Question

The point $A$ divides the line segment joining the points $P(-5,1)$ and $Q(3,5)$ in the ratio $k:1$. Find the two values of $k$ for which the area of $△ABC$ is equal to $2$ square units where $B$ is $(1,5)$ and $C(7,-2)$ .

Answer 1

It is given that point A divides the join of $P(-5,1)andQ(3,5)$ in the ratio of K:1.

Here, $x_1=-5,y_1=1,x_2=3,y_2=5$,$m=k$ and $n=1$

$P(X,Y)=(\frac{m{x}_1+n{x}_2}{m+n},\frac{m{y}_1+n{y}_2}{m+n})$

So, The Coordinate of A $=(\frac{3k-5}{k+1},\frac{5k+1}{k+1})$

we have,

Area of $△ABC=2$ Sq.units

$\Rightarrow \frac{1}{2}\left|\begin{array}{c}(\frac{3k-5}{k+1}\times 5-2+7\times \frac{5k-1}{k+1})-(\frac{5k+1}{k+1}\times 1+35-2\times \frac{3k-1}{k+1})\end{array}\right|=2$

$\Rightarrow \frac{1}{2}\left|\begin{array}{c}(\frac{15k-25}{k+1}-2+\frac{5k-1}{k+1})-(\frac{5k+1}{k+1}+35-\frac{6k-10}{k+1})\end{array}\right|=2$

$\Rightarrow \frac{1}{2}\left|\begin{array}{c}\frac{(15k-25-2k-2+35k+7)-(5k+1+35k+35-6k+10)}{k+1}\end{array}\right|=2$

$\Rightarrow \frac{1}{2}\left|\begin{array}{c}\frac{14k-66}{k+1}\end{array}\right|=2\Rightarrow \frac{7k-33}{k+1}=\pm 2$

$\Rightarrow 7k=-33=\pm (k+1)$

$\Rightarrow 7k-33=2k+2,7k-33=-2k-2$

$\Rightarrow 5k=35,9k=31\Rightarrow k=7,\frac{31}{9}$