Question

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

Explanation:

Step1: Find the mid - point of \(A(2,5)\) and \(A'(6,1)\)

The mid - point formula is \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\).
For \(x\) - coordinate: \(\frac{2 + 6}{2}=\frac{8}{2}=4\)
For \(y\) - coordinate: \(\frac{5+1}{2}=\frac{6}{2}=3\)
The mid - point \(M=(4,3)\) lies on line \(\ell\).

Step2: Find the slope of the line joining \(A(2,5)\) and \(A'(6,1)\)

The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\).
\(m_{AA'}=\frac{1 - 5}{6 - 2}=\frac{-4}{4}=-1\)
Since line \(\ell\) is the perpendicular bisector of \(AA'\), if the slope of \(AA'\) is \(m_1=-1\), and the slope of line \(\ell\) is \(m_2\), then \(m_1\times m_2=-1\). So \(m_2 = 1\) (because \((-1)\times m_2=-1\Rightarrow m_2 = 1\))

Step3: Use the point - slope form \(y - y_0=m(x - x_0)\)

Using the point \((x_0,y_0)=(4,3)\) and \(m = 1\)
\(y-3=1\times(x - 4)\)
\(y-3=x - 4\)
\(y=x-4 + 3\)
\(y=x - 1\)

Answer:

\(y=x - 1\)