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

###### CBSE Solved Question Paper 3 Mark | Coordinate Geometry

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.

## NCERT Solution to Class 10 Maths

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##### 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.

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