Question

1. graph the image of the quadrilateral below using a scale factor of k = 3/2.

w (__, __)
x (__, __)
y (__, __)
z (__, __)

Explanation:

Step1: Assume coordinates

Let's assume the original coordinates of the vertices of the quadrilateral are \(W(x_1,y_1)\), \(X(x_2,y_2)\), \(Y(x_3,y_3)\), \(Z(x_4,y_4)\). To find the new - coordinates \(W'(x_1',y_1')\), \(X'(x_2',y_2')\), \(Y'(x_3',y_3')\), \(Z'(x_4',y_4')\) after dilation with a scale factor \(k = \frac{3}{2}\), we use the formula \(x'=k\times x\) and \(y'=k\times y\).

Step2: Calculate new coordinates for each vertex

For a vertex \((x,y)\) of the original quadrilateral, the new vertex \((x',y')\) has coordinates \(x'=\frac{3}{2}x\) and \(y'=\frac{3}{2}y\). For example, if \(W\) has coordinates \((a,b)\), then \(W'\) has coordinates \((\frac{3}{2}a,\frac{3}{2}b)\). You need to first identify the original coordinates of \(W\), \(X\), \(Y\), \(Z\) from the graph (let's say \(W(x_1,y_1)\), \(X(x_2,y_2)\), \(Y(x_3,y_3)\), \(Z(x_4,y_4)\)). Then \(W'(\frac{3}{2}x_1,\frac{3}{2}y_1)\), \(X'(\frac{3}{2}x_2,\frac{3}{2}y_2)\), \(Y'(\frac{3}{2}x_3,\frac{3}{2}y_3)\), \(Z'(\frac{3}{2}x_4,\frac{3}{2}y_4)\).

Answer:

You need to first determine the original coordinates of \(W\), \(X\), \(Y\), \(Z\) from the graph. Then multiply each \(x\) - coordinate and \(y\) - coordinate of the vertices by \(\frac{3}{2}\) to get the coordinates of \(W'\), \(X'\), \(Y'\), \(Z'\). For example, if \(W=(2,2)\), then \(W'=(3,3)\) (since \(2\times\frac{3}{2}=3\) for both \(x\) and \(y\) coordinates). Without the actual coordinates of \(W\), \(X\), \(Y\), \(Z\) from the graph, we cannot give the exact numerical answers. But the general method is to use the transformation \((x,y)\to(\frac{3}{2}x,\frac{3}{2}y)\) for each vertex.