Find the shortest distance between a line and a point
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OpenStudy (swissgirl):
perprendicualar
OpenStudy (ahsome):
Yes, the perpendicular line.
OpenStudy (swissgirl):
y=5
OpenStudy (ahsome):
The line is ment to be straight, BTW
OpenStudy (swissgirl):
5
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OpenStudy (ahsome):
Thanks. What is the way to show that with working though?
OpenStudy (swissgirl):
pythagorean thm shows that all other lines not perpendicular must have greater length
OpenStudy (ahsome):
Therefore, @swissgirl?
OpenStudy (ahsome):
@swissgirl ?
OpenStudy (swissgirl):
hello
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OpenStudy (mathmath333):
d=|ax+by+c|/sqrt{a^2+b2}
OpenStudy (ahsome):
Is this it?
\[d=\frac{|ax+by+c|}{\sqrt{a^2+b2}}\]
OpenStudy (ahsome):
Huh, @mathmath333
OpenStudy (mathmath333):
yes
OpenStudy (ahsome):
I have NO idea what that is, Could you explain please?
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OpenStudy (mathmath333):
|dw:1412759323290:dw|
OpenStudy (ahsome):
Whoops, wrote the wrong question. Sorry :(
Find a line that is perpendicular to x=10 and (4,5)
OpenStudy (ahsome):
Do you know, @mathmath333 ?
OpenStudy (mathmath333):
|dw:1412759730379:dw|
OpenStudy (ahsome):
Yup. So what is that line? y=5?
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OpenStudy (mathmath333):
yes
OpenStudy (ahsome):
K, thanks. But how would you do the working for that though?
OpenStudy (mathmath333):
u can find the equation of line f,as slope m = 0 thus y=5
y-y1=m(x-x1)
OpenStudy (ahsome):
So you would use the equation:
\[y-y_1=m(x-x_1)\]
To find the slope of the equation y=5?
OpenStudy (mathmath333):
yes this would work
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OpenStudy (ahsome):
So would you sub in the values, \((4,5)\) for that equation?
OpenStudy (swissgirl):
-.-
OpenStudy (ahsome):
@swissgirl ?
OpenStudy (mathmath333):
yes of cos
OpenStudy (ahsome):
I am still confused though. When I look on the video on KhanAcademy, the equation was in y=mx+c form. Now, it is in x=c form. I am not sure what to do from there?
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OpenStudy (mathmath333):
y-y1=m(x-x1)
y-y1=mx-mx1
y=mx-mx1+y1
OpenStudy (mathmath333):
-mx1+y1=c
OpenStudy (mathmath333):
u know what m is?
OpenStudy (ahsome):
The gradient?
OpenStudy (mathmath333):
how do u find it
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OpenStudy (mathmath333):
m=(y2-y1)/(x2-x1)
OpenStudy (ahsome):
\[m=\frac{y_2-y_1}{x_2-x_1}\]
OpenStudy (mathmath333):
put the points (10.5) and (4,5) in the equation
y=mx-mx1+y1
u will find y=5
OpenStudy (ahsome):
how did you get 10.5?
OpenStudy (mathmath333):
thats obvious through vision|dw:1412761094388:dw|
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OpenStudy (ahsome):
That's the issue. I need to do ALL of this algebraicly, without making a graph :(
OpenStudy (ikram002p):
never leave a question without graphing
OpenStudy (ahsome):
I am not allowed to for this question. I asked
OpenStudy (ikram002p):
--____--
w.e
OpenStudy (ahsome):
?
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OpenStudy (ahsome):
Still confused :(
OpenStudy (gorv):
shortest distance is perpendicular distance between from point to line
OpenStudy (gorv):
that is draw perpendicular (or u can say direct distance ) fromthe point on line