# Math Help - Finding the minimum distance from a point to a line-Having trouble

1. ## Finding the minimum distance from a point to a line-Having trouble

Hey!
I am having trouble finding the minimum distance from a point to a line. Here is the question:

For each of the following, find the distance from Q(3,-2) to each line.
a) 3x - 2y - 6 = 0

Solution
First I put into the form y = mx + b (because I am more familiar with this)
$y = \frac{3}{2}x - 3$

I now know that the slope of the line is $\frac {3}{2}$, which means that the line perpendicular to it has the slope of $\frac {-2}{3}$.
With this information I found the equation of a line with the slope $\frac{3}{-2}$. Which is $y = \frac{3}{-2}x + \frac{5}{2}$.

I then found where the equations $y = \frac{3}{-2}x + \frac{5}{2}$ and $y = \frac{3}{2}x - 3$ intersect by solving the system of equations. I got the point $(\frac{11}{6}, \frac{-1}{4})$.

So I now know the point on the equation closest to the point Q. So all I need to do is find the distance between them. So:
$\sqrt{((\frac{11}{6})(3))^2 + (-2)(\frac{1}{-4})^2}$
$= \sqrt{\frac{61}{2}}$
This is my final answer.

The book says the answer should be $\frac{7\sqrt{13}}{13}$.

I am doing exactly what the textbook is telling me to do. What am I doing wrong?

2. Originally Posted by Kakariki
Hey!
I am having trouble finding the minimum distance from a point to a line. Here is the question:

For each of the following, find the distance from Q(3,-2) to each line.
a) 3x - 2y - 6 = 0

Solution
First I put into the form y = mx + b (because I am more familiar with this)
$y = \frac{3}{2}x - 3$

I now know that the slope of the line is $\frac {3}{2}$, which means that the line perpendicular to it has the slope of $\frac {-2}{3}$. correct
With this information I found the equation of a line with the slope $\frac{3}{-2}$. Which is $y = \frac{3}{-2}x + \frac{5}{2}$. you changed your mind here though,

I then found where the equations $y = \frac{3}{-2}x + \frac{5}{2}$ and $y = \frac{3}{2}x - 3$ intersect by solving the system of equations. I got the point $(\frac{11}{6}, \frac{-1}{4})$.

So I now know the point on the equation closest to the point Q. So all I need to do is find the distance between them. So:
$\sqrt{((\frac{11}{6})(3))^2 + (-2)(\frac{1}{-4})^2}$
$= \sqrt{\frac{61}{2}}$
This is my final answer.

The book says the answer should be $\frac{7\sqrt{13}}{13}$.

I am doing exactly what the textbook is telling me to do. What am I doing wrong?

Bobak

3. Originally Posted by bobak
Bobak
Oh dear! What a silly mistake I made!!! Thank you for pointing it out.

4. Originally Posted by Kakariki
Hey!
I am having trouble finding the minimum distance from a point to a line. Here is the question:

For each of the following, find the distance from Q(3,-2) to each line.
a) 3x - 2y - 6 = 0

Solution
First I put into the form y = mx + b (because I am more familiar with this)
$y = \frac{3}{2}x - 3$

I now know that the slope of the line is $\frac {3}{2}$, which means that the line perpendicular to it has the slope of $\frac {-2}{3}$.
With this information I found the equation of a line with the slope $\frac{3}{-2}$. Which is $y = \frac{3}{-2}x + \frac{5}{2}$.

I then found where the equations $y = \frac{3}{-2}x + \frac{5}{2}$ and $y = \frac{3}{2}x - 3$ intersect by solving the system of equations. I got the point $(\frac{11}{6}, \frac{-1}{4})$.

So I now know the point on the equation closest to the point Q. So all I need to do is find the distance between them. So:
$\sqrt{((\frac{11}{6})(3))^2 + (-2)(\frac{1}{-4})^2}$
$= \sqrt{\frac{61}{2}}$
This is my final answer.

The book says the answer should be $\frac{7\sqrt{13}}{13}$.

I am doing exactly what the textbook is telling me to do. What am I doing wrong?
$3x - 2y - 6 = 0$

$2y = 3x - 6$

$y = \frac{3}{2}x - 3$.

So on this line, any point can be written as

$\left(x, \frac{3}{2}x - 3\right)$.

The distance from this arbitrary point to $Q(3, -2)$ is

$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$= \sqrt{\left(x - 3\right)^2 + \left[\frac{3}{2}x - 3 - (-2)\right]^2}$

$= \sqrt{(x - 3)^2 + \left(\frac{3}{2}x - 1\right)^2}$

$= \sqrt{x^2 - 6x + 9 + \frac{9}{4}x^2 - 3x + 1}$

$= \sqrt{\frac{13}{4}x^2 - 9x + 10}$

$= \left(\frac{13}{4}x^2 - 9x + 10\right)^{\frac{1}{2}}$

So to find the minimum distance, you need to differentiate the function, set it $= 0$ and solve for $x$.

Let $u = \frac{13}{4}x^2 - 9x + 10$ so that $D = u^{\frac{1}{2}}$.

$\frac{du}{dx} = \frac{13}{2}x - 9$.

$\frac{dD}{du} = \frac{1}{2}u^{-\frac{1}{2}}$

$= \frac{1}{2\sqrt{\frac{13}{4}x^2 - 9x + 10}}$.

So $\frac{dD}{dx} = \frac{\frac{13}{2}x - 9}{2\sqrt{\frac{13}{4}x^2 - 9x + 10}}$

$= \frac{13x - 18}{4\sqrt{\frac{13}{4}x^2 - 9x + 10}}$

$= \frac{13x - 18}{2\sqrt{13x^2 - 36x + 40}}$.

Setting this equal to $0$...

$\frac{13x - 18}{2\sqrt{13x^2 - 36x + 40}} = 0$

$13x - 18 = 0$

$13x = 18$

$x = \frac{18}{13}$.

When $x = \frac{18}{13}, y = \frac{3}{2}\left(\frac{18}{13}\right) - 3$

$= \frac{27}{13} - 3$

$= -\frac{12}{13}$.

So the point where the distance between the point $Q(3, -2)$ and the line is minimised is at $(x, y) = \left(\frac{18}{13}, -\frac{12}{13}\right)$. The distance between them is

$D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$= \sqrt{\left(\frac{18}{13} - 3\right)^2 + \left[-\frac{12}{13} - (-2)\right]^2}$

$= \sqrt{\left(-\frac{21}{13}\right)^2 + \left(\frac{14}{13}\right)^2}$

$= \sqrt{\frac{441}{169} + \frac{196}{169}}$

$= \sqrt{\frac{637}{169}}$

$= \frac{7\sqrt{13}}{13}$.