origin?Show answer
$36 + 64 = 10$ units.
5.Find the midpoint of $(- 3, 8)$ and $(7, - 2)$ .Show answer
$(\frac{- 3 + 7}{2}, \frac{8 - 2}{2}) = (2, 3)$
6.In the section formula, which coordinate does $m_{1}$ multiply?Show answer
The coordinate of the second point B, giving $\frac{m _{1} x _{2} + m _{2} x _{1}}{m _{1} + m _{2}}$ .
7.A quadrilateral has four equal sides but unequal diagonals. What is it?Show answer
A rhombus, not a square.
8.What condition on distances shows that A, B, C are collinear with B in the middle?Show answer
$A B + B C = A C$

Best order to study it

Revise the Cartesian plane first: quadrant signs, points on the axes, distance from each axis and mirror images.
Learn the distance formula through its Pythagoras derivation, then practise sides, perimeters and unknown coordinates from a given distance.
Apply distances to proofs: types of triangles, squares, rhombuses, rectangles and parallelograms, and the collinearity test.
Learn the section formula with its cross pairing, then the midpoint and points of trisection as special cases.
Practise the k : 1 method for unknown ratios, and the parallelogram midpoint condition for missing vertices.
Finish with mixed questions that use both formulas together, including case studies set on a grid.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you write the distance formula and explain how Pythagoras theorem gives it?
Can you find a point on the x-axis equidistant from two given points in under three minutes?
Can you prove four points form a square, remembering to check the diagonals?
Can you test three points for collinearity, simplifying surds before adding?
Can you write the section formula for internal division with $m_{1}$ paired correctly?
Can you find both points of trisection of a segment and check them by equal steps?
Can you find the ratio in which the y-axis divides a segment using $k : 1$ ?
Can you find the fourth vertex of parallelogram ABCD using midpoints of diagonals?
Can you get two values of x from $(x - a)^{2} = c$ and decide whether to keep both?
Can you state the distance of $(x, y)$ from each axis without mixing them up?

Coordinate Geometry: frequently asked questions

Are area of a triangle from coordinates and external division in the 2026-27 syllabus?

No. Both were removed in the NCERT rationalisation and are not in the 2026-27 CBSE syllabus. The section formula is needed for internal division only.

Do I need to learn the derivation of the distance formula?

It is worth understanding how it comes from Pythagoras theorem, using the right triangle formed by horizontal and vertical lines. Knowing why it works helps you remember it correctly. It also helps with special cases like points on the same horizontal line. Most questions ask you to apply the formula.

In the section formula, how do I decide which point is the first point?

The first point is the one the ratio starts from. If P divides AB in the ratio 2 : 3, then AP : PB = 2 : 3, so A is the first point and B is the second. The first ratio number then multiplies the coordinates of B.

Should I leave lengths as surds or convert them to decimals?

Leave them in simplest surd form, such as 3 root 5 units, unless the question asks for a decimal. You are not allowed a calculator, and surds keep your comparisons exact. This matters most in collinearity and shape proofs.

How do I show that four points form a rhombus and not a square?

Find all four sides and show they are equal, then find both diagonals. If the diagonals are unequal, the figure is a rhombus but not a square. If they are equal, it is a square.

Based on: CBSE Mathematics syllabus 2026-27; NCERT Mathematics, Textbook for Class X (Reprint 2026-27)