Question

Show that the points (–2, 3), (8, 3) and (6, 7) are the vertices of a right-angled triangle.

Answer 1

The given points are A (−2, 3), B (8, 3) and C (6, 7).

By, distance formula, AB `sqrt{{8-(-2)}^2+(3-3)^2`

`rArr AB^2=10^2+0`

`rArrBC^2=100`

`BC=sqrt((6-8)^2+(7-3)^2)`

`rArrBC^2=(-2)^2+4^2`

`rArrBC^2=4+16`

`rArrBC^2=20`

`BC=sqrt((-2-6)^2+(3-7)^2)`

`rArrCA^2=(-8)^2+(-4)^2`

`rArrCA^2=64+16`

`rArrCA^2=80`

It is observed that BC² + CA² = 20 + 80 = 100 = AB²

So, by the converse of Pythagoras Theorem, ΔABC is right angled at C.