Question

You are setting up a zip line in your yard. You map out your yard in a coordinate plane. An equation of the line representing the zip line is y=−4/5x+5. There is a tree in your yard at the point (4, 10). Each unit in the coordinate plane represents 1 foot. Approximately how far away is the tree from the zip line? Round your answer to the nearest tenth.

Answer 1

Answer:

The tree is approximately 6.4 feet away from the zip line.

Step-by-step explanation:

Equation of the line representing the zip line is  $y=-\frac{4}{5}x+5$

First we need to convert the above equation into $ax+by+c=0$ form. So........

$y=-\frac{4}{5}x+5\\ \\ 5y=5(-\frac{4}{5}x+5)\\ \\ 5y=-4x+25\\ \\ 4x+5y-25=0$

Thus,  $a=4, b=5$ and $c=-25$

The formula for distance$(d)$ from a point $(x_{0}, y_{0})$ to the line $ax+by+c=0$ is.........

$d=\frac{|a(x_{0})+b(y_{0})+c|}{\sqrt{a^2+b^2}}$

Given that, there is a tree in your yard at the point (4, 10). So here,  $x_{0}=4$ and $y_{0}=10$

Thus, the distance will be:   $d=\frac{|4(4)+5(10)-25|}{\sqrt{4^2+5^2}}= \frac{|16+50-25|}{\sqrt{16+25}}= \frac{41}{\sqrt{41}}=\sqrt{41}= 6.403... \approx 6.4$  (Rounding to the nearest tenth)

So, the tree is approximately 6.4 feet away from the zip line.