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

Answer:

the area of ∆ABC = 9

Step-by-step explanation:

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

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

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

$x_{H}=\frac{-1+5}{2} =2\\y_{H}=\frac{2+2}{2} =2$

Then

H(2 , 2)

therefore,

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

Finally,

$\text{the area of triangle ABC} =\frac{\text{CH} \times \text{AB} }{2} =\frac{3\times 6}{2} =9$