virtus
perpendicular distance from point B(0,5) to the line x+3y-5=0
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ash2326
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@virtus Are you here?
virtus
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yes!
ash2326
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@virtus There maybe a direct formula for this, but I don't remember that. I'll start from the basics. Do you have time for this?
virtus
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sure, thanks ash2326
waterineyes
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Can I give the formula??
ash2326
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Thanks virtus, let's begin. I recommend that you understand the problem, you'll remember it always. What do you want?
waterineyes
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Go ahead @ash2326 ..
ash2326
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|dw:1341645333105:dw|
Let AB be our line and C is the point for which we need to find the perpendicular distance.
Do you understand the diagram?
virtus
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yes
ash2326
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You can see that the perpendicular line from the point C is intersecting the line AB at a point. Let it be D
If we could find the coordinates of D, then we can easily find the distance.
do you have any ideas what should we do?
virtus
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POI
ash2326
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POI???
virtus
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find the Point Of Intersection
waterineyes
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Ha ha ha ha..
ash2326
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Yeah:) @virtus
For that we need to find the equation of the line CD, do you have any idea how to do that?
virtus
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well you know what point C is and the gradient of line CD would be -1/ gradient of line AB
virtus
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oh wait do we know what C is ?
ash2326
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Awesome:D Could you do that?
We have AB
as
\(x+3y-5=0\)
Yeah we know C as (0, 5)
virtus
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x+3y-15=0
ash2326
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Oops, You made a mistake
I can see that perpendicular line's slope is same as AB, it's to be -1/ slope AB
Check again :)
virtus
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oh silly me
ash2326
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No problem, :D
find it again?
virtus
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y-5 =3(x-0)
y=3x +5
3x-y+5=0
ash2326
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Correct:D Now find the intersection of the line (3x-y+5=0) and (x+3y-5=0)
this will give us D
virtus
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D (-1,2)
ash2326
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Great work:D
now find the distance between (0, 5) and (-1, 2) using distance formula.
ash2326
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That will be the perpendicular distance:D
virtus
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ngawwwwwww! I SEE!!!! so that's how you do it ;)
oh btw i got square root 10
ash2326
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Yeah, that's correct:D Great work btw:D
virtus
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thanks so much for your invaluble help @ash2326 :D
ash2326
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@virtus you did all the work. I just guided you:D You are a good student:D
dpaInc
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you can also use the formula from analytic geometry that the distance from a given point to a given line in general form \(\large Ax+By+C=0 \) is given by
Distance = \(\large \frac{Ax_0+By_0+C}{\pm \sqrt{A^2+B^2}} \) where \(\ (x_0, y_0) \) are the coordinates of the given point.