# Find the reflection of point A(3, -2) through the mirror line of y = \frac{1}{2}x-1

## Question:

Find the reflection of point A(3, -2) through the mirror line of {eq}y = \frac{1}{2}x-1. {/eq}

## Reflecting Objects Along a Line:

In the coordinate plane, reflecting an object along a line means flipping the object along the given line such that its distance from the line is maintained. If a point {eq}(x_1,y_1) {/eq} is reflected about a line {eq}ax+by+c=0 {/eq}, we can find the mirror image of the point along the given line using the following relation:{eq}\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=-2\dfrac{ax_1+by_1+c}{a^2+b^2} {/eq}

In the given case, point {eq}p=(x_1,y_1)=(3,-2) {/eq}

The line along which the point is reflected is given by:

{eq}y=\dfrac{1}{2}x-1 {/eq}

{eq}2y=x-2 {/eq}

{eq}x-2y-2=0 {/eq}

Hence, a = 1, b = -2, c = -2

To find the x-coordinate of the reflected point Q:

{eq}\dfrac{x-x_1}{a}=-2\dfrac{ax_1+by_1+c}{a^2+b^2} {/eq}

Putting the values:

{eq}\dfrac{x-3}{1}=-2\times \dfrac{3-4-2}{1^2+(-2)^2} {/eq}

{eq}x-3=-2\times \dfrac{-3}{5}=\dfrac{6}{5} {/eq}

{eq}5x-15=6 {/eq}

{eq}5x=6+15=21 {/eq}

{eq}x=\dfrac{21}{5} {/eq}

To find the y-coordinate:

{eq}\dfrac{y-y_1}{b}=-2\dfrac{ax_1+by_1+c}{a^2+b^2} {/eq}

{eq}\dfrac{y+2}{-2}=-2 \times \dfrac{3-4-2}{1^2+(-2)^2} {/eq}

{eq}\dfrac{y+2}{2}=2\times \dfrac{-3}{5}=\dfrac{-6}{5} {/eq}

{eq}5y+10=-12 {/eq}

{eq}5y=-12-10 {/eq}

{eq}y=\dfrac{-22}{5} {/eq}

Hence, the coordinates of the reflected point are {eq}(\dfrac{21}{5},\dfrac{-22}{5}) {/eq}