Question

write the coordinates of the vertices after a reflection over the x-axis.
j((
k(
l(
m(

Explanation:

Step1: Recall the reflection rule

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

Step2: Find coordinates of \(J\)

The original coordinates of \(J\) are \((- 5,-10)\). Using the reflection rule \((x,y)\to(x, - y)\), for \(J(-5,-10)\), we have \(x=-5\) and \(y = - 10\). Then \(J'(-5,10)\) since \(-y=-(-10) = 10\).

Step3: Find coordinates of \(K\)

The original coordinates of \(K\) are \((-5,-5)\). Using the reflection rule \((x,y)\to(x, - y)\), for \(K(-5,-5)\), we have \(x=-5\) and \(y=-5\). Then \(K'(-5,5)\) since \(-y=-(-5)=5\).

Step4: Find coordinates of \(L\)

The original coordinates of \(L\) are \((-2,-1)\). Using the reflection rule \((x,y)\to(x, - y)\), for \(L(-2,-1)\), we have \(x = - 2\) and \(y=-1\). Then \(L'(-2,1)\) since \(-y=-(-1)=1\).

Step5: Find coordinates of \(M\)

The original coordinates of \(M\) are \((-2,-6)\). Using the reflection rule \((x,y)\to(x, - y)\), for \(M(-2,-6)\), we have \(x=-2\) and \(y = - 6\). Then \(M'(-2,6)\) since \(-y=-(-6)=6\).

Answer:

\(J'(-5,10)\), \(K'(-5,5)\), \(L'(-2,1)\), \(M'(-2,6)\)