Ask your own question, for FREE!
Mathematics 10 Online
OpenStudy (ahsome):

Find the shortest distance between a line and a point

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

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

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?

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?

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

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?

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

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|

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):

?

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

OpenStudy (gorv):

|dw:1412766145053:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!