Question

Let $L$denote the line in the $xy$-plane with $x$ and$y$ intercepts as $3$ and $1$ respectively. Then the image of the point $(-1,-4)$in this line is

• A. $\left(\frac{11}{5},\frac{28}{5}\right)$
• B. $\left(\frac{8}{5},\frac{29}{5}\right)$
• C. $\left(\frac{29}{5},\frac{11}{5}\right)$
• D. $\left(\frac{29}{5},\frac{8}{5}\right)$

Answer 1

The correct option is A

$\left(\frac{11}{5},\frac{28}{5}\right)$

Explanation for the correct option:

Finding the image point:

Given that the line $L$ in the $xy$-plane with $x$ and$y$ intercepts as $3$ and $1$ respectively.

Therefore, the equation of line be $\frac{x}{a}+\frac{y}{b}=1$. Substituting the values we get,

$$\begin{gathered} \Rightarrow \frac{x}{3}+\frac{y}{1}=1 \\ \Rightarrow x+3y-3=0...\left(i\right) \end{gathered}$$

We know that the image of a point $(h,k)$ in the line $ax+by-c=0$

$\Rightarrow \frac{x-h}{a}=\frac{y-k}{b}=-2\left(\frac{ah+bk-c}{\sqrt{a^2+b^2}}\right)$

Therefore, we have $a=1,b=3\&c=3$

And the given point $(h,k)$ $=(-1,-4)$

$\begin{array}{rcl}\frac{x+1}{1}& =& \frac{y+4}{3}=-2\left(\frac{-1-12-3}{\sqrt{1^2+3^2}}\right)\\ \frac{x+1}{1}& =& \frac{y+4}{3}=\frac{16}{5}\\ \frac{x+1}{1}& =& \frac{16}{5}\&\frac{y+4}{3}=\frac{16}{5}\\ x& =& \frac{11}{5}\&y=\frac{28}{5}\end{array}$

Hence, option (A) is the correct answer.