Question

Find the area of the triangle formed by the mid point of side of the triangle whose vertices are $(2,1),(-2,3),(4,-3)$

• A. $1.5sq.units$
• B. $3sq.units$
• C. $6sq.units$
• D. $12sq.units$

Answer 1

The correct option is A $1.5sq.units$

Let the vertices of the triangle be $A(2,1),B(-2,3)$ and $C(4,-3)$.

Let the mid-points of $AB,BC$ and $AC$ be $P.Q$ and $R$. respectively.

Here,

$P(x_1,y_1)$

$Q(x_2,y_2)$

$R(x_3,y_3)$

Now,

$P(x_1,y_1)=(\frac{2-2}{2},\frac{1+3}{2})=(0,2)$

$Q(x_1,y_1)=(\frac{-2+4}{2},\frac{3-3}{2})=(1,0)$

$R(x_3,y_3)=(\frac{4+2}{2},\frac{-3+1}{2})=(3,-1)$

Therefore, area of $\Delta PQR$,

$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$=\frac{1}{2}[0(0+1)+1(-1-2)+3(2-0)]$

$=\frac{1}{2}\times 3$

$=\frac{3}{2}$

Hence, area of the required triangle is $1.5sq.units$.