Question

dana reflects point a(2,5) across line $ell$ to get image point a(6,1). what is an equation for line $ell$? the equation for line $ell$ is

Explanation:

Step1: Find the mid - point of AA'

The mid - point formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. For $A(2,5)$ and $A'(6,1)$, the mid - point $M$ is $(\frac{2 + 6}{2},\frac{5+1}{2})=(4,3)$.

Step2: Find the slope of AA'

The slope formula for two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For $A(2,5)$ and $A'(6,1)$, the slope of $AA'$ is $m_{AA'}=\frac{1 - 5}{6 - 2}=\frac{-4}{4}=-1$.

Step3: Find the slope of line $\ell$

Since line $\ell$ is the perpendicular bisector of $AA'$, if the slope of $AA'$ is $m_1$ and the slope of line $\ell$ is $m_2$, then $m_1\times m_2=-1$. Given $m_{AA'}=-1$, the slope of line $\ell$ is $m = 1$.

Step4: Find the equation of line $\ell$

Using the point - slope form of a line $y - y_0=m(x - x_0)$, where $(x_0,y_0)=(4,3)$ and $m = 1$. We have $y-3=1\times(x - 4)$, which simplifies to $y=x - 1$.

Answer:

$y=x - 1$