# Let p = (5,1), and let q be the reflection of p over the line y = 1/2x + 2. Find the coordinates...

## Question:

Let p = (5,1), and let q be the reflection of p over the line {eq}y = \frac{1}{2}x + 2 {/eq}. Find the coordinates of q.

## Reflection of a Point about a Line:

If a point {eq}P(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 P along the given line. We can find the reflection of the point, we use 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)=(5,1) {/eq}

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

{eq}\dfrac{1}{2}x-y+2=0 {/eq}

{eq}x-2y+4=0 {/eq}

Hence, a = 1, b = -2, c = 4

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-5}{1}=-2\times \dfrac{5-2+4}{1^2+(-2)^2} {/eq}

{eq}x-5=-2\times \dfrac{7}{6}=\dfrac{-7}{3} {/eq}

{eq}3x-15=-7 {/eq}

{eq}3x=-7+15=8 {/eq}

{eq}x=\dfrac{8}{3} {/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-1}{-2}=-2 \times \dfrac{5-2+4}{1^2+(-2)^2} {/eq}

{eq}\dfrac{y-1}{2}=2\times \dfrac{7}{6}=\dfrac{7}{3} {/eq}

{eq}3y-3=14 {/eq}

{eq}3y=17 {/eq}

{eq}y=\dfrac{17}{3} {/eq}

Hence, the coordinates of the reflected point Q are {eq}(\dfrac{8}{3},\dfrac{17}{3}) {/eq} 