Question

Prove that the points A(-5,4), B(-1.-2) and C(5,2) are the vertices of an isosceles right angled triangle. Also, find the coordinates of D, so that ABCD is a square.

Answer 1

we have to prove that A(-5,4), B(-1, -2) and C(5,2) are the vertices of an isosceles right angled triangle.

use distance formula,

AB = $\sqrt{(-5+1)^2+(4+2)^2}=\sqrt{16+36}=2\sqrt{13}$

BC = $\sqrt{(-1-5)^2+(-2-2)^2}=\sqrt{36+16}=2\sqrt{13}$

CA=$\sqrt{(5+5)^2+(2-4)^2}=\sqrt{100+4}=2\sqrt{26}$

here, it is clear that,

AB² + BC² = 52 + 52 = 104 = CA²

from Pythagoras theorem, we know any triangle will be a right angled triangle when sides of triangle follow above condition.

so, ABC is a right angled triangle.

Let D (x, y) such that ABCD is a square.

we know, a square is also a parallelogram.

so, midpoint of diagonal of AC =midpoint of diagonal BD

{(5 - 5)/2, (4 + 2)/2} = {(x - 1)/2, (y - 2)/2}

or, (0, 3) = {(x - 1)/2, (y - 2)/2}

(x - 1)/2 = 0 => x = 1

and (y - 2)/2 = 3 => y = 8

hence, D = (1, 8)