Question

Find the image of (–3, 4) with respect to the line y = x.​

Answer 1

$\textbf{To find:}$

$\text{The image of (-3,4) with respect to x-y=0}$

$\textbf{Solution:}$

$\text{The image of the point (x_1,y_1) with respect}$

$\text{to the line ax+by+c=0 is given by}$

$\boxed{\bf\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-2(ax_1+by_1+c)}{a^2+b^2}}$

$\text{The image of (-3,4) with respect to x-y=0 is}$

$\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{-2(ax_1+by_1+c)}{a^2+b^2}$

$\dfrac{x+3}{1}=\dfrac{y-4}{-1}=\dfrac{-2(1(-3)+(-1)(4)+0}{1^2+(-1)^2}$

$x+3=-y+4=\dfrac{-2(1(-7))}{2}$

$x+3=-y+4=7$

$x+3=7\implies\,x=4$

$-y+4=7\implies\,y=-3$

$\therefore\textbf{The image of (-3,4) is (4,-3)}$

Find more:

The image of a point with respect to the line 2x-y+6=0 assuming the line to be a mirror is (6,6) find the point also find the equation of line joining this point and it's range

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