# Thread: distance of a point from a line

1. ## distance of a point from a line

Hello !! This is a simple problem, but I'm not getting the right results. I am getting 9/5 but the text says that the correct answer is 13/5.

Problem:
Determine the distance of the point (2,3) from the line joining the points (-3,-4) and (1,-1).

Are you getting the same answer as mine ?

2. I get 13/5.

Using vector arithmetic
(1,-1) - (-3,-4) = (4,3) perpendicular is (-3,4)
(2,3) - (1,-1) = (1,4)

(-3,4)*(1,4) / 5 = (-3 + 16) / 5 = 13/5

3. Did you try it on graph paper...just to see?!

4. I assume you are trying to find the minimum distance from that point to the line.

First, find the equation of the line:

$\displaystyle (x_1,y_1) = (-3,-4)$ and $\displaystyle (x_2, y_2) = (1,-1)$.

So $\displaystyle m = \frac{y_2-y_1}{x_2-x_1} = \frac{-1-(-4)}{1-(-3)} = \frac{3}{4}$ and

$\displaystyle y - y_1 = m(x-x_1)$

$\displaystyle y - (-4) = \frac{3}{4}[x -(-3)]$

$\displaystyle y + 4 = \frac{3}{4}(x+3)$

$\displaystyle y + \frac{16}{4} = \frac{3}{4}x + \frac{9}{4}$

$\displaystyle y = \frac{3}{4}x - \frac{7}{4}$.

The minimum distance from the point $\displaystyle (2,3)$ to the line will occur when that distance is perpendicular to the line.

Drawing this distance gives another straight line. Since it is perpendicular to the other line, their gradients multiply to give $\displaystyle -1$.

So $\displaystyle \frac{3}{4}m = -1$

$\displaystyle m = -\frac{4}{3}$.

You know that $\displaystyle (2,3)$ lies on your new line, so

$\displaystyle y - 3 = -\frac{4}{3}(x - 2)$

$\displaystyle y - \frac{9}{3} = -\frac{4}{3}x + \frac{8}{3}$

$\displaystyle y = -\frac{4}{3}x + \frac{17}{3}$.

Now to find where the point where the two lines cross:

$\displaystyle \frac{3}{4}x - \frac{7}{4} = -\frac{4}{3}x + \frac{17}{3}$

$\displaystyle 3(3x - 7) = 4(-4x + 17)$

$\displaystyle 9x - 21 = -16x + 68$

$\displaystyle 25x = 89$

$\displaystyle x = \frac{89}{25}$.

When $\displaystyle x = \frac{89}{25}$,

$\displaystyle y = \frac{3}{4}\left(\frac{89}{25}\right) - \frac{7}{4}$

$\displaystyle = \frac{267}{100} - \frac{175}{100}$

$\displaystyle = \frac{23}{25}$.

So the two lines cross at $\displaystyle (x,y) = \left(\frac{89}{25}, \frac{23}{25}\right)$.

Now all that's left is to find the distance between $\displaystyle (2, 3)$ and $\displaystyle \left(\frac{89}{25}, \frac{23}{25}\right)$.

$\displaystyle D = \sqrt{\left(\frac{89}{25} - 2\right)^2 + \left(\frac{23}{25} - 3\right)^2}$

$\displaystyle = \sqrt{\left(\frac{39}{25}\right)^2 + \left(-\frac{52}{25}\right)^2}$

$\displaystyle = \sqrt{\frac{1521}{625} + \frac{2704}{625}}$

$\displaystyle = \sqrt{\frac{4225}{625}}$

$\displaystyle = \frac{13}{5}$.

So the minimum distance from the line joining $\displaystyle (-3,-4)$ and $\displaystyle (1,-1)$ to the point $\displaystyle (2,3)$ is $\displaystyle \frac{13}{5}$ units.

5. i see. got it now. just didn't analyzed the question well. thanks guys !!!