Question

graph the image of trapezoid defg after a translation 8 units left and 11 units down.

Explanation:

Step1: Recall translation rule

For a translation \(h\) units left and \(k\) units down, the rule for a point \((x,y)\) is \((x - h,y - k)\). Here \(h = 8\) and \(k=11\).

Step2: Find new coordinates of point \(D\)

Assume \(D=(4,1)\). New \(x\) - coordinate: \(4-8=-4\), new \(y\) - coordinate: \(1 - 11=-10\). So new \(D'=(-4,-10)\).

Step3: Find new coordinates of point \(E\)

Assume \(E=(8,1)\). New \(x\) - coordinate: \(8 - 8=0\), new \(y\) - coordinate: \(1-11=-10\). So new \(E'=(0,-10)\).

Step4: Find new coordinates of point \(F\)

Assume \(F=(6,9)\). New \(x\) - coordinate: \(6 - 8=-2\), new \(y\) - coordinate: \(9-11=-2\). So new \(F'=(-2,-2)\).

Step5: Find new coordinates of point \(G\)

Assume \(G=(-2,9)\). New \(x\) - coordinate: \(-2-8=-10\), new \(y\) - coordinate: \(9 - 11=-2\). So new \(G'=(-10,-2)\).

Step6: Graph the new trapezoid

Plot the points \(D'(-4,-10)\), \(E'(0,-10)\), \(F'(-2,-2)\) and \(G'(-10,-2)\) and connect them to form the new trapezoid.

Answer:

Graph the trapezoid with vertices \(D'(-4,-10)\), \(E'(0,-10)\), \(F'(-2,-2)\) and \(G'(-10,-2)\)