Find the exact perpendiuclar distance between two parallel lines.
a)y=3x-2 and y=3x+3
please explain your answer
You can easily use the "formula" for the distance between a point and a line. For example, you have $\displaystyle y = 3x-2$ so you know that the point (0,-2) is on the line. So use the formula to find the distance between (0,-2) and the line $\displaystyle y = 3x-2$.
Or, you can skip the formula and draw the graph. Use geometry.
Hi roger1505,
1 graph the lines
2 erect a vertical line @x=1. Apoint (1,1) on line y=3x-2 is produced
3@(1,1) erect a perpendicular to y=3x+3
4 write an equation for the perpendicular
5 solve for intersection of 4 and y=3x +3
6 use distance formula to find question d
d^2 = delta y^2 + delta x^2 (slope diagram between (1,1) and solution of 5 above
There is a simple solution using trig and the slope angle of the parallel lines
Hello, roger1505!
If you are allowed to use Trigonometry, here is another solution.
Find the exact perpendicular distance between two parallel lines: .$\displaystyle \begin{array}{ccc}y &=& 3x-2 \\ y &=& 3x + 3 \end{array}$
The two lines have y- intercepts at 3 and -2, and identical slopes.Code:| | / | / / | / / |/ / A * / : |@* / : | * / : | * C 5 | / : + - / - - - - : | / : | / : |/ @ B * - - - - E |
$\displaystyle AB \,=\,5.$
Draw $\displaystyle AC$ perpendicular to the line through $\displaystyle B.$
Let $\displaystyle \theta = \angle CBE$
Then the slope of $\displaystyle BC \,=\,\tan\theta \,=\,3$
Note that $\displaystyle \angle CAB \,=\, \theta.$
We have: .$\displaystyle \tan\theta \:=\:\frac{3}{1}\:=\:\frac{\text{opp}}{\text{adj}}$
$\displaystyle \theta$ is in a right triangle with: $\displaystyle opp = 3,\;adj = 1$
Hence: .$\displaystyle hyp \,=\, \sqrt{10} \quad\Rightarrow\quad \cos\theta \,=\,\tfrac{1}{\sqrt{10}}$
In right triangle $\displaystyle ACB\!:\;\;\cos\theta \,=\,\frac{AC}{5} \quad\Rightarrow\quad AC \:=\:5\cos\theta \:=\:5\left(\tfrac{1}{\sqrt{10}}\right)$
Therefore: .$\displaystyle AC \:=\:\frac{\sqrt{10}}{2}$
As long as we are into spoon feeding, look at reply #8.
Rewrite line #1 as $\displaystyle 3x-y-2=0$. Note that $\displaystyle (1,6)$ is on $\displaystyle y=3x+3$.
Apply the formula from reply #8: $\displaystyle \frac{|3(1)-1(6)-2|}{\sqrt{(3)^2+(-1)^2}}=\frac{5}{\sqrt{10}}=\frac{\sqrt{10}}{2}.$