Question

Image of the point $\left(-1,3\right)$ w.r.t. the line $y=2x$ is

• A. $\left(-1,-5\right)$
• B. $\left(1,3\right)$
• C. $\left(3,1\right)$
• D. $\left(5,1\right)$

Answer 1

The correct option is C

$\left(3,1\right)$

Explanation for the correct option.

Find the coordinates of the image.

If the image of the point $\left(x_1,y_1\right)$ with respect to line $ax+by+c=0$ is $\left(h,k\right)$ then:

$\frac{h-x_1}{a}=\frac{k-y_1}{b}=-2\frac{\left|a{x}_1+b{y}_1+c\right|}{a^2+b^2}$

Here the line is $y=2x$ and it can be written as: $-2x+y=0$.

So $a=-2,b=1,c=0$.

The point is $\left(-1,3\right)$. So $x_1=-1,y_1=3$.

Thus the coordinates of the image $\left(h,k\right)$ is given by substituting all the values in the formula

$$\begin{gathered} \frac{h-\left(-1\right)}{-2}=\frac{k-3}{1}=-2\frac{\left|(-2)\times \left(-1\right)+1\times 3+0\right|}{2^2+{\left(-1\right)}^2} \\ \Rightarrow \frac{h+1}{-2}=\frac{k-3}{1}=-2\frac{\left|2+3\right|}{4+1} \\ \Rightarrow \frac{h+1}{-2}=\frac{k-3}{1}=-2\times \frac{5}{5} \\ \Rightarrow \frac{h+1}{-2}=\frac{k-3}{1}=-2 \end{gathered}$$

First solving for $h$.

$$\begin{gathered} \frac{h+1}{-2}=-2 \\ \Rightarrow h+1=4 \\ \Rightarrow h=3 \end{gathered}$$

Now, solving for $k$.

$$\begin{gathered} \frac{k-3}{1}=-2 \\ \Rightarrow k-3=-2 \\ \Rightarrow k=1 \end{gathered}$$

So, the coordinates of the image is $\left(3,1\right)$.

Thus, the image of the point $\left(-1,3\right)$ w.r.t. the line $y=2x$ is $\left(3,1\right)$.

Hence, the correct option is C.