Solve this following

Question:

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

 

  1. $\left(\frac{8}{5}, \frac{29}{5}\right)$

  2. $\left(\frac{29}{5}, \frac{11}{5}\right)$

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

  4. $\left(\frac{29}{5}, \frac{8}{5}\right)$


Correct Option: , 3

Solution:

$\mathrm{L}: \frac{\mathrm{x}}{3}+\frac{\mathrm{y}}{1}=1 \Rightarrow \mathrm{x}+3 \mathrm{y}-3=0$

Image of point $(-1,-4)$

$\frac{x+1}{1}=\frac{y+4}{3}=-2\left(\frac{-1-12-3}{10}\right)$

$\frac{x+1}{1}=\frac{y+4}{3}=\frac{16}{5}$

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

 

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