Find the exact perpendiuclar distance between two parallel lines.
a)y=3x2 and y=3x+3
please explain your answer
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Find the exact perpendiuclar distance between two parallel lines.
a)y=3x2 and y=3x+3
please explain your answer
can you please do it. mainly because its a parallel, and i dont know how to sub it in
Your teacher did not show how...or you missed classes?
i wasnt in class when they taught this, but does anyone know?
You can easily use the "formula" for the distance between a point and a line. For example, you have $\displaystyle y = 3x2$ 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 $\displaystyle y = 3x2$.
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=3x2 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.
Quote:
Find the exact perpendicular distance between two parallel lines: .$\displaystyle \begin{array}{ccc}y &=& 3x2 \\ y &=& 3x + 3 \end{array}$
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 $\displaystyle 3xy2=0$. Note that $\displaystyle (1,6)$ is on $\displaystyle y=3x+3$.
Apply the formula from reply #8: $\displaystyle \frac{3(1)1(6)2}{\sqrt{(3)^2+(1)^2}}=\frac{5}{\sqrt{10}}=\frac{\sqrt{10}}{2}.$
http://latex.codecogs.com/png.latex?...%7D%7D%7B2%7D.
This is the best and simplest way of finding the perpendicular distance between two parallel lines. Hope you get it right. (Clapping)
What is your point LJ? Already answered 2 months ago...