MathsCBSE Class 10 · Coordinate Geometry

Coordinate Geometry Basics and the Cartesian Plane for Class 10

Updated 14 October 2026 · Fact-checked

Coordinate geometry places every point in a plane as an ordered pair (x, y) measured from two perpendicular axes. The board tests reading quadrants from signs, writing points on the axes, distance of a point from each axis, mirror images, and lengths of horizontal or vertical segments, usually as MCQs or as the first step of longer problems.

The x-value 3 is the distance from the y-axis, the y-value 4 is the distance from the x-axis, and flipping in the x-axis changes only the sign of y.

Illustration drawn for QuestionBuddy.

Understand Coordinate Geometry Basics and Cartesian Plane

The Cartesian plane is made of two number lines that cross at right angles. The horizontal one is the x-axis and the vertical one is the y-axis. They meet at the origin $O (0, 0)$ . Every point P is named by an ordered pair $(x, y)$ . Order matters: $(2, 5)$ and $(5, 2)$ are different points.

The x-coordinate (abscissa) is the signed distance of P from the y-axis. The y-coordinate (ordinate) is the signed distance of P from the x-axis. This feels backwards at first, but think of it this way: to reach $(3, 4)$ you move 3 units sideways, away from the y-axis, then 4 units up, away from the x-axis. So the distance of $(x, y)$ from the x-axis is $∣ y ∣$ , and from the y-axis is $∣ x ∣$ .

The axes split the plane into four quadrants, counted anticlockwise from the top right. The signs give the quadrant: I $(+, +)$ , II $(-, +)$ , III $(-, -)$ , IV $(+, -)$ . A point on an axis is in no quadrant. Every point on the x-axis has $y = 0$ , so it looks like $(a, 0)$ . Every point on the y-axis looks like $(0, b)$ .

Two more facts come straight from the signs. The mirror image of $(x, y)$ in the x-axis is $(x, - y)$ , and in the y-axis it is $(- x, y)$ . Reflecting in the origin changes both signs: $(- x, - y)$ . Also, if two points share the same y-coordinate, the segment joining them is horizontal and its length is $∣ x_{2} - x_{1} ∣$ . If they share the same x-coordinate, the segment is vertical and its length is $∣ y_{2} - y_{1} ∣$ .

These basics hold the whole chapter together. The distance formula is built from a horizontal and a vertical segment using Pythagoras theorem. In equidistance and ratio questions, your first move is to write the unknown point as $(x, 0)$ or $(0, y)$ . And a quadrant named in the question tells you which root of an equation to keep. Get the signs right here and the rest of the chapter becomes arithmetic.

Coordinate Geometry Basics and Cartesian Plane: formulas to remember

Distance of a point from the axes: $from x-axis = ∣ y ∣, from y-axis = ∣ x ∣$; For a point $(x, y)$ . The ordinate gives the distance from the x-axis.
Points on the axes: $(a, 0) on the x-axis, (0, b) on the y-axis$; Write unknown points in this form first.
Quadrant signs: $I (+, +), II (-, +), III (-, -), IV (+, -)$; Points on an axis lie in no quadrant.
Mirror images: $(x, y) \to (x, - y) (x-axis), (- x, y) (y-axis), (- x, - y) (origin)$; Only the sign of the coordinate across the mirror changes.
Horizontal and vertical segments: $A B = ∣ x_{2} - x_{1} ∣ (same y), A B = ∣ y_{2} - y_{1} ∣ (same x)$; No square roots needed when points share a coordinate.

How to solve Coordinate Geometry Basics and Cartesian Plane questions

Every basics question is about reading signs and sizes of coordinates correctly. Use this routine.

1Write every named point as an ordered pair, with unknowns where needed, e.g. a point on the x-axis as $(a, 0)$ .
2Use the quadrant to fix signs: decide the sign of x and of y before writing numbers.
3Use distance from the axes: distance from x-axis sets $∣ y ∣$ , distance from y-axis sets $∣ x ∣$ .
4For reflections, change only the sign of the coordinate across the mirror line.
5If two points share an x- or y-coordinate, find the segment length as a difference of the other coordinate.
6For shapes made of horizontal and vertical sides, find lengths this way, then perimeter or area.
7Check: does each point sit in the quadrant or on the axis the question says?

Coordinate Geometry Basics and Cartesian Plane in real life

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.

