MathsICSE Class 9 · Chapter 17

Coordinate Geometry ICSE Class 9: Plotting, Graphical Solutions and the Distance Formula

Q: Are the section and mid-point formulas in Class IX?

No. The Class IX syllabus covers plotting, graphical solutions and the distance formula. Section and mid-point formulas, and the equation of a line, are in Class X.

Q: Do I need graph paper for the graphical method?

Yes. Draw accurately on graph paper with a clear scale; the answer is read from the intersection.

Q: What if the lines meet at a point that is not on a grid crossing?

Read it as accurately as you can and check by substitution. Exam questions are usually set so the solution has whole-number coordinates.

Q: Why is the distance formula related to Pythagoras?

The horizontal and vertical differences between the points form the legs of a right triangle, and the distance is its hypotenuse.

Q: Can a distance be negative?

No. A distance is a square root of a sum of squares, so it is always zero or positive.

Updated 15 October 2026 · Fact-checked

Coordinate geometry places points on a plane using ordered pairs (x, y). ICSE Class 9 covers dependent and independent variables, plotting points in the Cartesian plane, solving two linear equations by drawing their graphs and reading the intersection, and the distance formula √((x₂ − x₁)² + (y₂ − y₁)²).

What Coordinate Geometry is about

The Cartesian plane is formed by two perpendicular number lines, the x-axis (horizontal) and the y-axis (vertical), meeting at the origin $O (0, 0)$ . Every point is described by an ordered pair $(x, y)$ : $x$ is the abscissa (distance along the x-axis) and $y$ the ordinate.

The CISCE 2028 Class IX syllabus covers: (a) dependent and independent variables; (b) ordered pairs, coordinates and plotting points; (c) solving simultaneous linear equations graphically; and (d) the distance formula.

In the school exam expect short questions on quadrants and plotting, 3-mark distance questions (find a length, an unknown coordinate, or a point equidistant from two points), 4-mark proofs that points form an isosceles, right-angled or square figure, and a 4 to 5-mark graph question on graph paper.

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Mathematics. Class IX is examined by your school: one 80-mark paper of 3 hours plus 20 marks of internal assessment. Coordinate Geometry is in the 8. Coordinate Geometry part of the 2028 syllabus.Source: CISCE ICSE 2028 Mathematics syllabus, Class IX

Key concepts

Dependent and independent variables

In a relation such as $y = 2 x + 3$ , we choose values of $x$ freely and then calculate $y$ . So $x$ is the independent variable and $y$ is the dependent variable: its value depends on $x$ .

In real situations: the cost of petrol depends on the number of litres bought (litres independent, cost dependent); the distance travelled at a fixed speed depends on time.

On a graph, the independent variable is usually plotted on the x-axis and the dependent variable on the y-axis.

Worked example

A taxi charges ₹50 plus ₹15 per km. Write the fare $F$ in terms of the distance $d$ , name the dependent and independent variables, and find $F$ for $d = 8$ .

Try it first, then show the solutionSolution

$F = 50 + 15 d$ .
$d$ is independent; $F$ is dependent.
$F = 50 + 120 = ₹170$ .

Answer: F = 50 + 15d; d independent, F dependent; ₹170

Ordered pairs, quadrants and plotting

The axes divide the plane into four quadrants:

I: $(+, +)$ ; II: $(-, +)$ ; III: $(-, -)$ ; IV: $(+, -)$ .

Points on the x-axis have $y = 0$ ; points on the y-axis have $x = 0$ .

The order matters: $(2, 5)$ and $(5, 2)$ are different points. To plot $(a, b)$ , move $a$ units along the x-axis (right if positive), then $b$ units parallel to the y-axis (up if positive).

On graph paper, choose and state a scale (e.g. 1 cm = 1 unit) and label both axes.

Worked example

State the quadrant or axis of each point: $A (- 3, 4)$ , $B (5, - 2)$ , $C (0, - 6)$ , $D (- 1, - 7)$ .

Try it first, then show the solutionSolution

$A (-, +)$ : quadrant II. $B (+, -)$ : quadrant IV.
$C$ has $x = 0$ : on the y-axis (below the origin).
$D (-, -)$ : quadrant III.

Answer: A: II; B: IV; C: y-axis; D: III

Graph of a linear equation

Every solution $(x, y)$ of a linear equation $a x + b y = c$ lies on a straight line, and every point on that line is a solution.

