Extra Questions: Class 10 Maths Chapter 7 Coordinate Geometry

These go beyond Exercises 7.1-7.2 into genuine HOTS territory — a general proof, a reverse-engineering collinearity problem, an assertion-reason question, a fresh-numbers median/centroid problem, a reverse section-formula problem, and a quadrilateral-area problem (extending the removed Exercise 7.3/7.4 content). See the standard-level Chapter 7 Solutions first if you haven’t already. These Class 10 Mathematics Chapter 7 important questions are handy for last-minute exam practice.

Last Updated: October 24, 2026

How to Approach the HOTS Questions in This Chapter

Extra Questions (HOTS Level): Coordinate Geometry (Class 10 Maths Chapter 7)

Q1. (General proof) Prove that if a point P(x,y) is equidistant from two points A(x₁,y₁) and B(x₂,y₂), then P lies on the perpendicular bisector of AB.
PA=PB ⇒ PA²=PB² ⇒ (x-x₁)²+(y-y₁)²=(x-x₂)²+(y-y₂)². Expanding and cancelling the x²,y² terms gives 2x(x₂-x₁)+2y(y₂-y₁)=x₂²-x₁²+y₂²-y₁² — a linear equation in x,y, i.e. a straight line. Since this is exactly the locus of every point equidistant from A and B, and the perpendicular bisector of AB is by definition that same locus, P must lie on the perpendicular bisector of AB. This holds for any A, B — not tied to specific numbers.

Q2. (Reverse-engineering, fresh numbers) Find the value of k such that the points (7,-2), (5,1) and (3,k) are collinear.
Collinear ⇒ area=0: ½|7(1-k)+5(k+2)+3(-2-1)|=0 ⇒ 7-7k+5k+10-9=0 ⇒ 8-2k=0 ⇒ k=4.

Q3. (Assertion-Reason) Assertion (A): The distance between (0,0) and (36,15) is 39 units, and a right triangle can be formed using the third point (0,15), with legs 36 and 15. Reason (R): In a right triangle, hypotenuse² equals the sum of the squares of the other two sides (Pythagoras theorem), and 36²+15²=39².
36²+15²=1296+225=1521=39², confirming the right angle at (0,15). Both A and R are true, and R is the correct explanation of A.

Q4. (Fresh-numbers word problem) The vertices of a triangular park are A(2,1), B(10,1) and C(6,7). Find (i) the length of the median from C to AB, and (ii) the coordinates of the centroid.
Midpoint of AB = (6,1). (i) Median length = √((6-6)²+(7-1)²) = 6 units. (ii) Centroid = ((2+10+6)/3, (1+1+7)/3) = (6, 3).

Q5. (Reverse section-formula problem, fresh numbers) Find the ratio in which the point (4, m) divides the segment joining A(2,-4) and B(6,8). Also find m.
Let the ratio be λ:1 from A. (6λ+2)/(λ+1)=4 ⇒ 6λ+2=4λ+4 ⇒ λ=1, so ratio 1:1 (the midpoint). y=(8-4)/2 ⇒ m=2.

Q6. (Quadrilateral area, extends removed Ex 7.3/7.4 content) Find the area of the quadrilateral whose vertices, taken in order, are (-4,-2), (-3,-5), (3,-2) and (2,3).
Using the shoelace extension of the triangle-area formula: Area=½|x₁(y₂-y₄)+x₂(y₃-y₁)+x₃(y₄-y₂)+x₄(y₁-y₃)|=½|(-4)(-8)+(-3)(0)+3(8)+2(0)|=½|32+24|=28 sq units.

See also: Revision Notes | Formulas Handbook

Frequently Asked Questions

How is the section formula used to find a point that divides a line segment in a given ratio?
The section formula calculates the coordinates of a point dividing a line segment joining two given points in a specific ratio, by taking a weighted average of the x and y coordinates based on that ratio.

How can the distance formula be used to check whether three given points are collinear?
If the sum of the distances between two pairs of the three points equals the distance between the remaining pair, the three points lie on the same straight line and are collinear.

Written by Satish

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