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
$tan \theta _{1}=\frac{tan ⁡ 45 ^\circ + tan ⁡ \theta _{2}}{1 - tan ⁡ 45 ^\circ tan ⁡ \theta _{2}}=\frac{1 + \frac{1}{2}}{1 - \frac{1}{2}}=3$ (where, $tan \theta _{2}$ is the slope of $AB$ )
Equation of the required line is $\left(y - 4\right)=3\left(x - 5\right)\Rightarrow 3x-y-11=0$