Question

Show that the points A(-1,2), B (5,2) and C (2,5) are the vertices of an isosceles triangle. Find the area of ∆ABC

Answer 1

the area of ∆ABC = 9

Explanation:

`CA=\sqrt{\left( -1-2\right)^(2) +\left( 2-5\right)^(2) } = √(18) =3√(2)`

`CB=\sqrt{\left( 5-2\right)^(2) +\left( 2-5\right)^(2) } = √(18) =3√(2)`

Since CA = CB then A(-1,2), B (5,2) and C (2,5) are the vertices of an isosceles triangle.

Area calculation:

Let The point H be the midpoint of side AB

`x_(H)=(-1+5)/(2) =2\\y_(H)=(2+2)/(2) =2`

Then

H(2 , 2)

therefore,

`CH=\sqrt{\left( 2-2{}\right)^(2) +\left( 2-5\right)^(2) } =3`

Finally,

`\text{the area of triangle ABC} =\frac{\text{CH} * \text{AB} }{2} =(3* 6)/(2) =9`