To draw it:
1. Rewrite as $y = \dots$ if convenient.
2. Make a table of at least three values (two fix the line; the third is a check).
3. Plot the points and join them with a ruler, extending the line beyond them.

Useful points: put $x = 0$ to get the y-intercept and $y = 0$ to get the x-intercept.

Worked example

Make a table of values for $2 x + y = 7$ using $x = 0, 1, 2$ .

Try it first, then show the solutionSolution

$y = 7 - 2 x$ .
$x = 0$ : $y = 7$ ; $x = 1$ : $y = 5$ ; $x = 2$ : $y = 3$ .
Points $(0, 7)$ , $(1, 5)$ , $(2, 3)$ lie on one straight line.

Answer: (0, 7), (1, 5), (2, 3)

Solving simultaneous equations graphically

Draw both lines on the same axes, with the same scale. The point where they intersect satisfies both equations, so its coordinates are the solution.

Lines intersect at one point: one solution.
Lines are parallel (same slope, different intercepts): no solution.
Lines coincide: infinitely many solutions.

Always check the solution by substituting into both equations, and write it clearly, e.g. ' $x = 2$ , $y = 3$ '.

Worked example

Solve graphically: $2 x + y = 7$ and $x - 2 y = - 4$ .

Try it first, then show the solutionSolution

Line 1: $(0, 7)$ , $(1, 5)$ , $(2, 3)$ . Line 2 ( $y = \frac{x + 4}{2}$ ): $(0, 2)$ , $(2, 3)$ , $(4, 4)$ .
Plot both on the same axes; they meet at $(2, 3)$ .
Check: $4 + 3 = 7$ ✓; $2 - 6 = - 4$ ✓.

Answer: x = 2, y = 3

The distance formula

The distance between $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ is

$P Q = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ .

It comes from Pythagoras: the horizontal gap $(x_{2} - x_{1})$ and vertical gap $(y_{2} - y_{1})$ are the legs of a right triangle whose hypotenuse is $P Q$ .

Distance from the origin: $O P = x^{2} + y^{2}$ . The order of the points does not matter, because the differences are squared.

Worked example

Find the distance between $(- 1, 4)$ and $(5, - 4)$ .

Try it first, then show the solutionSolution

$x_{2} - x_{1} = 6$ , $y_{2} - y_{1} = - 8$ .
Distance $= 36 + 64 = 100 = 10$ .

Answer: 10 units

Using distances to identify figures

Find all the relevant distances, then use geometry:

Collinear points: the two shorter distances add up to the longest.
Isosceles triangle: two sides equal; equilateral: all three equal.
Right-angled triangle: the square of the longest side equals the sum of the squares of the other two (converse of Pythagoras).
Square: four equal sides and equal diagonals; rhombus: four equal sides, unequal diagonals; rectangle: opposite sides equal and equal diagonals; parallelogram: opposite sides equal.

To find a point equidistant from $A$ and $B$ , set the squared distances equal (no square roots needed).

Worked example

Find the point on the x-axis equidistant from $A (2, - 5)$ and $B (- 2, 9)$ .

Try it first, then show the solutionSolution

Let the point be $P (x, 0)$ : $P A^{2} = P B^{2}$ .
$(x - 2)^{2} + 25 = (x + 2)^{2} + 81$ .
$- 4 x + 29 = 4 x + 85 \Rightarrow - 8 x = 56 \Rightarrow x = - 7$ .

Answer: (−7, 0)

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.

Distance on a grid map

Suppose a town map uses a grid in kilometres, with the school at $(1, 2)$ and the library at $(7, 10)$ .

How far apart are they in a straight line?

Try it first, then show the solutionSolution

Differences: 6 km east and 8 km north.
Distance $= 36 + 64 = 10$ km.

Answer: 10 km.

Memory hook: Grid map distance = Pythagoras on the differences.

Best meeting point on a road

Imagine a straight road runs along the x-axis, and two friends live at $A (2, - 5)$ and $B (- 2, 9)$ (in km). They want to meet at a bus stop on the road that is the same distance from both homes.

Where should the stop be?

Try it first, then show the solutionSolution

Stop $P (x, 0)$ : $(x - 2)^{2} + 25 = (x + 2)^{2} + 81$ .
$x = - 7$ : the stop is at $(- 7, 0)$ .

Answer: At (−7, 0).

Memory hook: Equidistant: equal squared distances.

Graphs as solutions

Suppose two mobile plans cost $C = 100 + 2 m$ and $C = 160 + m$ rupees for $m$ minutes of calls in a month.