Coordinate Geometry Basics and Cartesian Plane

Imagine your school hall is set up for the board exam, and the invigilator marks the floor as a grid in metres with the teacher's desk at the origin. The front wall runs along the x-axis and the central aisle along the y-axis. Riya's desk R is 3 m to the left of the aisle and 4 m in front of the wall, and Arjun's desk A is the mirror image of R in the central aisle.

Riya's desk R is in the second quadrant. Write the coordinates of R and A, find the distance RA, and find the perimeter of the rectangle with vertices R, A, the foot of A on the x-axis, and the foot of R on the x-axis.

Try it first, then show the solutionSolution

R is 3 m from the y-axis on the left, so $x = - 3$ ; 4 m from the x-axis, so $y = 4$ : $R (- 3, 4)$ .
Mirror in the y-axis changes sign of x: $A (3, 4)$ .
Same y-coordinate, so $R A = ∣3 - (- 3) ∣ = 6$ m.
Feet on the x-axis: $A^{'} (3, 0)$ and $R^{'} (- 3, 0)$ .
$A A^{'} = ∣4 - 0∣ = 4$ m, so the rectangle is 6 m by 4 m.
Perimeter $= 2 (6 + 4) = 20$ m.

Answer: $R (- 3, 4)$ , $A (3, 4)$ , $R A = 6$ m, perimeter = 20 m

Memory hook: Exam hall grid: left of the aisle means negative x, the aisle is the mirror.

Tricks to remember Coordinate Geometry Basics and Cartesian Plane

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

Opposite axis rule
x tells how far you walk sideways, so x is the distance from the y-axis. y tells how far you walk up or down, so y is the distance from the x-axis.
Use it when: Distance-from-axis MCQs.
Quadrant sign walk
Go anticlockwise from the top right. I: both plus $(+, +)$ . II: x turns minus $(-, +)$ . III: y also turns minus $(-, -)$ . IV: x returns to plus, y stays minus $(+, -)$ .
Use it when: Fixing signs from a named quadrant.
Mirror flips what crosses it
Reflect in the x-axis: y crosses, so y flips. Reflect in the y-axis: x flips.
Use it when: Image questions.
Zero tells the axis
Zero in the y-place: on the x-axis. Zero in the x-place: on the y-axis.
Use it when: Writing unknown points.

Coordinate Geometry Basics and Cartesian Plane: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Quadrant and distance from axes (MCQ) · 1 mark · MCQBoard level

A point P lies in the fourth quadrant at a distance of 7 units from the y-axis and 2 units from the x-axis. Its reflection in the origin is:
(a) $(7, - 2)$
(b) $(- 7, 2)$
(c) $(- 2, 7)$
(d) $(2, - 7)$

Try it first, then show the solutionSolution

Distance from y-axis is $∣ x ∣ = 7$ ; distance from x-axis is $∣ y ∣ = 2$ .
Fourth quadrant: $x > 0$ , $y < 0$ , so $P (7, - 2)$ .
Reflection in the origin changes both signs: $(- 7, 2)$ .

Answer: (b) $(- 7, 2)$

Example 2: Points from conditions on axes · 3 marks · Short answerBoard level

A point A lies on the x-axis at a distance of 5 units from the origin, on the negative side. A point B lies on the y-axis with ordinate 12. C is the image of B in the x-axis. Find the coordinates of A, B and C, and the length BC. Also find the image A' of A in the y-axis, the length AA', and the distance of A from the line BC.

Try it first, then show the solutionSolution

A on the negative x-axis, 5 from O: $A (- 5, 0)$ . (½)
B on the y-axis with ordinate 12: $B (0, 12)$ . (½)
Image in x-axis flips y: $C (0, - 12)$ . (½)
B and C share $x = 0$ , so $B C = ∣12 - (- 12) ∣ = 24$ units. (½)
Image in y-axis flips x: $A^{'} (5, 0)$ ; A and A' share $y = 0$ , so $A A^{'} = ∣5 - (- 5) ∣ = 10$ units. (½)
Line BC is the y-axis, so the distance of A from it is $∣ x ∣ = ∣ - 5∣ = 5$ units. (½)

Answer: $A (- 5, 0)$ , $B (0, 12)$ , $C (0, - 12)$ ; $B C = 24$ units; $A^{'} (5, 0)$ , $A A^{'} = 10$ units; distance of A from BC = 5 units

Example 3: Unknown point with a quadrant condition · 3 marks · Short answerHarder board level

The point $P (2 a - 3, a + 1)$ is equidistant from both axes and lies in the second quadrant. Find a and P. Also find the distance of P from the point $Q (4, a + 1)$ .

