Question

Use the distance formula, show that the points (4,0), (2,1), and (-1,-5) form the vertices of a right triangle

Answer 1

Step-by-step explanation:

The distance formula between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the coordinates of the points.

$A(4,\ 0),\ B(2,\ 1),\ C(-1,\ -5)\\\\AB=\sqrt{(2-4)^2+(1-0)^2}=\sqrt{(-2)^2+1^2}=\sqrt{4+1}=\sqrt5\\\\AC=\sqrt{(-1-4)^2+(-5-0)^2}=\sqrt{(-5)^2+(-5)^2}=\sqrt{25+25}=\sqrt{50}\\\\BC=\sqrt{(-1-2)^2+(-5-1)^2}=\sqrt{(-3)^2+(-6)^2}=\sqrt{9+36}=\sqrt{45}$

If a ≤ b < c are the sides of the right triangle, then

a² + b² = c²

$\sqrt5$

used $(\sqrt{a})^2=a$ for a ≥ 0.

$AB^2+BC^2=AC^2$ therefore ΔABC is a right triangle.