Question

remember
for problems 12–15, solve for x.

1. $5(x + 12) = 10$
2. $4(x - 8) = 60$
3. $6(x + 3) = -30$
4. $10(x - 4) = -100$
5. $\triangle abc$ is shown in the coordinate plane.

a. graph the image of $\triangle abc$ under a dilation centered at the origin with scale factor 2. label the image $\triangle abc$.
b. what are the coordinates of the vertices of $\triangle abc$?
c. describe how you located the coordinates of the vertices of $\triangle abc$.

Explanation:

Step1: Solve 5(x+12)=10, isolate term

Divide both sides by 5:
$\frac{5(x+12)}{5}=\frac{10}{5}$
$x+12=2$
Subtract 12:
$x=2-12$

Step2: Solve 6(x+3)=-30, isolate term

Divide both sides by 6:
$\frac{6(x+3)}{6}=\frac{-30}{6}$
$x+3=-5$
Subtract 3:
$x=-5-3$

Step3: Solve 4(x-8)=60, isolate term

Divide both sides by 4:
$\frac{4(x-8)}{4}=\frac{60}{4}$
$x-8=15$
Add 8:
$x=15+8$

Step4: Solve 10(x-4)=-100, isolate term

Divide both sides by 10:
$\frac{10(x-4)}{10}=\frac{-100}{10}$
$x-4=-10$
Add 4:
$x=-10+4$

Step5: Identify original triangle vertices

From the graph:
$A=(1,-1)$, $B=(3,-2)$, $C=(2,-4)$

Step6: Dilate vertices by scale factor 2

Multiply each coordinate by 2:
$A'=(1\times2, -1\times2)=(2,-2)$
$B'=(3\times2, -2\times2)=(6,-4)$
$C'=(2\times2, -4\times2)=(4,-8)$

Answer:

1. $x=-10$
2. $x=23$
3. $x=-8$
4. $x=-6$

16b. $A'(2, -2)$, $B'(6, -4)$, $C'(4, -8)$
16c. Multiply the $x$- and $y$-coordinates of each original vertex ($A$, $B$, $C$) by the scale factor of 2 to get the coordinates of the dilated vertices $A'$, $B'$, $C'$.
(Note: For 16a, plot points $A'(2, -2)$, $B'(6, -4)$, $C'(4, -8)$ on the coordinate plane and connect them to form $\triangle A'B'C'$.)