Coordinate Geometry Test Class 10 Maths — 40 Marks
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PERIODIC TEST — Coordinate Geometry (2026-27) | CLASS 10 | MATHEMATICS
Max Marks: 40 · Time: 1 hour 30 minutes (Time left: 90:00)
| Section | Marks | Questions | Time | Strategy |
|---|---|---|---|---|
| A | 12 | 12 MCQ | 15 min | Recall the distance and midpoint formulas |
| B | 8 | 4 Short | 20 min | Substitute the coordinates carefully |
| C | 12 | 4 Long | 30 min | Show every substitution step |
| D | 8 | 2 Case | 20 min | Sketch the points on a rough grid |
| Review | — | — | 5 min | Recheck the signs of the coordinates |
General Instructions: All questions are compulsory. Section A has 12 MCQs of 1 mark each. Internal choice is given in Sections B, C and D — attempt any ONE of the two options. Leave irrational answers in surd form in lowest terms. A rough sketch is enough for case-study parts.
What is this paper? A printable 40-mark Class 10 Maths chapter test on Coordinate Geometry with a full per-question marking scheme, internal choice and practice/test modes. It is built to the current CBSE 2026-27 syllabus and is board-pattern practice, not a prediction of any examination paper.
d = √[(x2 − x1)2 + (y2 − y1)2]
M = (x1 + x22, y1 + y22)
P = (m x2 + n x1m + n, m y2 + n y1m + n)
These are the distance formula, the mid-point formula, and the section formula for a point dividing a segment internally in the ratio m : n.
Section A — Multiple Choice (12 × 1 = 12 marks)
Choose the single best option.
The distance between the points (0, 0) and (3, 4) is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Distance formula
d = √[(3 − 0)2 + (4 − 0)2]
= √(9 + 16)
= √25
= 5
= √(9 + 16)
= √25
= 5
Answer: (a) 5
The mid-point of the line segment joining (2, 4) and (6, 8) is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Mid-point formula
M = ((2 + 6)/2, (4 + 8)/2)
= (8/2, 12/2)
= (4, 6)
= (8/2, 12/2)
= (4, 6)
Answer: (a) (4, 6)
The distance of the point (5, 12) from the origin is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Distance from the origin
d = √(52 + 122)
= √(25 + 144)
= √169
= 13
= √(25 + 144)
= √169
= 13
Answer: (a) 13
The distance between the points (2, 3) and (2, 7) is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Same x-coordinate
d = √[(2 − 2)2 + (7 − 3)2]
= √(0 + 16)
= 4
= √(0 + 16)
= 4
Answer: (a) 4
The coordinates of the point dividing the segment joining (1, 2) and (4, 5) internally in the ratio 1 : 2 are
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Section formula (internal), m : n = 1 : 2
x = (1×4 + 2×1)/(1 + 2) = 6/3 = 2
y = (1×5 + 2×2)/(1 + 2) = 9/3 = 3
Point = (2, 3)
y = (1×5 + 2×2)/(1 + 2) = 9/3 = 3
Point = (2, 3)
Answer: (a) (2, 3)
A point on the x-axis has coordinates of the form
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
y is zero on the x-axis
Every point on the x-axis has its distance from the x-axis equal to 0, so y = 0.
Form: (x, 0)
Form: (x, 0)
Answer: (b) (x, 0)
The perimeter of the triangle with vertices (0, 0), (3, 0) and (0, 4) is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Find the three sides
(0,0) to (3,0): 3
(0,0) to (0,4): 4
(3,0) to (0,4): √(32 + 42) = 5
Perimeter = 3 + 4 + 5 = 12
(0,0) to (0,4): 4
(3,0) to (0,4): √(32 + 42) = 5
Perimeter = 3 + 4 + 5 = 12
Answer: (a) 12
The distance between the points (−1, 2) and (3, −1) is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Substitute
d = √[(3 − (−1))2 + (−1 − 2)2]
= √(42 + (−3)2)
= √(16 + 9) = √25 = 5
= √(42 + (−3)2)
= √(16 + 9) = √25 = 5
Answer: (a) 5
The point (−2, 3) lies in the quadrant
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Signs decide the quadrant
x is negative and y is positive.
That is Quadrant II.
That is Quadrant II.
Answer: (b) II
If the distance between the points (x, 0) and (−3, 0) is 5, then x is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Distance on the x-axis
|x − (−3)| = 5
|x + 3| = 5
x + 3 = 5 or x + 3 = −5
x = 2 or x = −8
|x + 3| = 5
x + 3 = 5 or x + 3 = −5
x = 2 or x = −8
Answer: (a) 2 or −8
The mid-point of the segment joining (a, b) and (−a, −b) is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Mid-point formula
M = ((a + (−a))/2, (b + (−b))/2)
= (0/2, 0/2)
= (0, 0)
= (0/2, 0/2)
= (0, 0)
Answer: (b) (0, 0)
The distance of the point (p, q) from the origin is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Apply the distance formula with (0, 0)
d = √[(p − 0)2 + (q − 0)2]
= √(p2 + q2)
= √(p2 + q2)
Answer: (c) √(p2 + q2)
Section B — Short Answer (4 × 2 = 8 marks)
Find the distance between the points (2, 3) and (−1, −1).
Marking Scheme: Applying the distance formula 1 mark · Simplified answer 1 mark.
Total: 2 Marks
Total: 2 Marks
Substitute the coordinates
d = √[(−1 − 2)2 + (−1 − 3)2]
= √[(−3)2 + (−4)2]
= √(9 + 16) = √25
= √[(−3)2 + (−4)2]
= √(9 + 16) = √25
Simplify
d = 5 units
Answer: 5 units
Find the coordinates of the point which divides the segment joining (−1, 7) and (4, −3) internally in the ratio 2 : 3.
Marking Scheme: Correct section formula with m : n = 2 : 3 1 mark · Both coordinates 1 mark.
Total: 2 Marks
Total: 2 Marks
Apply the section formula
x = (2×4 + 3×(−1))/(2 + 3) = (8 − 3)/5 = 5/5 = 1
y = (2×(−3) + 3×7)/5 = (−6 + 21)/5 = 15/5 = 3
y = (2×(−3) + 3×7)/5 = (−6 + 21)/5 = 15/5 = 3
Point
Point = (1, 3)
Answer: (1, 3)
Find the value(s) of x if the distance between (x, 3) and (2, −1) is 5.
Marking Scheme: Setting up the equation 1 mark · Both values of x 1 mark.
Total: 2 Marks
Total: 2 Marks
Use the distance formula
(x − 2)2 + (3 − (−1))2 = 52
(x − 2)2 + 16 = 25
(x − 2)2 = 9
(x − 2)2 + 16 = 25
(x − 2)2 = 9
Solve
x − 2 = 3 or x − 2 = −3
x = 5 or x = −1
x = 5 or x = −1
Answer: x = 5 or x = −1
Answer either Option 1 or Option 2.
Option 1 — Point on the y-axis
Marking Scheme: Letting the point be (0, y) and equating distances 1 mark · Value of y 1 mark.
Total: 2 Marks
Total: 2 Marks
Let the point be (0, y)
Distance to (6, 5): (0 − 6)2 + (y − 5)2
Distance to (−4, 3): (0 + 4)2 + (y − 3)2
Set equal: 36 + (y − 5)2 = 16 + (y − 3)2
Distance to (−4, 3): (0 + 4)2 + (y − 3)2
Set equal: 36 + (y − 5)2 = 16 + (y − 3)2
Solve
36 + y2 − 10y + 25 = 16 + y2 − 6y + 9
61 − 10y = 25 − 6y
36 = 4y → y = 9
61 − 10y = 25 − 6y
36 = 4y → y = 9
Answer: (0, 9)
— OR —
Option 2
Find the mid-point of the segment joining (7, −3) and (−1, 9).
Marking Scheme: Mid-point formula applied 1 mark · Point 1 mark.
Total: 2 Marks
Total: 2 Marks
Apply the mid-point formula
M = ((7 + (−1))/2, (−3 + 9)/2)
= (6/2, 6/2)
= (3, 3)
= (6/2, 6/2)
= (3, 3)
Answer: (3, 3)
[Attempt any ONE of the two options of Q16]
Section C — Long Answer (4 × 3 = 12 marks)
Find the value(s) of y for which the distance between the points P(2, −3) and Q(10, y) is 10.
Marking Scheme: Correct distance equation 1 mark · Simplifying to a quadratic/linear condition 1 mark · Both values of y 1 mark.
Total: 3 Marks
Total: 3 Marks
Use the distance formula
PQ = 10
(10 − 2)2 + (y + 3)2 = 100
64 + (y + 3)2 = 100
(10 − 2)2 + (y + 3)2 = 100
64 + (y + 3)2 = 100
Simplify
(y + 3)2 = 36
Solve
y + 3 = 6 or y + 3 = −6
y = 3 or y = −9
y = 3 or y = −9
Answer: y = 3 or y = −9
Find the coordinates of the points of trisection of the line segment joining A(2, −2) and B(−7, 4).
Marking Scheme: Identifying the ratios 1 : 2 and 2 : 1 1 mark · First trisection point 1 mark · Second trisection point 1 mark.
Total: 3 Marks
Total: 3 Marks
Trisection ratios
The two points divide AB in the ratios 1 : 2 and 2 : 1.
Point P (ratio 1 : 2)
P = ((1×(−7) + 2×2)/3, (1×4 + 2×(−2))/3)
= ((−7 + 4)/3, (4 − 4)/3)
= (−1, 0)
= ((−7 + 4)/3, (4 − 4)/3)
= (−1, 0)
Point Q (ratio 2 : 1)
Q = ((2×(−7) + 1×2)/3, (2×4 + 1×(−2))/3)
= ((−14 + 2)/3, (8 − 2)/3)
= (−4, 2)
= ((−14 + 2)/3, (8 − 2)/3)
= (−4, 2)
Answer: (−1, 0) and (−4, 2)
Answer either Option 1 or Option 2.
Option 1 — Parallelogram
Marking Scheme: Using “diagonals bisect each other” 1 mark · Equation for x 1 mark · Equation for y 1 mark.
Total: 3 Marks
Total: 3 Marks
Diagonals bisect each other
For ABCD in order, mid-point of AC = mid-point of BD.
AC mid = ((1 + x)/2, (2 + 6)/2) = ((1 + x)/2, 4)
BD mid = ((4 + 3)/2, (y + 5)/2) = (7/2, (y + 5)/2)
AC mid = ((1 + x)/2, (2 + 6)/2) = ((1 + x)/2, 4)
BD mid = ((4 + 3)/2, (y + 5)/2) = (7/2, (y + 5)/2)
Equate x-coordinates
(1 + x)/2 = 7/2 → 1 + x = 7 → x = 6
Equate y-coordinates
4 = (y + 5)/2 → y + 5 = 8 → y = 3
Answer: x = 6, y = 3
— OR —
Option 2
Find the distance between the points (a, b) and (−a, −b).
Marking Scheme: Substitution in the distance formula 1 mark · Simplification 1 mark · Final surd form 1 mark.
Total: 3 Marks
Total: 3 Marks
Substitute
d = √[(−a − a)2 + (−b − b)2]
= √[(−2a)2 + (−2b)2]
= √(4a2 + 4b2)
= √[(−2a)2 + (−2b)2]
= √(4a2 + 4b2)
Simplify
= √[4(a2 + b2)]
= 2√(a2 + b2)
= 2√(a2 + b2)
Answer: 2√(a2 + b2)
[Attempt any ONE of the two options of Q19]
Answer either Option 1 or Option 2.
Option 1 — Point on the x-axis
Marking Scheme: Letting the point be (x, 0) and equating distances 1 mark · Solving for x 1 mark · The point 1 mark.
Total: 3 Marks
Total: 3 Marks
Let the point be (x, 0)
Distance to (2, −5): (x − 2)2 + 25
Distance to (−2, 9): (x + 2)2 + 81
Set equal: (x − 2)2 + 25 = (x + 2)2 + 81
Distance to (−2, 9): (x + 2)2 + 81
Set equal: (x − 2)2 + 25 = (x + 2)2 + 81
Solve
x2 − 4x + 4 + 25 = x2 + 4x + 4 + 81
−4x + 29 = 4x + 85
−8x = 56 → x = −7
−4x + 29 = 4x + 85
−8x = 56 → x = −7
The point
(−7, 0)
Answer: (−7, 0)
— OR —
Option 2
Find the coordinates of the point which divides the segment joining (4, −3) and (8, 5) internally in the ratio 3 : 1.