Try it first, then show the solutionSolution

Equidistant from axes: $∣2 a - 3∣ = ∣ a + 1∣$ . (½)
Second quadrant: $2 a - 3 < 0$ and $a + 1 > 0$ , so $- (2 a - 3) = a + 1$ . (½)
$3 - 2 a = a + 1 \Rightarrow a = \frac{2}{3}$ . (½)
Check: $2 a - 3 = - \frac{5}{3} < 0$ , $a + 1 = \frac{5}{3} > 0$ . Valid. (½)
$P (- \frac{5}{3}, \frac{5}{3})$ , $Q (4, \frac{5}{3})$ : same y-coordinate. (½)
$P Q = 4 - (- \frac{5}{3}) = \frac{17}{3}$ units. (½)

Answer: $a = \frac{2}{3}$ , $P (- \frac{5}{3}, \frac{5}{3})$ , $P Q = \frac{17}{3}$ units

Example 4: Rectangle with sides parallel to the axes · 5 marks · Long answerHarder board level

ABCD is a rectangle with sides parallel to the axes. $A (- 4, - 1)$ and $C (2, 3)$ are opposite vertices. (i) Find B and D, taking B on the same horizontal line as A. (ii) Name the quadrant of each vertex. (iii) Find the perimeter and area. (iv) Find the points where side AB and side BC meet the axes, if any, and the distance of D from each axis.

Try it first, then show the solutionSolution

B shares y with A and x with C: $B (2, - 1)$ . (½)
D shares x with A and y with C: $D (- 4, 3)$ . (½)
Quadrants: A $(-, -)$ III, B $(+, -)$ IV, C $(+, +)$ I, D $(-, +)$ II. (1)
$A B = ∣2 - (- 4) ∣ = 6$ units, $B C = ∣3 - (- 1) ∣ = 4$ units. (1)
Perimeter $= 2 (6 + 4) = 20$ units; area $= 6 \times 4 = 24$ sq units. (1)
AB is the line $y = - 1$ ; it crosses the y-axis at $(0, - 1)$ . BC is the line $x = 2$ ; it crosses the x-axis at $(2, 0)$ . (½)
D $(- 4, 3)$ : 3 units from the x-axis, 4 units from the y-axis. (½)

Answer: $B (2, - 1)$ , $D (- 4, 3)$ ; quadrants III, IV, I, II; perimeter 20 units, area 24 sq units; AB meets y-axis at $(0, - 1)$ , BC meets x-axis at $(2, 0)$ ; D is 3 from x-axis, 4 from y-axis

Example 5: Case study: a sports day grid · 4 marks · Case-basedHOTS / competency

Suppose for sports day your school marks the ground as a grid in metres with the flag post at the origin. The starting line of a race is the segment from $S_{1} (- 6, - 8)$ to $S_{2} (6, - 8)$ . The finishing line is the image of the starting line in the x-axis. A first-aid tent T is found by reflecting $S_{2}$ first in the y-axis and then reflecting that image in the x-axis. A water stall W has coordinates $(p - 4, 2 p + 1)$ ; it is equidistant from both axes and lies in the second quadrant.
(i) Write the end points of the finishing line. (1)
(ii) Write the coordinates of T and its quadrant. (1)
(iii) Find p and W. Hence find the distance of W from the finishing line and from the vertical track $x = - 6$ . (2)

Try it first, then show the solutionSolution

(i) Reflect in x-axis: $F_{1} (- 6, 8)$ , $F_{2} (6, 8)$ . (1)
(ii) $S_{2} (6, - 8)$ in the y-axis gives $(- 6, - 8)$ ; that in the x-axis gives $T (- 6, 8)$ , quadrant II (it coincides with $F_{1}$ ). (1)
(iii) Equidistant from axes: $∣ p - 4∣ = ∣2 p + 1∣$ . Case $p - 4 = 2 p + 1$ gives $p = - 5$ , point $(- 9, - 9)$ in quadrant III, rejected. (½)
Case $p - 4 = - (2 p + 1)$ gives $3 p = 3$ , $p = 1$ , so $W (- 3, 3)$ , which is in quadrant II. Accepted. (½)
Finishing line is $y = 8$ : distance $= ∣8 - 3∣ = 5$ m. (½)
Track $x = - 6$ : distance $= ∣ - 3 - (- 6) ∣ = 3$ m. (½)

