Question

write the coordinates of the vertices after a reflection over the line $y=-x$.
$a(square,square)$
$b(square,square)$
$c(square,square)$
$d(square,square)$

Explanation:

Step1: Recall the rule for reflection over \(y = -x\)

The rule for reflecting a point \((x,y)\) over the line \(y=-x\) is \((x,y)\to(-y, -x)\).

Step2: Find the coordinates of \(A\)

The original coordinates of \(A\) are \((- 8,-10)\). Using the rule \((x,y)\to(-y,-x)\), we substitute \(x=-8\) and \(y = - 10\). Then \(A'=(-(-10),-(-8))=(10,8)\).

Step3: Find the coordinates of \(B\)

The original coordinates of \(B\) are \((4,-10)\). Using the rule \((x,y)\to(-y,-x)\), we substitute \(x = 4\) and \(y=-10\). Then \(B'=(-(-10),-4)=(10,-4)\).

Step4: Find the coordinates of \(C\)

The original coordinates of \(C\) are \((4,-6)\). Using the rule \((x,y)\to(-y,-x)\), we substitute \(x = 4\) and \(y=-6\). Then \(C'=(-(-6),-4)=(6,-4)\).

Step5: Find the coordinates of \(D\)

The original coordinates of \(D\) are \((-8,-6)\). Using the rule \((x,y)\to(-y,-x)\), we substitute \(x=-8\) and \(y=-6\). Then \(D'=(-(-6),-(-8))=(6,8)\).

Answer:

\(A'(10,8)\), \(B'(10,-4)\), \(C'(6,-4)\), \(D'(6,8)\)