Question

write the coordinates of the vertices after a reflection over the line x = 1.

Explanation:

Step1: Find the distance from each x - coordinate to the line \(x = 1\)

For a point \((x,y)\), the distance from \(x\) to \(x = 1\) is \(d=\vert x - 1\vert\). After reflection over the line \(x = 1\), the new \(x\) - coordinate is \(x'=1+(1 - x)=2 - x\) (if \(x<1\)) or \(x'=1-(x - 1)=2 - x\) (if \(x\geq1\)). The \(y\) - coordinate remains the same.

• For point \(K(3, - 1)\):

The new \(x\) - coordinate is \(x'=2-3=-1\), and \(y\) - coordinate \(y'=-1\). So the reflected point \(K'\) is \((-1,-1)\)

• For point \(N(3,6)\):

The new \(x\) - coordinate is \(x'=2 - 3=-1\), and \(y\) - coordinate \(y'=6\). So the reflected point \(N'\) is \((-1,6)\)

• For point \(L(8,-1)\):

The new \(x\) - coordinate is \(x'=2 - 8=-6\), and \(y\) - coordinate \(y'=-1\). So the reflected point \(L'\) is \((-6,-1)\)

• For point \(M(8,6)\):

The new \(x\) - coordinate is \(x'=2 - 8=-6\), and \(y\) - coordinate \(y'=6\). So the reflected point \(M'\) is \((-6,6)\)

Answer:

\(K'(-1,-1)\), \(N'(-1,6)\), \(L'(-6,-1)\), \(M'(-6,6)\)