Marking Scheme: Correct section formula 1 mark · x-coordinate 1 mark · y-coordinate and point 1 mark.
Total: 3 Marks
Total: 3 Marks
Apply the section formula
x = (3×8 + 1×4)/(3 + 1) = (24 + 4)/4 = 28/4 = 7
y = (3×5 + 1×(−3))/4 = (15 − 3)/4 = 12/4 = 3
y = (3×5 + 1×(−3))/4 = (15 − 3)/4 = 12/4 = 3
Answer: (7, 3)
[Attempt any ONE of the two options of Q20]
Section D — Case Study (2 × 4 = 8 marks)
A park is planned on a coordinate grid. Three lamp posts are fixed at A(1, 1), B(7, 1) and C(7, 9).
Marking Scheme: Part (a) 1 mark · Part (b) 1.5 marks · Part (c) 1.5 marks.
Total: 4 Marks
Total: 4 Marks
(a) Distance AB
AB = 7 − 1 = 6 units
(b) Distance BC
BC = 9 − 1 = 8 units
(c) Distance AC and the type of triangle
AC = √[(7 − 1)2 + (9 − 1)2] = √(36 + 64) = √100 = 10
Check: AB2 + BC2 = 36 + 64 = 100 = AC2
So the triangle is right-angled at B.
Check: AB2 + BC2 = 36 + 64 = 100 = AC2
So the triangle is right-angled at B.
Answer: (a) 6 units (b) 8 units (c) AC = 10 units; right-angled at B
Answer either Option 1 or Option 2.
Option 1 — A walk on a grid
Riya’s house is at (0, 0) and a shop is at (5, 12). She then walks to a point (5, 0).
Marking Scheme: Part (a) 1 mark · Part (b) 1.5 marks · Part (c) 1.5 marks.
Total: 4 Marks
Total: 4 Marks
(a) House to shop
d = √(52 + 122) = √(25 + 144) = √169 = 13 units
(b) Shop to (5, 0)
d = |12 − 0| = 12 units
(c) Mid-point and comparison
Mid-point of house and shop = ((0 + 5)/2, (0 + 12)/2) = (2.5, 6)
13 > 12, so the shop is farther from the house than from (5, 0).
13 > 12, so the shop is farther from the house than from (5, 0).
Answer: (a) 13 units (b) 12 units (c) (2.5, 6); yes, 13 > 12
— OR —
Option 2
A square has vertices A(0, 0), B(4, 0), C(4, 4) and D(0, 4).
Marking Scheme: Part (a) 1 mark · Part (b) 1.5 marks · Part (c) 1.5 marks.
Total: 4 Marks
Total: 4 Marks
(a) Length of a side
AB = 4 − 0 = 4 units
(b) Length of the diagonal AC
AC = √[(4 − 0)2 + (4 − 0)2] = √(16 + 16) = √32 = 4√2 units
(c) Mid-point of AC and the other diagonal
Mid-point of AC = (4/2, 4/2) = (2, 2)
BD = √[(0 − 4)2 + (4 − 0)2] = √32 = 4√2
Both diagonals are equal.
BD = √[(0 − 4)2 + (4 − 0)2] = √32 = 4√2
Both diagonals are equal.
Answer: (a) 4 units (b) 4√2 units (c) (2, 2); both diagonals equal 4√2
[Attempt any ONE of the two options of Q22]
Self-assessment for Sections B, C and D (28 marks)
The MCQ section is scored for you. For the written questions, tick each one you got right out of the marks shown, and add them up honestly.
Written total: 0 / 28
Keep going with these
Written and checked by Pandey Sir · Pandey Sir’s BEST Classes, Dammaiguda, Hyderabad. Every answer on this paper is worked out and checked before publishing.
This is board-pattern practice for Coordinate Geometry, built to the current CBSE Class 10 Maths syllabus 2026-27. It is not a prediction of any examination paper.
Curriculum scope checked against cbseacademic.nic.in. Spotted a mistake? Tell us here.