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=-6/7x+7. There is a tree in your yard at the point (6,14). 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

The correct answer is:

9.2 ft.

Step-by-step explanation:

The distance from a point (m,n) to a line Ax+By+C=0 is given by the formula:

`d=(|Am+Bn+C|)/(√(A^2+B^2))`

We first need to write our equation in the form Ax+By+C=0.
y=(-6/7)x+7

First we will add 6/7x to each side:
y+6/7x=(-6/7x)+7+(6/7x)
y+6/7x=7

Now we will subtract 7 from each side:
y+6/7x-7=7-7
y+6/7x-7=0

It will be easier to work with this equation if we do not have fractions. We can accomplish this by multiplying everything by the denominator of the fraction, 7:
y(7)+(6/7x)(7)-7(7)=0
7y+(42/7)x-49=0
7y+6x-49=0

Now we rearrange the terms to the x term is in front:
6x+7y-49=0

This is in the form Ax+By+C=0, where A=6, B=7 and C=-49.

Substituting these into our formula above along with our coordinates from our point (m,n)=(6, 14) we have:

`d=(|Am+Bn+C|)/(√(A^2+B^2)) \\ \\d=(|6(6)+7(14)-49|)/(√(6^2+7^2)) \\ \\d=(|36+98-49|)/(√(36+49)) \\ \\d=(|85|)/(√(85))=(85)/(√(85))=9.2`