Question

problem 4 here is triangle abc on the coordinate plane. using the origin as the center and a scale factor of 2, drag the points to make triangle def, the dilation of triangle abc. draw triangle

Explanation:

Step1: Recall dilation formula

If the origin \((0,0)\) is the center of dilation and the scale - factor is \(k\), and a point \((x,y)\) is dilated, the new point \((x',y')\) is given by \((x',y')=(k\cdot x,k\cdot y)\). Here \(k = 2\).

Step2: Assume coordinates of \(\triangle ABC\)

Let's assume \(A=(x_1,y_1)\), \(B=(x_2,y_2)\), \(C=(x_3,y_3)\) from the graph.

Step3: Calculate coordinates of \(\triangle DEF\)

For point \(D\) (corresponding to \(A\)), \(D=(2x_1,2y_1)\); for point \(E\) (corresponding to \(B\)), \(E=(2x_2,2y_2)\); for point \(F\) (corresponding to \(C\)), \(F=(2x_3,2y_3)\).

Step4: Plot the points

Plot the points \(D\), \(E\), and \(F\) on the coordinate - plane and connect them to form \(\triangle DEF\).

Since we don't have the actual coordinates of \(A\), \(B\), and \(C\) given numerically in the description:
If we assume \(A=(x_A,y_A)\), \(B=(x_B,y_B)\), \(C=(x_C,y_C)\), then \(D=(2x_A,2y_A)\), \(E=(2x_B,2y_B)\), \(F=(2x_C,2y_C)\). Plot \(D\), \(E\), \(F\) and connect them to get \(\triangle DEF\).

Answer:

Plot points \(D\), \(E\), \(F\) which are the dilated versions of \(A\), \(B\), \(C\) respectively with a scale factor of 2 centered at the origin, and connect them to form \(\triangle DEF\). The coordinates of \(D\), \(E\), \(F\) are obtained by multiplying the \(x\) and \(y\) - coordinates of \(A\), \(B\), \(C\) by 2.