Question

Please help me Show your work 10 points

Answer 1

Answer

After the dilation $\frac{5}{3}$ around the center of dilation (2, -2), our triangle will have coordinates:

$R'=(2,3)$

$S'=(2,-2)$

$T'=(-3,-2)$

Explanation

First, we are going to translate the center of dilation to the origin. Since the center of dilation is (2, -2) we need to move two units to the left (-2) and two units up (2) to get to the origin. Therefore, our first partial rule will be:

$(x,y)$→$(x-2, y+2)$

Next, we are going to perform our dilation, so we are going to multiply our resulting point by the dilation factor $\frac{5}{3}$. Therefore our second partial rule will be:

$(x,y)$→$\frac{5}{3} (x-2,y+2)$

$(x,y)$→$(\frac{5}{3} x-\frac{10}{3} ,\frac{5}{3} y+\frac{10}{3} )$

Now, the only thing left to create our actual rule is going back from the origin to the original center of dilation, so we need to move two units to the right (2) and two units down (-2)

$(x,y)$→$(\frac{5}{3} x-\frac{10}{3}+2,\frac{5}{3} y+\frac{10}{3}-2)$

$(x,y)$→$(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})$

Now that we have our rule, we just need to apply it to each point of our triangle to perform the required dilation:

$R=(2,1)$

$R'=(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})$

$R'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(1)+ \frac{4}{3})$

$R'=(\frac{10}{3} -\frac{4}{3} ,\frac{5}{3}+ \frac{4}{3})$

$R'=(2,3)$

$S=(2,-2)$

$S'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})$

$S'=(\frac{10}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})$

$S'=(2,-2)$

$T=(-1,-2)$

$T'=(\frac{5}{3} (-1)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})$

$T'=(-\frac{5}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})$

$T'=(-3,-2)$

Now we can finally draw our triangle: