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.
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