Using graphs, at how many minutes do the plans cost the same?

Try it first, then show the solutionSolution

Draw both lines with $m$ on the x-axis and $C$ on the y-axis.
They meet where $100 + 2 m = 160 + m$ , i.e. $m = 60$ , $C = 220$ .

Answer: At 60 minutes, both cost ₹220.

Memory hook: Where lines cross, both equations agree.

Topics in this chapter

Coordinate Geometry formula sheet

Distance formula: $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$; Order of points does not matter.
Distance from origin: $O P = x^{2} + y^{2}$; Special case with (0, 0).
Points on the axes: $(x, 0) on x-axis; (0, y) on y-axis$; Use as the unknown point.
Collinearity test: $A B + B C = A C$; AC the longest distance.
Equidistant point: $P A^{2} = P B^{2}$; Square both distances; no roots.

Coordinate Geometry: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Graphical solution of two linear equations

Solve two linear equations graphically on graph paper; 4 to 5 marks.

Draw

Make a table of three values for each equation.
Draw the x- and y-axes, mark the origin and state the scale (e.g. 2 cm = 1 unit on both axes).
Plot the points of the first equation and join them with a ruler; label the line with its equation.
Plot and draw the second line on the same axes; label it.
Mark the point of intersection and drop dotted lines to both axes to read its coordinates.

Label

x-axis, y-axis, O
scale
each line labelled with its equation
intersection point with coordinates

Watch out: A table, a scale, labelled lines and the solution written as x = …, y = … each carry marks.

How the school exam asks Coordinate Geometry

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

Question type	How it is asked	Method
Quadrants and plotting	State the quadrant of given points, plot points and name the figure, or find a missing vertex of a rectangle or square on graph paper.	Signs of x and y; plot carefully with a stated scale.
Variables	Identify dependent and independent variables in a formula or situation; complete a table.	The variable you choose freely is independent.
Graphical solution	Solve two linear equations by drawing graphs; sometimes find the area of the triangle formed with an axis.	Tables, same axes, intersection; area from base on the axis and height from the intersection.
Distance between points	Find a distance or a perimeter; find an unknown coordinate given a distance.	Substitute into the formula; square both sides for unknowns.
Prove the type of figure	Show that points form an isosceles or right triangle, a square, rhombus or rectangle, or are collinear.	Find all sides (and diagonals); compare.
Equidistant points	Find a point on an axis equidistant from two given points.	Use (x, 0) or (0, y); set squared distances equal.

Tricks to remember Coordinate Geometry

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

Along the corridor, then up the stairs
Plot x first (along), then y (up or down).
Use it when: Plotting ordered pairs.
Quadrant signs go round
Starting top right and going anticlockwise: (+, +), (−, +), (−, −), (+, −).
Use it when: Naming quadrants.
Distance is Pythagoras
Across difference squared plus up difference squared, then root.
Use it when: Distance formula.
Three points, one check
Plot three points per line: two to draw, one to catch mistakes.
Use it when: Drawing graphs.

Coordinate Geometry: exam-level solved questions

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

Example 1: Distance from the origin · 1 mark · MCQExam level

The distance of the point $(- 6, 8)$ from the origin is:
(a) 2
(b) 10
(c) 14
(d) 100

Try it first, then show the solutionSolution

$36 + 64 = 10$ .

Answer: (b)

Example 2: Unknown coordinate · 3 marks · Short answerExam level

Find $x$ if the distance between $(x, 2)$ and $(3, - 2)$ is 5 units.

Try it first, then show the solutionSolution

$(x - 3)^{2} + 16 = 25$ .
$(x - 3)^{2} = 9$ , so $x - 3 = \pm 3$ .
$x = 6$ or $x = 0$ .

Answer: x = 6 or 0

Example 3: Graphical solution · 4 marks · Long answerExam level

Solve graphically: $x + y = 5$ and $x - y = 1$ . Hence find the area of the triangle formed by these lines and the y-axis.

Try it first, then show the solutionSolution

Line 1: $(0, 5)$ , $(2, 3)$ , $(5, 0)$ . Line 2: $(0, - 1)$ , $(1, 0)$ , $(4, 3)$ .
The lines meet at $(3, 2)$ : $x = 3$ , $y = 2$ .
They meet the y-axis at $(0, 5)$ and $(0, - 1)$ : base 6 units on the y-axis.
Height $=$ distance of $(3, 2)$ from the y-axis $= 3$ . Area $= \frac{1}{2} \times 6 \times 3 = 9$ square units.

