Question

$P(3,1),Q(6,5)$ and $R(x,y)$ are three points such that the angle $PRQ$ is a right angle and the area of $△RQP=7$, then 4x-3y+5 is equals to

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is A $0$

We have given that

$ar\left(\Delta RQP\right)=7$

$P=\left(3,1\right)$

$Q=\left(6,5\right)$

And,

$R=\left(x,y\right)$

Now,

$ar\left(\Delta RQP\right)=\frac{1}{2}|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)|$

$\Rightarrow 7=\frac{1}{2}\left|3\left(5-y\right)+6\left(y-1\right)+x\left(1-5\right)\right|$ [On putting the value of points]

$\Rightarrow 7\times 2=15-3y+6y-6+x-5x$

$\Rightarrow 14=9+3y-4x$

$\Rightarrow 4x-3y=9-14$

$\Rightarrow 4x-3y=-5$

$\therefore 4x-3y+5=0$

Hence, the option $\left(A\right)$ is correct.