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