Question

△rs is dilated with the rule d_t1/1 (x, y), where the center of dilation is t(3, -2)
the distance between the x - coordinates of r and t is
the distance between the y - coordinates of r and t is
r is from t, so the coordinates of r are

Explanation:

Step 1: Find the coordinates of \(R\) and \(T\)

From the graph, \(R=(1,4)\) and \(T=(3, - 2)\)

Step 2: Calculate the distance between \(x -\)coordinates

The formula for the distance between two \(x -\)coordinates \(x_1\) and \(x_2\) is \(|x_1 - x_2|\). Here, \(x_1 = 1\) ( \(x -\)coordinate of \(R\)) and \(x_2=3\) ( \(x -\)coordinate of \(T\)). So \(|1 - 3|=2\)

Step 3: Calculate the distance between \(y -\)coordinates

The formula for the distance between two \(y -\)coordinates \(y_1\) and \(y_2\) is \(|y_1 - y_2|\). Here, \(y_1 = 4\) ( \(y -\)coordinate of \(R\)) and \(y_2=-2\) ( \(y -\)coordinate of \(T\)). So \(|4-(-2)|=|4 + 2|=6\)

Step 4: Use the dilation rule \(D_{T,\frac{1}{2}}\)

If we have a dilation with center \(T(x_T,y_T)\) and scale factor \(k=\frac{1}{2}\), and a point \(P(x,y)\), the formula for the dilated point \(P'(x',y')\) is \(x'=x_T + k(x - x_T)\) and \(y'=y_T + k(y - y_T)\)

For \(R=(1,4)\) and \(T=(3,-2)\) and \(k = \frac{1}{2}\)

\(x'=3+\frac{1}{2}(1 - 3)=3+\frac{1}{2}\times(- 2)=3-1 = 2\)

\(y'=-2+\frac{1}{2}(4+2)=-2 + 3=1\)

So the distance between the \(x -\)coordinates of \(R\) and \(T\) is \(2\), the distance between the \(y -\)coordinates of \(R\) and \(T\) is \(6\), \(R'\) is \(\frac{1}{2}\) times the distance from \(T\) (in \(x\) and \(y\) directions) and the coordinates of \(R'\) are \((2,1)\)

Answer:

The distance between the \(x -\)coordinates of \(R\) and \(T\) is \(2\). The distance between the \(y -\)coordinates of \(R\) and \(T\) is \(6\). \(R'\) is \(1\) unit from \(T\) in the \(x -\)direction and \(3\) units from \(T\) in the \(y -\)direction, so the coordinates of \(R'\) are \((4,1)\)