Question

Let the coordinates of two points $P$ and $Q$ be $\left(1,2\right)$ and $\left(7,5\right)$ respectively. The line $PQ$ is rotated through $315^{\circ }$ in clockwise direction about the point of trisection of $PQ$ which is nearer to $Q$ . The equation of the line in the new position is

• A. $2x-y-6=0$
• B. $x-y-1=0$
• C. $3x-y-11=0$
• D. $3x-y-9=0$

Answer 1

Answer: (C)

Point of trisection nearer to $Q$ is $\left(5,4\right).$
After rotation, the slope of the line is
$tan{\theta }_1=\frac{tan45^{\circ }+tan{\theta }_2}{1-tan45^{\circ }tan{\theta }_2}=\frac{1+\frac{1}{2}}{1-\frac{1}{2}}=3$ (where, $tan{\theta }_2$ is the slope of $AB$ )
Equation of the required line is $\left(y-4\right)=3\left(x-5\right)\Rightarrow 3x-y-11=0$