Question

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

Answer 1

Answer:

Let us calulate the length of the triangle sides:

AB = sqrt[(2-2)^2 + (5+1)^2]= sqrt(36) = 6

BC = sqrt[(5-2)^2 + (2-5)^2]= sqrt18 = 3sqrt2

AC = sqrt[(5-2)^2 + (2+1)^2 = sqrt18 = 3sqrt2

We notice that BC = AC = 3sqrt2

Then ABC is an isoscale triangle.

$area = \frac{1}{2} \times b \times ( {a}^{2} - \frac{ {b}^{2} }{4} ) \\ = \frac{1}{2} \times 3 \sqrt{2} \times (36 - \frac{18}{4} ) \\ = \frac{1}{2} \times 3 \sqrt{2} \times (36 - \frac{9}{2} ) \\ = \frac{1}{4} \times 3 \sqrt{2} \times 63 = \frac{189 \sqrt{2} }{4}$