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

Explanation:

The distance formula between two points:

`d=√((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=√((2-4)^2+(1-0)^2)=√((-2)^2+1^2)=√(4+1)=\sqrt5\\\\AC=√((-1-4)^2+(-5-0)^2)=√((-5)^2+(-5)^2)=√(25+25)=√(50)\\\\BC=√((-1-2)^2+(-5-1)^2)=√((-3)^2+(-6)^2)=√(9+36)=√(45)`

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

a² + b² = c²

`\sqrt5<√(45)<√(50)\\\\(\sqrt5)^2+(√(45))^2=5+45=50\\\\(√(50))^2=50\\\\\bold{CORRECT}`

used
`(√(a))^2=a` for a ≥ 0.

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