Find Distance between 2 Lines
- Thread starter HCL
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mmm4444bot
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The two lines have the same slope, so they are parallel.
This question wants you to calculate the shortest distance between the lines; that requires measuring along a line that is perpendicular to both lines.
The given line y=3x passes through the origin. Can you write the equation for the line which is perpendicular to y=3x, also passing through the origin? (Remember, perpendicular slopes are negative reciprocals of each other.)
Now find the intersection point of this perpendicular line with the line y=3x+4, and then use those (x,y) coordinates and the origin (0,0) in the distance formula.
Let me know, if I wrote anything that you do not understand.
Cheers :cool:
This question wants you to calculate the shortest distance between the lines; that requires measuring along a line that is perpendicular to both lines.
The given line y=3x passes through the origin. Can you write the equation for the line which is perpendicular to y=3x, also passing through the origin? (Remember, perpendicular slopes are negative reciprocals of each other.)
Now find the intersection point of this perpendicular line with the line y=3x+4, and then use those (x,y) coordinates and the origin (0,0) in the distance formula.
Let me know, if I wrote anything that you do not understand.
Cheers :cool:
yea so I got y=-1/3x but when i find the intersecting with =3x+4, I'm getting (0,4) and the origin (0,0), but when I place that in the distance formula, I'm getting 4 as my answer but it says the answer is (2*square root10)/5??
Hello, HCL!
We have: .\(\displaystyle \begin{Bmatrix}y \:=\:3x+4 \\ y \:=\:3x\end{Bmatrix}\)
The graph looks like this:
We want the distance \(\displaystyle d = PQ\)
. . where the line \(\displaystyle L\) through \(\displaystyle P\) and \(\displaystyle Q\) is perpendicular to both lines.
The slope of the lines is \(\displaystyle 3.\)
The slope of line \(\displaystyle L\) is \(\displaystyle \text{-}\frac{1}{3}\)
The equation of line \(\displaystyle L\) is: .\(\displaystyle y \:=\:\text{-}\frac{1}{3}x + 4\)
Find the intersection of \(\displaystyle L\) and \(\displaystyle y = 3x\)
. . \(\displaystyle 3x \:=\:\text{-}\frac{1}{3}x + 4 \quad\Rightarrow\quad \frac{10}{3}x \:=\:4 \quad\Rightarrow\quad x =\frac{6}{5} \)
Then: .\(\displaystyle y \,=\,3\left(\frac{6}{5}\right) \quad\Rightarrow\quad y = \frac{18}{5}\)
Find the distance from \(\displaystyle P(0,4)\) to \(\displaystyle Q\left(\frac{6}{5},\frac{18}{5}\right)\)
. . \(\displaystyle d^2 \:=\:\left(\frac{6}{5}-0\right)^2 + \left(\frac{18}{5} - 4\right)^2 \;=\;\left(\frac{6}{5}\right)^2 + \left(\text{-}\frac{2}{5}\right)^2 \)
. . \(\displaystyle d^2 \:=\:\frac{36}{25} + \frac{4}{25} \;=\;\frac{40}{25}\)
Therefore: .\(\displaystyle d \;=\;\sqrt{\dfrac{40}{25}} \:=\:\dfrac{2\sqrt{10}}{5}\)
We have: .\(\displaystyle \begin{Bmatrix}y \:=\:3x+4 \\ y \:=\:3x\end{Bmatrix}\)
The graph looks like this:
Code:
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|. . where the line \(\displaystyle L\) through \(\displaystyle P\) and \(\displaystyle Q\) is perpendicular to both lines.
The slope of the lines is \(\displaystyle 3.\)
The slope of line \(\displaystyle L\) is \(\displaystyle \text{-}\frac{1}{3}\)
The equation of line \(\displaystyle L\) is: .\(\displaystyle y \:=\:\text{-}\frac{1}{3}x + 4\)
Find the intersection of \(\displaystyle L\) and \(\displaystyle y = 3x\)
. . \(\displaystyle 3x \:=\:\text{-}\frac{1}{3}x + 4 \quad\Rightarrow\quad \frac{10}{3}x \:=\:4 \quad\Rightarrow\quad x =\frac{6}{5} \)
Then: .\(\displaystyle y \,=\,3\left(\frac{6}{5}\right) \quad\Rightarrow\quad y = \frac{18}{5}\)
Find the distance from \(\displaystyle P(0,4)\) to \(\displaystyle Q\left(\frac{6}{5},\frac{18}{5}\right)\)
. . \(\displaystyle d^2 \:=\:\left(\frac{6}{5}-0\right)^2 + \left(\frac{18}{5} - 4\right)^2 \;=\;\left(\frac{6}{5}\right)^2 + \left(\text{-}\frac{2}{5}\right)^2 \)
. . \(\displaystyle d^2 \:=\:\frac{36}{25} + \frac{4}{25} \;=\;\frac{40}{25}\)
Therefore: .\(\displaystyle d \;=\;\sqrt{\dfrac{40}{25}} \:=\:\dfrac{2\sqrt{10}}{5}\)
No, so they are parallel or they are the same line. \(\displaystyle \ \ \ \ \ \ \) If two lines have the same slope, then they might be parallel. If two lines are parallel, then they definitely have the same slope.
ok, thanks, I get it, but last question, why can't I multiply both sides by 3 to get rid of the fraction?? that was my mistake
mmm4444bot
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I'm not sure why you posted this, lookagain.
The two lines have differing y-intercepts. They are clearly not the same line. :???:
mmm4444bot
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We can always multiply both sides of an equation by 3.
Please show us what you were trying to do; I'm not sure at which point in working this exercise that you felt you could not multiply by 3.
Again, we can't see your mistakes, if you don't show your work.
In the long run, posting your efforts at the start of the thread is best because it will save you time.
Cheers :cool:
again, i think i see where i messed up but i was multiplying by the 3 when 3x=-1x/3+4 and when i got 9x=-x+ 12, i would put the 10x=12, and here is the embarassing mistake, when i divided by 12/10... i don't know... i'm getting confused on simple things because it's in test format and the simplest thing becomes lost in the overall problem of finding the distance. thanks again