karatechopper
How to find distance between two lines...
y=3x
y=3x+10
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zonazoo
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the vertical distance, or horizontal distance or what?
karatechopper
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|dw:1350957364620:dw|
zonazoo
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that perpendicular line between the two is the distance you want?
asnaseer
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Work out the equation of the perpendicular line. To make t easy, work out the equation of the line that is perpendicular to y=3x and passes through (0,0).
Then find the intersection point of this perpendicular line and the other line (y=3x + 10).
Then use distance formula.
asnaseer
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Hope that makes sense?
zonazoo
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yeah, thats what i would do
karatechopper
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I am very confused papa. Things are not going well in my house. So can you restart from baby steps? And walk me through the perpendicular lines?
asnaseer
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ok, step 1 - do you know the relationship of the slopes of two lines that are perpendicular to one another?
karatechopper
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I know that the perpendicular slope is the opposite reciprocal of the slope.
asnaseer
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almost - the product of the slopes of two perpendicular lines is always -1.
asnaseer
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so - what would be the slope of the line that is perpendicular to y=3x?
karatechopper
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-1/3x
asnaseer
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perfect - slope would be -1/3, and its equation would be of the form:
y = -(1/3)x + c
where c is the y-intercept
asnaseer
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agreed?
karatechopper
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Correct. I got that part.
asnaseer
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good, now if we take this perpendicular line such that it passes through the origin, then what would be the value for c?
asnaseer
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remember we have:
y = -(1/3)x + c
and we know y=0 when x=0, therefore c=?
asnaseer
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are you stuck?
karatechopper
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yes..
karatechopper
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C=0
asnaseer
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yes - correct
karatechopper
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Yay!
asnaseer
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so now we know that the line perpendicular to y=3x and that passes through the origin has the equation:
y = -(1/3)x
karatechopper
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Ok, so i drew that line in on my graph..
karatechopper
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|dw:1350958743362:dw|
asnaseer
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|dw:1350951532761:dw|
karatechopper
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Yea! That!
asnaseer
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so next you want to find the point of intersection of:
y = -(1/3)x
and:
y = 3x + 10
karatechopper
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Oh ok !
karatechopper
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How do i do that...
asnaseer
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can you find the point of intersection of these two lines?
asnaseer
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just substitue y=-(1/3)x from 1st equation into 2nd equation and solve for x.
asnaseer
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then use 1st equation to find the y value once you have the x value
karatechopper
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Ah ok!
karatechopper
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So like....the answer would be for x=3?
asnaseer
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make sure you have the correct sign - does it look like x would be positive here?
karatechopper
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aahhhh!!! negative negative!!
asnaseer
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:)
asnaseer
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now find y value
asnaseer
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use the first equation for that, i.e. y = -(1/3)x to find y. remember you just found x=-3
karatechopper
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Oh i know i am working it out!!!
karatechopper
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y=19
asnaseer
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how???
asnaseer
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x=-3, therefore:\[y=-\frac{1}{3}\times x=-\frac{1}{3}\times (-3) = ?\]
karatechopper
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Oh ok i was using a dif equation.
karatechopper
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y=-1
asnaseer
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please be more careful in your calculations - try again...
karatechopper
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gAAAA
karatechopper
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Ohhhhhhhh y=1
asnaseer
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good. so now you know the point of intersection is (-3, 1):
|dw:1350952783534:dw|
asnaseer
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so just use the distance formula to find the distance between (-3,1) and (0,0).
That will be the distance between these two lines.
karatechopper
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Alright thanks!
asnaseer
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yw :)
karatechopper
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Now SLEEP! I am ordering my papa to sleep:)