Question

The point $A$ divides the line segment joining the points $(-5,1)$ and $(3,5)$ in the ratio $k:1$. The coordinates of points $B$ and $C$ are $(1,5)$ and $(7,-2)$ respectively. If the area of $△ABC$ is $2$ square units, then the number of values of $k$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Given points are $(-5,1)$ and $(3,5)$ and the ratio is $k:1$,
So, the point $A=(\frac{3k-5}{k+1},\frac{5k+1}{k+1})$

Point $B=(1,5),C=(7,-2)$
Now, the area of $△ABC$
$2=\frac{1}{2}\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|$
$\Rightarrow 2=\frac{1}{2}\left|\frac{3k-5}{k+1}\right(5+2)+1(-2-\frac{5k+1}{k+1})+7(\frac{5k+1}{k+1}-5)|\Rightarrow 4=|\frac{21k-35-5k-1+35k+7}{k+1}-2-35|\Rightarrow \frac{51k-29}{k+1}-37=\pm 4\Rightarrow \frac{14k-66}{k+1}=\pm 4\Rightarrow 14k-66=\pm (4k+4)\Rightarrow 14k-66=4k+4\text{or}14k-66=-4k-4\therefore k=7,\frac{31}{9}$

Hence, the number of values of $k$ is $2$.