Question

Find the area of $△ABC$ with A(1, -4) and midpoints of sides through A being (2, -1) and (0, -1)

Answer 1

Let $x_2,y_2$ and $x_3,y_3$ be the co-ordinates of B and C respectively.

Since, the co-ordinates of A (1, -4) hence let us name midpoint of AB be D = $x_2,y_2$ and midpoint of AC be E= $x_3,y_3$ .

Now D ( $2,-1)$=$\frac{(1+x_2)}{2}$, $\frac{(1+y_2)}{2}$

$\to$ 2 =$\frac{(1+x_2)}{2}$,

$\to 4=1+x_2$

$\to 3=x_2$

$\to x_2=3$

Again $-1=\frac{(-4+y_2)}{2}$

$-4+y_2=-2$

$\to y_2=-2+4=2$

Similarly Now E $0,-1$

$\to 0=\frac{(1+x_3)}{2}$

$\to 0=-1+x_3$

$\to -1=x_3$

$\to x_3=-1$

From E $0,-1$

-1=$\frac{(-4+y_3)}{2}$

$\to y_3=2$

Let A $\{x_1,y_1)=A(1,-4)$
Let B $\{x_2,y_2)=B(3,2)$
Let C $\{x_3,y_3)=C(-1,2)$

Now

Area of $△ABC$

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

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

= $\frac{1}{2}\times \left[1\right(0)+3(6)-1(-6\left)\right]$

= $\frac{1}{2}\times [0+18+6]$

= $\frac{1}{2}\times \left[24\right]$

= 12 sq.unit

Thus,Area of $△ABC$ = 12 sq.units