Answer: x = 3, y = 2; area 9 sq. units

Example 4: Right isosceles triangle · 3 marks · Short answerHarder exam level

Show that $A (- 2, 2)$ , $B (1, 6)$ and $C (5, 3)$ are the vertices of a right-angled isosceles triangle.

Try it first, then show the solutionSolution

$A B = 9 + 16 = 5$ ; $B C = 16 + 9 = 5$ ; $A C = 49 + 1 = 50$ .
$A B = B C$ : isosceles.
$A B^{2} + B C^{2} = 50 = A C^{2}$ : right angle at $B$ .

Answer: Proved; right angle at B.

Example 5: A square from coordinates · 4 marks · Long answerHarder exam level

Prove that $(1, 7)$ , $(4, 2)$ , $(- 1, - 1)$ and $(- 4, 4)$ , taken in order, are the vertices of a square.

Try it first, then show the solutionSolution

Sides: each is $9 + 25 = 34$ (check all four pairs of consecutive points).
Diagonals: $(1, 7)$ to $(- 1, - 1)$ : $4 + 64 = 68$ ; $(4, 2)$ to $(- 4, 4)$ : $64 + 4 = 68$ .
Four equal sides and equal diagonals: a square.

Answer: Proved.

Example 6: Collinear points · 3 marks · Short answerHarder exam level

Show that $P (1, 1)$ , $Q (4, 5)$ and $R (7, 9)$ are collinear.

Try it first, then show the solutionSolution

$P Q = 9 + 16 = 5$ ; $QR = 9 + 16 = 5$ ; $P R = 36 + 64 = 10$ .
$P Q + QR = P R$ , so the points lie on one line ( $Q$ is the mid-point).

Answer: Collinear.

Example 7: Point on the y-axis · 4 marks · Long answerHOTS / competency

Find the point on the y-axis equidistant from $(- 5, - 2)$ and $(3, 2)$ , and its distance from each.

Try it first, then show the solutionSolution

Let $P (0, y)$ : $25 + (y + 2)^{2} = 9 + (y - 2)^{2}$ .
$25 + 4 y + 4 = 9 - 4 y + 4$ , so $8 y = - 16$ and $y = - 2$ .
$P (0, - 2)$ ; distance to $(- 5, - 2)$ $= 5$ ; to $(3, 2)$ $= 9 + 16 = 5$ ✓.

Answer: (0, −2); 5 units

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 10-mark question split into 3 + 3 + 4 marks.

Planning a school garden

A school marks out a garden on a grid in metres. Corners of a flower bed are $A (1, 1)$ , $B (7, 1)$ , $C (7, 5)$ and $D (1, 5)$ . A tap is at $T (4, 9)$ . Two straight water pipes follow the lines $y = x$ and $x + y = 8$ .

(i) · 3 marks
What shape is $A B C D$ ? Find its perimeter and the length of a diagonal.
Answer
Answer: Rectangle; 20 m; √52 ≈ 7.21 m
$A B = 6$ , $B C = 4$ , $C D = 6$ , $D A = 4$ , all angles right (sides along grid lines): a rectangle. Perimeter $= 20$ m. $A C = 36 + 16 = 52 \approx 7.21$ m.
(ii) · 3 marks
Which corner is nearest to the tap, and how far is it?
Answer
Answer: C and D, 5 m each
$T C = 9 + 16 = 5$ ; $T D = 9 + 16 = 5$ ; $T A = T B = 9 + 64 = 73$ . Nearest: $C$ and $D$ , 5 m.
(iii) · 4 marks
Find, by drawing or by checking, where the two pipes cross, and say whether that point lies inside the bed.
Answer
Answer: (4, 4); inside
$y = x$ through $(0, 0)$ , $(2, 2)$ , $(4, 4)$ ; $x + y = 8$ through $(0, 8)$ , $(4, 4)$ , $(8, 0)$ . They meet at $(4, 4)$ . Since $1 < 4 < 7$ and $1 < 4 < 5$ , the point is inside the bed.