Answer: (i) $(- 6, 8)$ and $(6, 8)$ (ii) $T (- 6, 8)$ , quadrant II (iii) $p = 1$ , $W (- 3, 3)$ ; 5 m from the finishing line, 3 m from the track $x = - 6$

Where students lose marks in Coordinate Geometry Basics and Cartesian Plane

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

Mistake: Swapping distance from the axes
Students think the x-axis distance is the x-coordinate.
Wrong working
Distance of $(4, - 9)$ from the x-axis $= 4$
Right working
Distance from x-axis $= ∣ y ∣ = 9$ units
Do this: The x-axis is horizontal; height above or below it is y.
Usually costs: Full 1 mark in an MCQ.
Mistake: Placing axis points in a quadrant
Signs are read without noticing a zero.
Wrong working
$(- 3, 0)$ is in quadrant II
Right working
$(- 3, 0)$ lies on the negative x-axis, in no quadrant
Do this: Check for a zero coordinate first.
Usually costs: 1 mark.
Mistake: Flipping the wrong sign in a reflection
Confusing mirror line with coordinate changed.
Wrong working
Image of $(2, 5)$ in x-axis $= (- 2, 5)$
Right working
Image in x-axis $= (2, - 5)$
Do this: The coordinate that crosses the mirror changes sign.
Usually costs: 1 mark, and later parts carry the error.
Mistake: Subtracting negatives wrongly in segment lengths
Rushing $a - (- b)$ .
Wrong working
Length from $(- 3, 2)$ to $(5, 2) = 5 - 3 = 2$
Right working
$∣5 - (- 3) ∣ = 8$ units
Do this: Write the bracket with the negative number every time.
Usually costs: ½ to 1 mark per step.

Exam tips

Read the quadrant condition first; it decides signs and which value to keep.
Write unknown axis points as $(a, 0)$ or $(0, b)$ before anything else.
Always give units in lengths and sq units in areas.
Use horizontal/vertical shortcuts instead of the full distance formula when points share a coordinate.
In case studies, list every point's coordinates at the start.

