An Important Question of M.L Aggarwal book of class 10 Based on Equation of a Straight Line Chapter for ICSE BOARD.

Co-ordinates of two opposite vertices of a square are given. Find the equation of the diagonal AC.

This is the Question Number 36, Exercise 12.2 of M.L Aggarwal.

# The points B (1, 3) and D (6, 8) are two opposite vertices of a square ABCD. Find the equation of the diagonal AC.

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Given, points B (1, 3) and D (6, 8) are two opposite vertices of a square ABCD

Slope of BD is given by

m

_{1 }= (8 – 3)/ (6 – 1) = 5/5 = 1We know that, the diagonal AC is a perpendicular bisector of diagonal BD

So, the slope of AC (m

_{2}) will bem

_{1}x m_{2}= -11 x m

_{2}= -1m

_{2}= -1And, the co-ordinates of mid-point of BD and AC will be

((1 + 6)/2 , (3 + 8)/2) = (7/2, 11/2)

So, the equation of AC is

y – 11/2 = -1 (x – 7/2)

2y – 11 = -2x – 7

2x + 2y – 7 – 11 = 0 ⇒ 2x + 2y – 18 = 0

Thus, the equation of diagonal AC is x + y – 9 = 0.