Answer: 28√5
Step-by-step explanation:
The endpoints of EF are E(-3 , 10) and F(5 , 6)
Let x1 = -3, y1 = 10, x2 = 5, y2 = 6
So, the length of EF is,
[tex]\begin{aligned}&\mathrm{EF}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}} \\&\mathrm{EF}=\sqrt{\left(5-(-3)^{2}+(6-10)^{2}\right.} \\&\mathrm{EF}=\sqrt{(5+3)^{2}+(-4)^{2}} \\&\mathrm{EF}=\sqrt{64+16} \\&\mathrm{EF}=\sqrt{80} \\&\mathrm{EF}=4 \sqrt{5}\end{aligned}[/tex]
Scale factor = 7
Therefore,
[tex]\begin{aligned}&\mathrm{E}^{\prime} \mathrm{F}^{\prime}=7 \mathrm{EF} \\&\mathrm{E}^{\prime} \mathrm{F}^{\prime}=7 \cdot 4 \sqrt{5} \\&\mathrm{E}^{\prime} \mathrm{F}^{\prime}=28 \sqrt{5}\end{aligned}[/tex]
