Question

1. If EF has endpoints at E(-3, 10) and F(5.6) and is dilated about the origin by a factor of 7 which of the

following would be the length of its image, E'F'?

Answer 1

Answer: 28√5

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,

`\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}=√(64+16) \\&\mathrm{EF}=√(80) \\&\mathrm{EF}=4 √(5)\end{aligned}`

Scale factor = 7

Therefore,

`\begin{aligned}&\mathrm{E}^(\prime) \mathrm{F}^(\prime)=7 \mathrm{EF} \\&\mathrm{E}^(\prime) \mathrm{F}^(\prime)=7 \cdot 4 √(5) \\&\mathrm{E}^(\prime) \mathrm{F}^(\prime)=28 √(5)\end{aligned}`