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

Answer 1

The correct option is A $(\frac{11}{5},\frac{28}{5})$


Let $(x,y)$ be the image of the point $(-1,-4)$ in the line
$\frac{x}{3}+\frac{y}{1}=1$
$x+3y=3$
Image of a point $(x_1,y_1)$ in a line $ax+by+c=0$ is given by formula:
$\Rightarrow \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-2(a{x}_1+b{y}_1+c)}{a^2+b^2}$
$\Rightarrow \frac{x+1}{1}=\frac{y+4}{3}=\frac{-2(-1-12-3)}{1^2+3^2}$
$\Rightarrow \frac{x+1}{1}=\frac{y+4}{3}=\frac{16}{5}$
$\Rightarrow (x,y)\equiv (\frac{11}{5},\frac{28}{5})$