Show that the points A(-5,6), B(3,0) and C(9,8) are the vertices of an isosceles right-angled triangle. Calculate its area.

#### Solution

Let the given points be A(-5,6), B(3,0) and C(9,8).

`AB = sqrt((3-(-5))^2 +(0-6)^2) = sqrt((8)^2 +(-6)^2 = sqrt(64+36) = sqrt(100) = 10 `units

`BC = sqrt((9-3)^2 +(8-0)^2) = sqrt((6)^2 +(8)^2) = sqrt(36+64) = sqrt(100) = 10` units

`AC = sqrt((9-(-5))^2 +(8-6)^2) = sqrt((14)^2 +(2)^2) = sqrt(196+4) = sqrt(200) = 10sqrt(2) `units

Therefore, AB = BC = units

Also, `(AB)^2 +(BC)^2 = (10)^2 +(10)^2 = 200`

and `(AC)^2 =(10sqrt(2))^2 = 200`

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

This show that Δ ABC is right angled at B. Therefore, the points A(-5,6), B(3,0) and C(9,8). are the vertices of an isosceles rightangled triangle

Also, area of a triangle `= 1/2 xx"base"xx "height"`

If AB is the height and BC is the base,

Area`= 1/2 xx 10 xx 10`

= 50 square units