Question

graph the image of trapezoid jklm after a dilation with a scale factor of 5, centered at the origin.

Explanation:

Step1: Identify original coordinates

Let's assume the coordinates of trapezoid $JKLM$ are $J(- 2,-2)$, $K(0,-2)$, $L(2,2)$, $M(-2,2)$.

Step2: Apply dilation formula

For a dilation centered at the origin with scale - factor $k = 5$, the formula to find the new coordinates $(x',y')$ of a point $(x,y)$ is $(x',y')=(k\cdot x,k\cdot y)$.
For point $J(-2,-2)$: $J'=(5\times(-2),5\times(-2))=(-10,-10)$.
For point $K(0,-2)$: $K'=(5\times0,5\times(-2))=(0,-10)$.
For point $L(2,2)$: $L'=(5\times2,5\times2)=(10,10)$.
For point $M(-2,2)$: $M'=(5\times(-2),5\times2)=(-10,10)$.

Step3: Graph the new trapezoid

Plot the points $J'(-10,-10)$, $K'(0,-10)$, $L'(10,10)$, $M'(-10,10)$ on the coordinate - plane and connect them in order to get the dilated trapezoid.

Answer:

Graph the trapezoid with vertices $J'(-10,-10)$, $K'(0,-10)$, $L'(10,10)$, $M'(-10,10)$ on the given coordinate - plane.