Answer-
Approximately the tree is 8.5 feet away from the zip line.
Solution-
The line representing the zip line is,
[tex]\Rightarrow y=x-7[/tex]
[tex]\Rightarrow x-y-7=0[/tex]
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 [tex]Ax+By+C=0[/tex] is given by
[tex]d=\dfrac{|Am+Bn+C|}{\sqrt{A^2+B^2}}[/tex]
Putting the values,
[tex]d=\dfrac{|(1)(2)+(-1)(7)+(-7)|}{\sqrt{(1)^2+(-1)^2}}[/tex]
[tex]=\dfrac{|2-7-7|}{\sqrt{1+1}}[/tex]
[tex]=\dfrac{|-12|}{\sqrt{2}}[/tex]
[tex]=\dfrac{12}{\sqrt{2}}[/tex]
[tex]=6\sqrt{2}[/tex]
[tex]=8.5\ \text{feet}[/tex]
Therefore, approximately the tree is 8.5 feet away from the zip line.
Answer:
8.5 feet
Step-by-step explanation:
We are given that equation of the line representing the zip line is y=x-7
The equation can be written as
[tex]x-y-7=0[/tex]
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=[tex]\mid\frac{ax+by+c}{\sqrt{a^2+b^2}}\mid[/tex]
Using this formula
[tex]d=\mid \frac{x-y-7}{\sqrt{1+1}}\mid [/tex]
Substitute the value of x=2 and y=7
Then, we get
[tex]d=\mid\frac{2-7-7}{\sqrt2}\mid=6\sqrt 2 units[/tex]
If 1 unit represent 1 foot
Then , the distance, [tex]d=8.5 feet[/tex]
Hence, the approximate distance of tree from the zip line=8.5 feet
