Distance between point and a line

**AshleyT** · May 8th 2011, 05:33 AM

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

Any help is appreciated , thanks.

**abhishekkgp** · May 8th 2011, 05:47 AM

you need to use the formula for the length of the perpendicular from a point and a line.

let the line be $ax+by+c=0$
let the point be $(x_1,y_1)$
the perpendicular's length is $|ax_1+by_1+c|/(\sqrt{a^2+b^2})$ .

**AshleyT** · May 8th 2011, 06:12 AM

Originally Posted by abhishekkgp

you need to use the formula for the length of the perpendicular from a point and a line.

let the line be $ax+by+c=0$
let the point be $(x_1,y_1)$
the perpendicular's length is $|ax_1+by_1+c|/(\sqrt{a^2+b^2})$ .

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

**abhishekkgp** · May 8th 2011, 07:59 AM

Originally Posted by AshleyT

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

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).??

**AshleyT** · May 8th 2011, 08:45 AM

Originally Posted by abhishekkgp

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).??

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 =(

**AshleyT** · May 8th 2011, 09:32 AM

And....isn't their answer wrong?
Shouldn't it not be -21? But -18?

**abhishekkgp** · May 8th 2011, 09:36 AM

Originally Posted by AshleyT

And....isn't their answer wrong?
Shouldn't it not be -21? But -18?

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 $t_0$ 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.

**Plato** · May 8th 2011, 09:40 AM

Originally Posted by AshleyT

And....isn't their answer wrong?
Shouldn't it not be -21? But -18?

First of all and most important, distance is never negative.

The answer is $\frac{18}{\sqrt{34}}.$

**AshleyT** · May 8th 2011, 12:00 PM

Originally Posted by Plato

First of all and most important, distance is never negative.

The answer is $\frac{18}{\sqrt{34}}.$

Why do they set their answer as -t0 ? And not t0? I still don't quite understand how they are finding the distance in terms of being perpendicular when their x + 5t, y + 3t is parrallel to the line, not perpendicular?

Thankyou

.

**abhishekkgp** · May 8th 2011, 12:15 PM

Originally Posted by AshleyT

Why do they set their answer as -t0 ? And not t0? I still don't quite understand how they are finding the distance in terms of being perpendicular when their x + 5t, y + 3t is parrallel to the line, not perpendicular?

Thankyou

.

its not $(x+5t,y+3t)$ . its $(1+5t,2+3t)$ . this is the parametric equation of a line. and this is perpendicular to the line given in question.
try eliminatinf $t$ in the following equations:
$x=1+5t$
$y=2+3t$ .
what do you get?

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