Question

The point $A$ divides the join of the points $(5,1)$ and $(3,5)$ in the ratio $k:1$ and coordinates of points $B$ and $C$ are $(1,5)$ and $(7,2)$ respectively. If the area of $\Delta ABC$ be $2$ units, then $k$ equals

• A. $7,9$
• B. $6,7$
• C. $7,\frac{31}{9}$
• D. $9,\frac{31}{9}$

Answer 1

Answer: (C)

$7,\frac{31}{9}$

Let coordinates of point $A$ are $(x,y)$, then
$x=\frac{3k+5}{k+1},y=\frac{5k+1}{k+1}$

Now area of triangle is $2$.

$\Rightarrow \frac{1}{2}\left|\begin{array}{ccc}\frac{3k+5}{k+1}& \frac{5k+1}{k+1}& 1\\ 1& 5& 1\\ 7& 2& 1\end{array}\right|=2$

$\Rightarrow \frac{1}{k+1}\left|\begin{array}{ccc}3k+5& 5k+1& k+1\\ 1& 5& 1\\ 7& 2& 1\end{array}\right|=4$

Applying $R_2\to R_2-R_1,R_3\to R_3-R_1$
Then, $\left|\begin{array}{ccc}3k+1& 5k+1& 1\\ -3k& 4-5k& 0\\ 6-3k& 1-5k& 0\end{array}\right|=4(k+1)$

$\Rightarrow (6-3k)(4-5k)-\left(3k\right)(1-5k)=4(k+1)$

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