Q. If (-4,3) and (2,5) are pair of opposite vertices of a square. Find the equations of the diagonals of the square.

Soln:

The equation of diagonal BD is given by as,
y-y₁ = (y₂-y₁)/(x₂-x₁)*(x-x₁)
Here, x₁ = -4, y₁ = 3 and x₂ = 2, y₂ = 5

from above equation,                

y-3 = (5-3)/(2+4) * (x+4)

Simplifying we get,
x-3y+13 = 0. This is equation of the diagonal BD.
To find equation of another diagonal AC, we have to find the midpoint of the diagonal BD whose end points are B(-4,3) and D(2,5).

Mid-point of BD,
(x,y) = [(x₁+x₂)/2,(y₁+y₂)/2]

or, (x,y) = [(-4+2)/2, (3+5)/2]

or, (x,y) = (-1,4)

\ x = -1 and y = 4 .
Since, the diagonal are bisecting each other at right angle. The angle between them is 90⁰.

Slope of diagonal AC (m₂) is given by m₁*m₂ = -1 ………eqn(i)
(m₁*m₂ = -1, when two lines are perpendicular, the product of their slopes is equal to -1) 

Note that m₁= (y₂-y₁)/(x₂-x₁)

\ m₁= 1/3

From eqn(i), (1/3)* m₂ = -1, we get, m₂ = -3

For diagonal AC, x₁ = -1 and y₁ = 4, using one point formula, the equation of diagonal AC is

y-y₁= m₂*(x-x₁)

or, y-4 = -3*(x+1)

simplifying we get, 3x+y-1 = 0

Hence, the required equations of diagonals are: x-3y+13 = 0 and 3x+y-1 = 0