Hello student!

Let (x - y +c1) and (x+ y +c2) be the required perpendicular lines.

The equations will be a pair of perpendicular lines if it satisfies:-

(x -y)(x + y) + c2(x - y) +c1(x + y) + c1c2 =0

The given equation can thus be written as:-

x^2-y^2-x+3y-2=0-------------------------(i)

=>(x-y)(x+y) + c1(x+y) + c2(x-y) + c1c2=0

=>x^2 - y^2 + (c1+c2)x + (-c2+c1)y + c1c2=0----(ii)

From equation (i) and (ii), we have:-

c1 + c2 = -1and -c2 + c1 = 3

Solving the above two equations, we get:-

2C1 = 2 => c1 = 1 and c2 = -2.

Hence, x - y +1 and x + y - 2 are the required equations of the perpendicular lines.

Hope this would help you:)