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 perimeter of the triangle with vertices $(0, 4)$ , $(0, 0)$ and $(3, 0)$ . (3 marks)

Step	Marks
Sides along the axes: 4 and 3	1
Third side $= 9 + 16 = 5$	1
Perimeter $= 12$ units	1

Find the point on the x-axis that is equidistant from $(7, 6)$ and $(- 3, 4)$ . (4 marks)

Step	Marks
Let $P (x, 0)$ ; $P A^{2} = P B^{2}$	1
$(x - 7)^{2} + 36 = (x + 3)^{2} + 16$	1
$- 14 x + 85 = 6 x + 25 \Rightarrow 20 x = 60$	1
$x = 3$ : $P (3, 0)$	1

Solve graphically $2 x - y = 4$ and $x + y = 5$ , and find the coordinates of the points where the lines meet the x-axis. (5 marks)

Step	Marks
Tables: $2 x - y = 4$ : $(0, - 4)$ , $(2, 0)$ , $(4, 4)$ ; $x + y = 5$ : $(0, 5)$ , $(5, 0)$ , $(3, 2)$	1
Axes with a stated scale; both lines drawn with a ruler and labelled	1
Intersection read as $(3, 2)$	1
Solution $x = 3$ , $y = 2$ ; check $6 - 2 = 4$ , $3 + 2 = 5$	1
x-axis points: $(2, 0)$ and $(5, 0)$	1

Where students lose marks in Coordinate Geometry

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

Mistake: Swapping x and y when plotting
$(2, 5)$ plotted as $(5, 2)$ moves the point to the wrong place.
Do this: Say 'along, then up' as you plot.
Usually costs: Every point affected.
Mistake: No scale or unlabelled lines on the graph
Graph marks are given for the scale, labelled axes and labelled lines as well as the answer.
Do this: Write the scale in a corner and the equation beside each line.
Usually costs: 1 to 2 marks.
Mistake: Sign errors inside the distance formula
$(x_{2} - x_{1})$ with a negative coordinate: $5 - (- 1) = 6$ , not 4.
Do this: Put negative coordinates in brackets before subtracting.
Usually costs: The answer.
Mistake: Taking only one value when squaring
$(x - 3)^{2} = 9$ gives $x = 6$ or $x = 0$ ; students give only one.
Do this: Write $\pm$ when taking the square root.
Usually costs: 1 mark.
Mistake: Using only sides to prove a square
Four equal sides make a rhombus; the diagonals must also be equal for a square.
Do this: Find both diagonals as well.
Usually costs: 1 mark.

