CBSE Maths · Class 10
Chapter 7: Coordinate Geometry
Name: ________________________
Date: ______________
Marks: ____ / 10
Distance formula, section formula, area of a triangle from coordinates.
Questions
- Midpoint of (−1,4) and (5,−2).
- Area of triangle (0,0),(4,0),(0,3).
- Distance from origin to (7,24).
- Distance between (1,2) and (4,6).
- Section formula: (2,4) and (6,8) divided in 1:3.
- Are (1,2),(2,3),(3,4) collinear?
- Centroid of (1,2),(3,8),(5,5).
- Ratio in which (4,6) divides join of (2,4) and (8,10).
- Find k if (k,3),(2,k) and (6,5) are collinear.
- Find the point on the x-axis equidistant from (2,−5) and (−2,9).
Extra Practice (Optional)
- Verify: (2,−2),(14,10),(11,13),(−1,1) is a parallelogram.
- Find x so that (3,x) is equidistant from (3,6) and (−3,4).
- Coordinates of point dividing (−1,3) and (4,−7) in 2:3.
- Show A(1,7),B(4,2),C(−1,−1),D(−4,4) is a square.
- Find area of △ with vertices (2,3),(−1,0),(2,−4).
Answer Key (For Teachers)
- Answer: (2,1).
Working: Average of coordinates.
- Answer: 6.
Working: 21⋅4⋅3. - Answer: 25.
Working: 625. - Answer: 5.
Working: 9+16. - Answer: (3,5).
Working: Apply formula.
- Answer: Collinear.
- Answer: (3,5).
Working: Average vertices.
- Answer: 1:2.
Working: Set k:1, solve. - Answer: k=211±33.
Working: Area =0: k(k−5)+2(5−3)+6(3−k)=0. → k2−11k+22=0. - Answer: (−7,0).
Working: Let point (x,0): (x−2)2+25=(x+2)2+81. → −4x+25=4x+81⇒x=−7.
Gyaan Maths · CBSE Class 10 · Chapter 7