# Thread: Distance from a point to a line

1. ## Distance from a point to a line

Find the distance from P to L:

A. P(2,1) L: y = x + 2
B. P(4,6) L: 4x + 3y = 12

If I can get some guidance on these and steps to follow I should be able to do the rest of the problems. The books explanation on how to do these is a little fuzzy to me.

2. Originally Posted by VitaX
Find the distance from P to L:

A. P(2,1) L: y = x + 2
B. P(4,6) L: 4x + 3y = 12

If I can get some guidance on these and steps to follow I should be able to do the rest of the problems. The books explanation on how to do these is a little fuzzy to me.
There is a long way to get the result - and of course a shortcut.

A. Long way:
1. Calculate the equation of the perpendicular line to the line L: $\displaystyle y = -x+3$
2. Calculate the coordinates of the point of intersection of L and the normal to L: $\displaystyle Q\left(\dfrac12\ ,\ \dfrac52\right)$
3. Calculate the distance $\displaystyle \overline{QP} = \dfrac32 \sqrt{2}$

B. Shortcut:

Let $\displaystyle P(x_P, y_P)$ denote a point and $\displaystyle L:Ax+By+C=0$ a straight line then the distance d of P from L is calculated by:

$\displaystyle d(P,L)=\dfrac{Ax_P + By_P + C}{\sqrt{A^2+B^2}}$

With your problem: The line $\displaystyle L:y = x+2~\implies~-x+y-2=0$

Plug in the values of A, B, C and the coordinates of P and you'll get the same result.

I'll leave the second example for you.

3. Thanks for the help. I got the same answer as you did using the long method. I'll try the other one now.

Edit: For B. I get d = 4.4 units. Correct?

4. Originally Posted by VitaX
Thanks for the help. I got the same answer as you did using the long method. I'll try the other one now.

Edit: For B. I get d = 4.4 units. <<<< perfect!
...

5. I ran into some difficult problems in this section of the homework. The reason they are difficult is because the problems are all variables. So trying to solve them is more difficult (to me atleast)

Find the distance from P to L:

P(a,b) L: x = -1 Edit : I get d = a + 1 ... correct?
P(x1,y1) L: Ax + By = C

How would I go about solving these two?

6. Originally Posted by VitaX
I ran into some difficult problems in this section of the homework. The reason they are difficult is because the problems are all variables. So trying to solve them is more difficult (to me atleast)

Find the distance from P to L:

P(a,b) L: x = -1 Edit : I get d = a + 1 ... correct? <<<<<<< Yes
P(x1,y1) L: Ax + By = C

How would I go about solving these two?
In both cases you can use the formula I've posted in my previous post:

$\displaystyle L: x = -1~\implies~1\cdot x + 0\cdot y + 1 = 0$

Thus:

$\displaystyle d(P(a,b),L)=\dfrac{1 \cdot a+0\cdot b + 1}{\sqrt{1^2+0^2}}=a+1$

And the 2nd problem yields nearly the formula which I posted in post#2. (There is a slight difference with the constant C)