Question

write the coordinates of the vertices after a reflection over the y - axis.

Explanation:

Step1: Recall the reflection rule over y - axis

When reflecting a point \((x,y)\) over the \(y\) - axis, the rule is \((x,y)\to(-x,y)\).

Step2: Find the coordinates of \(J\)

The original coordinates of \(J\) are \((-6,-2)\). Using the reflection rule \((x,y)\to(-x,y)\), for \(x=-6\) and \(y = - 2\), we get \(-x=6\). So \(J'\) has coordinates \((6,-2)\).

Step3: Find the coordinates of \(K\)

The original coordinates of \(K\) are \((-6,0)\). Using the reflection rule \((x,y)\to(-x,y)\), for \(x=-6\) and \(y = 0\), we get \(-x = 6\). So \(K'\) has coordinates \((6,0)\).

Step4: Find the coordinates of \(L\)

The original coordinates of \(L\) are \((0,0)\). Using the reflection rule \((x,y)\to(-x,y)\), for \(x = 0\) and \(y=0\), we get \(-x=0\). So \(L'\) has coordinates \((0,0)\).

Step5: Find the coordinates of \(M\)

The original coordinates of \(M\) are \((0,-2)\). Using the reflection rule \((x,y)\to(-x,y)\), for \(x = 0\) and \(y=-2\), we get \(-x=0\). So \(M'\) has coordinates \((0,-2)\).

Answer:

\(J'(6,-2)\), \(K'(6,0)\), \(L'(0,0)\), \(M'(0,-2)\)