# Thread: Distance between point and a line

1. ## Distance between point and a line

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

Any help is appreciated , thanks.

2. 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 $\displaystyle |ax_1+by_1+c|/(\sqrt{a^2+b^2})$.

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

4. Originally Posted by AshleyT
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).??

5. 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).??
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 =(

Shouldn't it not be -21? But -18?

7. Originally Posted by AshleyT
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 $\displaystyle t_0$ is wrong. Our answer matches with the formula i had posted in my first post. don't worry what you have done is correct.

8. Originally Posted by AshleyT
Shouldn't it not be -21? But -18?
First of all and most important, distance is never negative.

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

9. Originally Posted by Plato
First of all and most important, distance is never negative.

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

10. 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 $\displaystyle (x+5t,y+3t)$. 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?