Pythagorean Theorem

Compute for the length of each side using the distance formula .

$a=\sqrt{[(4-(-2)]^2+(5-1)^2}=\sqrt{6^2+4^2}=\sqrt{36+16}=\sqrt{52}=\sqrt{4 \times13}=2\sqrt{13}$

$b=\sqrt{[1-(-8)]^2+(-2-4)^2}=\sqrt{9^2+6^2}=\sqrt{81+36}=\sqrt{117}=\sqrt{9 \times13}=3\sqrt{13}$

$c=\sqrt{(4-4)^2+[(5-(-8)]^2}=\sqrt{13^2}=13$

By the pythagorean theorem , we have

$c^2 = a^2+b^2$

$13^2=(2\sqrt{13})^2+(3\sqrt{13})^2$

$169=4(13)+9(13$

$169=52+117$

$169=169$ $\large\implies$ $\text{The statement is true.}$

$\boxed{\color{#D61F06}\large We~therefore~conclude~that~the~points~(4,5),(-2,1)~and~(4,-8)~are~vertices~of~a~right~triangle.}$

Solving for the distances using the formula, $\large d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ ,

$a = \sqrt{(4+2)^2 + (5-1)^2} = \sqrt{52}$

$b = \sqrt{(4+2)^2 + (-8-1)^2} = \sqrt{117}$

$c = \sqrt{(4-4)^2 + (-8-5)^2} = 13$

Now by Pythagorean theorem, check if it is a right triangle

$a^2 + b^2 = c^2$

$(\sqrt{52})^2 + (\sqrt{117})^2 = 13^2$

$52 + 117 = 169$

$169 = 169$ $\implies$ $true$

$\color{#3D99F6}\boxed{\therefore {It~is~a~right~triangle.}}$