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 · MCQExam level
The point $(0, - 3)$ lies:
(a) in quadrant III
(b) on the x-axis
(c) on the y-axis
(d) in quadrant IV
Hint
x = 0.
Answer
(c)
Q2 · 2 marks · Short answerExam level
Find the distance between $(2, 3)$ and $(5, 7)$ .
Hint
3 and 4.
Answer
5 units.
Q3 · 2 marks · Short answerExam level
Show that the triangle with vertices $(0, 0)$ , $(4, 0)$ and $(4, 3)$ is right-angled, and find its perimeter.
Hint
Sides 4, 3, 5.
Answer
Sides 4, 3, 5 with $16 + 9 = 25$ : right-angled at $(4, 0)$ . Perimeter $= 12$ .
Q4 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): The points $(1, 2)$ , $(3, 4)$ and $(5, 6)$ are collinear.
Reason (R): The distance between the first and last points equals the sum of the other two distances.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
2√2 + 2√2 = 4√2.
Answer
(c)
Q5 · 3 marks · Short answerHarder exam level
Find $y$ if the point $(4, y)$ is 10 units from $(- 2, 3)$ .
Hint
36 + (y − 3)² = 100.
Answer
$(y - 3)^{2} = 64$ , so $y = 11$ or $y = - 5$ .
Q6 · 3 marks · Short answerHarder exam level
Show that $A (3, 0)$ , $B (6, 4)$ , $C (- 1, 3)$ are the vertices of a right-angled isosceles triangle.
Hint
Find all three sides.
Answer
$A B = 5$ , $A C = 16 + 9 = 5$ , $B C = 49 + 1 = 50$ . $A B = A C$ and $25 + 25 = 50$ : right angle at $A$ .
Q7 · 4 marks · Long answerHarder exam level
Solve graphically $x - 2 y = 0$ and $3 x + 4 y = 20$ .
Hint
Tables: (0, 0), (2, 1), (4, 2) and (0, 5), (4, 2), (−4, 8).
Answer
The lines meet at $(4, 2)$ : $x = 4$ , $y = 2$ . Check: $4 - 4 = 0$ ✓, $12 + 8 = 20$ ✓.
Q8 · 4 marks · Long answerHarder exam level
Prove that the points $(- 2, - 1)$ , $(4, 0)$ , $(3, 3)$ and $(- 3, 2)$ , taken in order, form a parallelogram, and say whether it is a rectangle.
Hint
Opposite sides, then diagonals.
Answer
$A B = 37$ , $B C = 10$ , $C D = 37$ , $D A = 10$ : opposite sides equal, so a parallelogram. Diagonals: $A C = 25 + 16 = 41$ , $B D = 49 + 4 = 53$ : unequal, so not a rectangle.
Q9 · 4 marks · Long answerHarder exam level
The lines $y = 2 x$ and $x + y = 6$ meet at $P$ . Find $P$ graphically and the area of the triangle formed by the two lines and the x-axis.
Hint
x-axis points (0, 0) and (6, 0).
Answer
$P = (2, 4)$ . Base on the x-axis from $(0, 0)$ to $(6, 0)$ : 6; height 4. Area $= 12$ square units.
Q10 · 3 marks · Short answerHarder exam level
If $A (x, 2)$ is equidistant from $B (3, - 1)$ and $C (5, 1)$ , find $x$ .
Hint
AB² = AC².
Answer
$(x - 3)^{2} + 9 = (x - 5)^{2} + 1 \Rightarrow - 6 x + 18 = - 10 x + 26 \Rightarrow x = 2$ .
Q11 · 2 marks · Short answerExam level
In the formula $C = \frac{5}{9} (F - 32)$ converting temperatures, name the independent and dependent variables, and find $F$ when $C = 25$ .
Hint
Rearrange.
Answer
$F$ independent, $C$ dependent. $25 = \frac{5}{9} (F - 32) \Rightarrow F = 77$ .
Q12 · 4 marks · Long answerHOTS / competency
Three vertices of a rhombus are $A (2, - 1)$ , $B (5, 3)$ and $C (10, 3)$ . Find the fourth vertex $D$ by plotting, and verify using the distance formula that all sides are equal.
Hint
In a rhombus ABCD, D is found from A by the same move as from B to C.
Answer
From $B$ to $C$ : 5 right, 0 up; so $D = (7, - 1)$ . $A B = 9 + 16 = 5$ , $B C = 5$ , $C D = 9 + 16 = 5$ , $D A = 5$ ✓.

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.In which quadrant is (−4, −3)?Show answer
Quadrant III.
2.What is the ordinate of (5, −2)?Show answer
−2.
3.Write the distance formula.Show answer
√((x₂ − x₁)² + (y₂ − y₁)²).
4.Coordinates of a point on the y-axis?Show answer
(0, y).
5.In y = 3x + 1, which is the dependent variable?Show answer
y.
6.How is the solution shown on a graph of two lines?Show answer
As the point of intersection.
7.What does it mean if two lines are parallel?Show answer
The equations have no solution.
8.Distance of (3, 4) from the origin?Show answer
5.

Best order to study it

Learn axes, quadrants and ordered pairs; practise plotting.
Learn dependent and independent variables with examples.
Draw graphs of linear equations from tables.
Solve pairs of equations graphically on graph paper.
Practise the distance formula and figure-identification proofs.

Quick revision checklist

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

Can you name the quadrant from the signs?
Can you identify dependent and independent variables?
Do you use three points to draw each line?
Do you state a scale and label your lines?
Can you read the solution from the intersection and check it?
Can you find an unknown coordinate from a distance?
Can you prove a figure is a square, rhombus or right triangle?

Coordinate Geometry: frequently asked questions

Are the section and mid-point formulas in Class IX?

No. The Class IX syllabus covers plotting, graphical solutions and the distance formula. Section and mid-point formulas, and the equation of a line, are in Class X.

Do I need graph paper for the graphical method?

Yes. Draw accurately on graph paper with a clear scale; the answer is read from the intersection.

What if the lines meet at a point that is not on a grid crossing?

Read it as accurately as you can and check by substitution. Exam questions are usually set so the solution has whole-number coordinates.

Why is the distance formula related to Pythagoras?

The horizontal and vertical differences between the points form the legs of a right triangle, and the distance is its hypotenuse.

Can a distance be negative?

No. A distance is a square root of a sum of squares, so it is always zero or positive.

Based on: CISCE ICSE 2028 Mathematics (51) syllabus (Classes IX and X); CISCE Regulations for ICSE Examination Year 2028