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=x−7. There is a tree in your yard at the point (2, 7). 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

8.5 feet

Explanation:

We are given that equation of the line representing the zip line is y=x-7

The equation can be written as

`x-y-7=0`

There is a tree in your yard at the point (2,7).

We have to find the distance of the tree from the zip line.

The perpendicular distance of point (x,y) from the line ax+by+c=0 is given by

Distance =d=
`\mid(ax+by+c)/(√(a^2+b^2))\mid`

Using this formula

`d=\mid (x-y-7)/(√(1+1))\mid`

Substitute the value of x=2 and y=7

Then, we get

`d=\mid(2-7-7)/(\sqrt2)\mid=6\sqrt 2 units`

If 1 unit represent 1 foot

Then , the distance,
`d=8.5 feet`

Hence, the approximate distance of tree from the zip line=8.5 feet

Answer 2

Approximately the tree is 8.5 feet away from the zip line.

Solution-

The line representing the zip line is,

`\Rightarrow y=x-7`

`\Rightarrow x-y-7=0`

There is a tree in your yard at the point (2, 7)

The perpendicular distance from point to line formula will give us the result,

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

`d=(|Am+Bn+C|)/(√(A^2+B^2))`

Putting the values,

`d=(|(1)(2)+(-1)(7)+(-7)|)/(√((1)^2+(-1)^2))`

`=(|2-7-7|)/(√(1+1))`

`=(|-12|)/(√(2))`

`=(12)/(√(2))`

`=6√(2)`

`=8.5\ \text{feet}`

Therefore, approximately the tree is 8.5 feet away from the zip line.