Practice questions on Coordinate Geometry Basics and Cartesian Plane

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

Q1 · 2 marks · Very short answerBoard level
Find the point on the y-axis, below the x-axis, whose distance from the x-axis is twice the distance of $(- 3, 5)$ from the y-axis. Write its image in the x-axis.
Hint
Distance from the y-axis is $∣ x ∣$ .
Answer
Distance of $(- 3, 5)$ from the y-axis $= 3$ , so the required distance is 6; below the x-axis gives the point $(0, - 6)$ ; image in the x-axis $(0, 6)$ .
Q2 · 3 marks · Short answerHarder board level
The point $(3 - k, 2 k - 8)$ lies on the x-axis. Find k, the point, and its distance from $(- 4, 0)$ . In which quadrant is $(3 - k, k)$ ?
Hint
On the x-axis, y = 0.
Answer
$2 k - 8 = 0 \Rightarrow k = 4$ ; point $(- 1, 0)$ ; distance $∣ - 1 - (- 4) ∣ = 3$ units; $(- 1, 4)$ is in quadrant II.
Q3 · 4 marks · Long answerHOTS / competency
A rectangle ABCD has its sides parallel to the axes. Its opposite vertices are $A (1 - a, 2 b - 5)$ , which lies in quadrant III, and $C (2 a + 1, b + 4)$ , which lies in quadrant I. The side parallel to the x-axis is 9 units long and the side parallel to the y-axis is 8 units long. Find a and b, all four vertices, and the perimeter and area of the rectangle. Check that A and C lie in the stated quadrants.
Hint
The horizontal side is the difference of the x-coordinates of A and C; the vertical side is the difference of their y-coordinates.
Answer
Horizontal side: $(2 a + 1) - (1 - a) = 3 a = 9$ , so $a = 3$ . Vertical side: $(b + 4) - (2 b - 5) = 9 - b = 8$ , so $b = 1$ . So $A (- 2, - 3)$ and $C (7, 5)$ ; A is in quadrant III and C in quadrant I, as required. The other vertices share coordinates: $B (7, - 3)$ and $D (- 2, 5)$ . Perimeter $= 2 (9 + 8) = 34$ units; area $= 9 \times 8 = 72$ square units.
Q4 · 5 marks · Long answerBoard level
Points $A (2, - 3)$ , $B (2, 5)$ and $C (- 4, 5)$ are given. Find D so that ABCD is a rectangle with sides parallel to the axes. Find its perimeter, area, and the points where its sides cross the axes.
Hint
D shares x with C and y with A.
Answer
$D (- 4, - 3)$ ; $A B = 8$ , $B C = 6$ ; perimeter 28 units, area 48 sq units; AB ( $x = 2$ ) meets x-axis at $(2, 0)$ , CD ( $x = - 4$ ) at $(- 4, 0)$ , AD ( $y = - 3$ ) meets y-axis at $(0, - 3)$ , BC ( $y = 5$ ) at $(0, 5)$ .
Q5 · 1 mark · Assertion-reasonBoard level
Assertion (A): The point $(0, - 7)$ lies in the fourth quadrant.
Reason (R): A point with positive abscissa and negative ordinate lies in the fourth quadrant.
(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
Is abscissa 0 positive?
Answer
(d). $(0, - 7)$ is on the y-axis, in no quadrant, so A is false; R is true.
Q6 · 3 marks · Short answerHOTS / competency
The point $(k + 2, k^{2} - 9)$ lies on the negative x-axis. Find k, and the image of the point $(k, k + 5)$ in the y-axis. Hence find the length of the segment joining $(k, k + 5)$ to its image.
Hint
On the x-axis the ordinate is 0; on the negative side the abscissa is negative. Test both roots.
Answer
$k^{2} - 9 = 0 \Rightarrow k = 3$ or $k = - 3$ . For $k = 3$ , $k + 2 = 5 > 0$ (positive side), rejected; for $k = - 3$ , $k + 2 = - 1 < 0$ , so $k = - 3$ and the point is $(- 1, 0)$ . Then $(k, k + 5) = (- 3, 2)$ , image in the y-axis $(3, 2)$ ; length $= ∣3 - (- 3) ∣ = 6$ units.
Q7 · 3 marks · Short answerHarder board level
The point $A (k, 0)$ lies on the negative x-axis and $B (0, k + 3)$ lies on the positive y-axis. C is the image of B in the x-axis and D is the image of A in the y-axis. The diagonal AD of quadrilateral ABDC is 4 units long. Find k, all four points, the other diagonal BC, and the area of ABDC.
Hint
A and D both lie on the x-axis, so AD is the difference of their x-coordinates. Remember the conditions k < 0 and k + 3 > 0.
Answer
$D (- k, 0)$ , so $A D = - k - k = - 2 k = 4$ and $k = - 2$ ; this satisfies $k < 0$ and $k + 3 = 1 > 0$ . Points: $A (- 2, 0)$ , $B (0, 1)$ , $C (0, - 1)$ , $D (2, 0)$ . $B C = 2$ units. The diagonals lie along the axes and bisect each other at right angles, so ABDC is a rhombus with area $\frac{1}{2} \times 4 \times 2 = 4$ square units.

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.Distance of $(x, y)$ from the y-axis?Show answer
$∣ x ∣$
2.Signs in quadrant III?Show answer
$(-, -)$
3.General point on the y-axis?Show answer
$(0, b)$
4.Image of $(a, b)$ in the y-axis?Show answer
$(- a, b)$
5.Another name for the y-coordinate?Show answer
Ordinate

Coordinate Geometry Basics and Cartesian Plane: frequently asked questions

What is the difference between abscissa and ordinate?

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

Does a point on an axis belong to any quadrant?

No. Points on the axes, including the origin, lie on boundary lines and are not in any quadrant.

Is the area of a triangle from coordinates in the 2026-27 syllabus?

No, the area of a triangle using coordinates is not in the 2026-27 syllabus. External division has also been removed.

Why do I need basics if the board asks distance and section formula?

Almost every longer question starts by writing points correctly, such as a point on an axis or a point in a named quadrant. A sign error here spoils the whole answer.