Question

Find the image of the point ( 3, 2) with respect to the line x + 2y + 1 = 0

Answer 1

$Letimageof\left(3,2\right)inlinex+2y+1=0be\left(h,k\right)$

Now,

$\frac{h-3}{1}=\frac{k-2}{2}=\frac{-2\left(3\left(1\right)+2\left(2\right)+1\right)}{\left(1^2+2^2\right)}\frac{h-3}{1}=\frac{k-2}{2}=\frac{-2\left(8\right)}{5}h=3-\frac{16}{5}=\frac{-1}{5}k=2-\frac{32}{5}=\frac{-22}{5}$

$Hence,Imageof\left(3,2\right)inlinex+2y+1=0is\left(-\frac{1}{3},\frac{-22}{5}\right).$