Question

The point $P(2,1)$ is shifted by $3\sqrt{2}$ parallel to the line $x+y=1,$ in the direction of increasing ordinate, to reach $Q$.The image of $Q$ by the line $x+y=1$ is

• A. $(5,-2)$
• B. $(-1,4)$
• C. $(3,-4)$
• D. $(-3,2)$

Answer 1

Answer: (D)

$(-3,2)$

The point $P(2,1)$ if shifted by $3\sqrt{2}$ parallel to the line $x+y=1$, in the direction of increasing ordinate, to reach point $Q(\alpha ,\beta )$.

The equation of line produced from $P(2,1)$ to $Q(\alpha ,\beta )$ parallel to line $x+y=1$ in parametric form is,

$\Rightarrow \frac{\alpha -2}{cos\theta }=\frac{\beta -1}{sin\theta }=-3\sqrt{2}$ .....$\left(1\right)$

Where $\theta$ is slope of line $PQ$, which is parallel to line $x+y=1$

Hence Slope of $PQ$ $=-1$

$\Rightarrow \text{if}tan\theta =-1$

$\Rightarrow cos\theta =\frac{1}{\sqrt{2}}$

$\Rightarrow sin\theta =-\frac{1}{\sqrt{2}}$

Hence from equation $\left(1\right)$, $\alpha =-1,\beta =4$ and $Q(-1,4)$

Now image of $Q(-1,4)$ in the line $x+y=1$ is given by,

$\Rightarrow \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-2(a{x}_1+b{y}_1+c)}{a^2+b^2}$

$\Rightarrow \frac{x+1}{1}=\frac{y-4}{1}=\frac{-2(-1+4-1)}{1^2+1^2}$

Hence $x=-3$ and $y=2$

Correct option is $D$.