PYQs
Practise questions grouped by chapter and source year.
Practice questions are organised using the years and labels in our source collection. Work through the solutions with your teacher; these are not a complete official examination paper.
Practice question 1
Source year: 20253 marksFind point dividing (1,3) and (4,6) in ratio 2:1. [CBSE 2025]
Practice question 2
Source year: 20242 marksFind distance between A(5,3) and B(1,6). [CBSE 2024]
Practice question 3
Source year: 20253 marksFind coordinates of point which divides line segment joining (1,3) and (4,6) in ratio 2:1. [CBSE 2025]
Practice question 4
Source year: 20242 marksFind distance between points A(5,3) and B(1,6). [CBSE 2024]
Practice question 5
Source year: 20233 marksIf points (2,3), (4,k) and (6,-3) are collinear, find k. [CBSE 2023]
Practice question 6
Source year: 20223 marksFind area of triangle with vertices (1,1), (5,3) and (3,7). [CBSE 2022]
Practice question 7
Source year: 20212 marksFind midpoint of line segment joining (3,4) and (5,6). [CBSE 2021]
Formulae
Revisit important relationships and quick revision formulae.
Distance
d = √[(x₂-x₁)² + (y₂-y₁)²]
Section Formula
x = (mx₂+nx₁)/(m+n)
Distance Formula
Distance between (x₁,y₁) and (x₂,y₂):\nd = √[(x₂-x₁)² + (y₂-y₁)²]
Section Formula
Point dividing line in ratio m:n:\nx = (mx₂+nx₁)/(m+n), y = (my₂+ny₁)/(m+n)
Midpoint Formula
Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2)
Area of Triangle
Area = ½|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|
Derivations
Work through the reasoning behind key concepts.
Distance Formula
For points (x₁,y₁) and (x₂,y₂), form right triangle. Horizontal distance = |x₂-x₁|, Vertical distance = |y₂-y₁|. By Pythagoras: d² = (x₂-x₁)² + (y₂-y₁)². Therefore d = √[(x₂-x₁)² + (y₂-y₁)²].
Section Formula
Point P divides AB in ratio m:n. Use concept of weighted average. x-coordinate: x = (mx₂+nx₁)/(m+n). Similarly for y. This formula comes from AP:PB = m:n and using similar triangles.
Distance Formula Derivation
Using Pythagoras theorem on right triangle formed by points:\nHorizontal distance = |x₂-x₁|\nVertical distance = |y₂-y₁|\nd² = (x₂-x₁)² + (y₂-y₁)²\nd = √[(x₂-x₁)² + (y₂-y₁)²]
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