Question

Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are A(2,1), B(4,3) and C(2,5)

Answer 1

The vertices of the triangle are A (2,1), B ( 4,3) and C (2,5)

Co-ordinates of midpoint of AB = $P(x_1,y_1)=(\frac{2+4}{2},\frac{1+3}{2})=(3,2)$

Co-ordinates of midpoint of BC = $Q(x_2,y_2)=(\frac{4+2}{2},\frac{3+5}{2})=(3,4)$

Co-ordinates of midpoint of AC = $P(x_3,y_3)=(\frac{2+2}{2},\frac{1+5}{2})=(2,3)$

Now,

Area of ΔPQR = $\frac{1}{2}\left(x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right)$

$=\frac{1}{2}\left(3\right(4-3)+3(3–2)+2(2–4\left)\right)$

$=\frac{1}{2}\left(3\right(1)+3(1)+2(-2\left)\right)$

$=\frac{1}{2}(3+3–4)$

$=\frac{1}{2}\left(2\right)$

= 1 sq.unit

Hence, the area of the required triangle is 1 sq.unit