Question

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

Answer 1

$A(-5,4)B(-1,-2)C(5,2)$.

$AB=\sqrt{4^2+6^2}=\sqrt{52}$

$BC=\sqrt{6^2+4^2}=\sqrt{52}$

$AC=\sqrt{{10}^2+2^2}=\sqrt{104}$

It is isosceles triangle.

with $AB=BC$

For making square

$BD=AC=\sqrt{104}$

$\sqrt{(x+1{)}^2+(y+2{)}^2}=\sqrt{104}$------(1)

Also, slope of $AC$ is $\frac{-1}{slopeofBD}$

Slope of $BD=\frac{y+2}{x+1}=\frac{-1}{\frac{-2}{10}}$

$\frac{y+2}{x+1}=5$------(2)

from (1) & (2)

$x=1y=8$

$D(1,8)$ is new vertex of square.