Question

You have studied in Class IX, that a median of a triangle divides it into two triangles of equal areas. Verify this result for $△ABC$ whose vertices are $A(4,-6),B(3,-2)$ and $C(5,2)$

Answer 1

Let $AD$ be the median

$D$ is the midpoint of $BC$

$\Rightarrow D=(\frac{3+5}{2},\frac{-2+2}{2})=(4,0)$

We know that area of triangle is $=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3\left(y_1{y}_2\right)|$ ...$\left(1\right)$

Now consider the $△ABD$

Here $\left(x_{,}{y}_1\right)=(4,-6);(x_2,y_2)=(4,0);(x_3,y_3)=(3,-2)$

Substituting these values in $\left(1\right)$ we get

$\Rightarrow Area△ABD=\frac{1}{2}|4(0+2)+4(-2+6)+3(-6)|$

$\Rightarrow Area△ABD=\frac{1}{2}|6|$

$\Rightarrow Area△ABD=3unit{s}^2$

Now consider the $△ACD$

Here $\left(x_{,}{y}_1\right)=(4,-6);(x_2,y_2)=(4,0);(x_3,y_3)=(3,-2)$

Substituting these values in $\left(1\right)$ we get

$\Rightarrow Area△ACD=\frac{1}{2}|4(0-2)+4(2+6)+5(-6)|$

$\Rightarrow Area△ACD=\frac{1}{2}|6|$

$\Rightarrow Area△ACD=3unit{s}^2$

$\Rightarrow Area△ABD=Area△ACD$

Hence verified.