Question

The points A$(2,3)$, B$(4,-1)$ and C$(-1,2)$ are the vertices of $\Delta ABC$. Find the length of perpendicular from C on AB and hence find the area of $\Delta ABC$.

Answer 1

Consider the given points

$A(x_1,y_1)=(2,3)$

$B(x_2,y_2)=(4,-1)$

$C(x_3,y_3)=(-1,2)$

Let $D$ the middle point of $AB$

Then,

$D(x,y)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$D(x,y)=(\frac{2+4}{2},\frac{3-1}{2})$

$D(x,y)=(3,1)$ be the middle point of AB.

We know that,

$Areaoftriangle=\frac{1}{2}\times l\times b$

$l=CD=\sqrt{{(-1-3)}^2+{(2-1)}^2}$

$CD=\sqrt{16+1}=\sqrt{17}$

And,

$b=AB=\sqrt{{(2-4)}^2+{(3+1)}^2}$

$AB=\sqrt{4+16}$

$AB=\sqrt{20}$

Now,

$Areaoftriangle\left(ABC\right)=\frac{1}{2}\times CD\times AB$

$=\frac{1}{2}\times \sqrt{17}\times \sqrt{20}$

$=\frac{1}{2}\times \sqrt{17}\times 2\sqrt{5}$

$=\sqrt{85}squareunit.$

Hence, this is the answer.