Question

Prove that the points A(7, 10), B(–2, 5) and C(3, –4) are the vertices of an isosceles right triangle.

Show that the points A(7, 10), B(–2, 5) and C(3, –4) are the vertices of an isosceles right triangle.

Answer 1

The given points are A (7, 10), B(-2, 5) and C(3, -4).

`AB= sqrt((-2-7)^2 +(5-10)^2) = sqrt((-9)^2 +(-5)^2) = sqrt((81+25)) = sqrt(106)`

`BC = sqrt((3-(-2))^2 +(-4-5)^2) = sqrt((5)^2 +(-9)^2 )= sqrt((25+81) )= sqrt(106)`

`AC = sqrt((3-7)^2 +(-4-10)^2) = sqrt(( -4)^2 +(-14)^2) = sqrt(16+196) = sqrt(212)`

Since, AB and BC are equal, they form the vertices of an isosceles triangle

Also,`(AB)^2 + (BC)^2 = ( sqrt(106))^2 +( sqrt(106)^2) = 212`

and `(AC)^2 = (sqrt(212))^2 = 212.

`Thus , (AB)^2 + (BC)^2 = (AC)^2`

This show that  ΔABC is right- angled at B. Therefore, the pointsA (7, 10), B(-2, 5) and C(3, -4). are the vertices of an isosceles rightangled triangle.