Question

The point A divides the join of P(-5, 1) and Q(3, 5) in the ratio k:1. The values of k for which the area of $\Delta ABC$ where B(1, 5), C(7, -2) is 2 sq. units is

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

Answer 1

Answer: (A)

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

$A=(p,q)dividesthelengthPQintheratiom:n=k:1\therefore ThepointA=(x_1,y_1)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})=(\frac{k{x}_2+x_1}{k+1},\frac{k{y}_2+y_1}{k+1})=(\frac{3k-5}{k+1},\frac{5k+1}{k+1}).Nowar\Delta ABC=2\left(given\right)\therefore ar\Delta ABC=\frac{1}{2}\left|\begin{array}{ccc}\frac{3k-5}{k+1}& \frac{5k+1}{k+1}& 1\\ 1& 5& 1\\ 7& -2& 1\end{array}\right|=\pm 2\left(sincethe\Delta maylieontheeithersideoftheaxes\right)$
$\Rightarrow 14k-66=\pm 2(2k+2)$
Solving we get either $k=7$ or $k=31/9$