CBSE Previous Year Question Paper for Class 10 Maths
2018 Section C Question 15 A

Coordinate geometry question. Two key concepts tested:
1. Property of diagonals of a parallelogram. Applying the mid-point formula on this property
2. Computing length of a line segment joining two points
A medium difficulty 3-mark question. Exercise caution while computing values of 'a' and 'b' and the lengths of the sides.

Question 15A: A(-2, 1), B(a, 0), C(4, b) D(1, 2) are the vertices of a parallelogram ABCD. Find the values of a and b. Hence find the lengths of its sides.

Video Explanation

Explanatory Answer

Step 1: Compute 'a' and 'b'

A(-2, 1), B(a, 0), C(4, b) and D(1, 2) are vertices of parallelogram.
Key Property: In a parallelogram, diagonals bisect each other.

The diagonals are AC and BD.
Essentially, the coordinates of mid-point of AC = coordinates of mid-point of BD.

(\\frac{-2 + 4}{2}), \\frac{1 + b}{2})) = (\\frac{a + 1}{2}), \\frac{0 + 2}{2}))
(2 , \\frac{1 + b}{2})) = (\\frac{a + 1}{2}), 1)
2 = \\frac{a + 1}{2}) and \\frac{1 + b}{2}) = 1
4 = a + 1 and 1 + b = 2
a = 3 and b = 1

Step 2: Compute the length of the sides of the parallelogram ABCD

Opposite sides of a parallelogram are equal. So, AB = CD and BC = AD.
Length of line segment joining points (x₁, y₁) and (x₂, y₂) is \\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\\)
Length of AB = \\sqrt{(1 - (-2))^2 + (0- 1)^2}\\) = \\sqrt{3^2 + 1^2}\\) = \\sqrt{10}\\)
Length of BC = \\sqrt{(4 - 1)^2 + (1 - 0)^2}\\) = \\sqrt{3^2 + 1^2}\\) = \\sqrt{10}\\)

Note: Because, AB = BC, the parallelogram is a rhombus.