http://myasdcommunity.com/mathsquestion%202.jpeg

Any help is appreciated , thanks.

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- May 8th 2011, 05:33 AMAshleyTDistance between point and a line
http://myasdcommunity.com/mathsquestion%202.jpeg

Any help is appreciated , thanks. - May 8th 2011, 05:47 AMabhishekkgp
you need to use the formula for the length of the perpendicular from a point and a line.

let the line be $\displaystyle ax+by+c=0$

let the point be $\displaystyle (x_1,y_1)$

the perpendicular's length is $\displaystyle |ax_1+by_1+c|/(\sqrt{a^2+b^2})$. - May 8th 2011, 06:12 AMAshleyT
Thankyou very much for your reply :). Bit confused about the solutions given(not yours):

http://myasdcommunity.com/solutions%...o%20sense.jpeg

I don't understand how they get (x + 5t, y + 3t) ? If they are perpendicular, surely it should be (x - 3t, y + 5t) or something similar?

Thankyou - May 8th 2011, 07:59 AMabhishekkgp
their solution is a good one. i mean that's the derivation of the formula i posted. knowing the derivation is good. figuring out the derivation is even better. so i recommend that you should understand it.

i can help you understand that through a conversation. be sure to ask questions till you don't get the thing clearly.

here we go.

0) forget for a moment the solution they have given. we will come back to that later. produce your own solution.

1) our aim is to find the length of the perpendicular from (1,2) to the line 5x+3y+7=0.

2) now draw the point (1,2) and 5x+3y+7=0 on a graph paper or any regular paper but make the figure accurate enough.

3) now draw roughly a line*passing through (1,2) and perpendicular to 5x+3y+7=0*. call this line L'

4) L' will intersect 5x+3y+7=0 somewhere. (agreed?). Can you find the equation of line L'.??

5) call the point of intersection of these two lines as K.

6) Can you find K. further, can you find the distance between K and (1,2).?? - May 8th 2011, 08:45 AMAshleyT
Hey, thanks for the reply.

I was trying to use this method earlier.

I obtained L2 (perpendicular passing through (1, 2) as y = 3x/5 + 7/5)

I then re-arranged L1 and put L1 = L2 to obtain the point (-28/17, 7/17), call this point K.

Square root ((Kx - 1)^2 + (Ky - 2)^2)

Sorry, tried to use Latex but kept getting errors!

But i did not obtain the correct answer =( - May 8th 2011, 09:32 AMAshleyT
And....isn't their answer wrong?

Shouldn't it not be -21? But -18? - May 8th 2011, 09:36 AMabhishekkgp
yes!

i was about to post this. well you can also see that http://myasdcommunity.com/solutions%...o%20sense.jpeg is wrong. they have found out the parametric equation of the line. that's correct. but calculation of $\displaystyle t_0$ is wrong.*Our*answer matches with the formula i had posted in my first post. don't worry what you have done is correct.

reply. - May 8th 2011, 09:40 AMPlato
- May 8th 2011, 12:00 PMAshleyT
- May 8th 2011, 12:15 PMabhishekkgp
its not $\displaystyle (x+5t,y+3t)$. its $\displaystyle (1+5t,2+3t)$. this is the parametric equation of a line. and this

*is*perpendicular to the line given in question.

try eliminatinf $\displaystyle t$ in the following equations:

$\displaystyle x=1+5t$

$\displaystyle y=2+3t$.

what do you get?