Question

Find the area of the triangle formed by joining the mid points of the sides of the triangle formed with coordinates A (-4, 3), B (2, 3) and C (4, 5).

• A. 2 sq.units
• B. 0.5 sq. units
• C. 1 sq. units
• D. 1.5 sq. units

Answer 1

The correct option is D

1.5 sq. units

The mid-point of the coordinates $(x_1,y_1),(x_2,y_2)$ is given by $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
$\therefore$ The mid-points of AB, BC and CA are $(\frac{-4+2}{2}$, $\frac{3+3}{2})$, $(\frac{4+2}{2}$, $\frac{3+5}{2})$ and $(\frac{-4+4}{2}$, $\frac{5+3}{2})$ respectively. Let us denote these by D, E and F.
i.e., D(-1, 3), E(3, 4) and F(0, 4).

Area of the triangle formed by $(x_1,y_1),(x_2,y_2)(x_3,y_3)$ is given by
$\text{Area}=\frac{1}{2}$|$x_1$($y_2$-$y_3$)+$x_2$($y_3$-$y_1$)+$x_3$($y_1$-$y_2$)|

Now, let $x_1$ = -1, $y_1$ = 3, $x_2$ = 3, $y_2$ = 4 $x_3$ = 0 and $y_3$ = 4

$\Rightarrow \text{Required area}=\frac{1}{2}$| -1(4 - 4) + 3(4 - 3) + 0(3 - 4)| = 1.5 square units

$\therefore$ Area of the triangle = 1.5 square units