Question

The point $P(1,1)$ is translated parallel to $2x=y$ in the first quadrant through a unit distance . The coordinates of the point $P$ in new position are

• A. $(1\pm \frac{2}{\sqrt{5}},1\pm \frac{1}{\sqrt{5}})$
• B. $(1\pm \frac{1}{\sqrt{5}},1\pm \frac{2}{\sqrt{5}})$
• C. $\left(\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}\right)$
• D. $\left(\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}}\right)$

Answer 1

Answer: (B)

$(1\pm \frac{1}{\sqrt{5}},1\pm \frac{2}{\sqrt{5}})$

Given equation of line is $y=2x$.

Let the new position of point $P$ be $P^{\prime }(x_1,y_1)$

Equation of line passing through $P(1,1)$ and parallel to $y=2x$ is

$y-1=2(x-1)$

$\Rightarrow y=2x-1$

Since, $P^{\prime }(x_1,y_1)$ lies on this line

$\Rightarrow y_1=2x_1-1$ ...(1)

Also, given $P$ is translated to $P^{\prime }$ by a unit distance

$P{P}^{\prime }=1$

$\Rightarrow (x_1-1{)}^2+(y_1-1{)}^2=1$

$\Rightarrow (x_1-1{)}^2+(2x_1-2{)}^2=1$ (by (1))

$\Rightarrow 5x_1^2-10x_1+4=0$

$\Rightarrow x_1=\frac{10\pm \sqrt{20}}{10}$

$\Rightarrow x_1=\frac{5\pm \sqrt{5}}{5}$

$\Rightarrow x_1=1\pm \frac{1}{\sqrt{5}}$

Put this value in (1), we get

$y_1=2(1\pm \frac{1}{\sqrt{5}})-1$

$\Rightarrow y_1=1\pm \frac{2}{\sqrt{5}}$

Hence, the new position of $P$ is $(1\pm \frac{1}{\sqrt{5}},1\pm \frac{2}{\sqrt{5}})$