Find the Distance Between a Point and a Line - Pre-Calculus

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Find the minimum distance between the point and the following line:

$f(x)=-$\frac{1}{3}$x+2$

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The minimum distance from the point to the line would be found by drawing a segment perpendicular to the line directly to the point. Our first step is to find the equation of the new line that connects the point to the line given in the problem. Because we know this new line is perpendicular to the line we're finding the distance to, we know its slope will be the negative inverse of the line its perpendicular to. So if the line we're finding the distance to is:

$f(x)=-$\frac{1}{3}$x+2$

Then its slope is -1/3, so the slope of a line perpendicular to it would be 3. Now that we know the slope of the line that will give the shortest distance from the point to the given line, we can plug the coordinates of our point into the formula for a line to get the full equation of the new line:

$-4=3(-2)+b\rightarrow b=2$

Now that we know the equation of our perpendicular line, our next step is to find the point where it intersects the line given in the problem:

$3x+2=-$\frac{1}{3}$x+2$

$$\frac{10}{3}$x=0\rightarrow x=0$

So if the lines intersect at x=0, we plug that value into either equation to find the y coordinate of the point where the lines intersect, which is the point on the line closest to the point given in the problem and therefore tells us the location of the minimum distance from the point to the line:

So we now know we want to find the distance between the following two points:

and

Using the following formula for the distance between two points, which we can see is just an application of the Pythagorean Theorem, we can plug in the values of our two points and calculate the shortest distance between the point and line given in the problem:

$d=$\sqrt{(x_2-x_$1)^2$+(y_2-y_$1)^2$}$$

Which we can then simplify by factoring the radical:

$\sqrt{40}$=$\sqrt{4\cdot 10}$=2$\sqrt{10}$

What is the shortest distance between the line and the origin?

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The shortest distance from a point to a line is always going to be along a path perpendicular to that line. To be perpendicular to our line, we need a slope of $-$\frac{1}{3}$$ .

To find the equation of our line, we can simply use point-slope form, using the origin, giving us

$y-0=\left(-$\frac{1}{3}\right)(x-0)$ which simplifies to $y=-$\frac{x}{3}$$ .

Now we want to know where this line intersects with our given line. We simply set them equal to each other, giving us $-$\frac{x}{3}$=3x-4$ .

If we multiply each side by , we get .

We can then add to each side, giving us .

Finally we divide by , giving us $x=\frac{6}{5}$$ .

This is the x-coordinate of their intersection. To find the y-coordinate, we plug $\frac{6}{5}$ into $y=-$\frac{x}{3}$$ , giving us $y=-$\frac{2}{5}$$ .

Therefore, our point of intersection must be $\left($\frac{6}{5}$,-$\frac{2}{5}\right)$ .

We then use the distance formula using $\left($\frac{6}{5}$,-$\frac{2}{5}\right)$ and the origin.

This give us $d=$\sqrt{\left(\frac{36}{25}\right)+\left($\frac{4}{25}\right)}=$\sqrt{\frac{40}{25}$}=\frac{2\sqrt{10}$}{5}$ .

Find the distance from point $\left ( 1,-$\frac{1}{2}\right )$ to the line .

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Draw a line that connects the point and intersects the line at a perpendicular angle.

The vertical distance from the point $\left ( 1,-$\frac{1}{2}\right)$ to the line will be the difference of the 2 y-values.

The distance can never be negative.

$5-\left(-$\frac{1}{2}\right) = $\frac{10}{2}$+\frac{1}{2}$=\frac{11}{2}$$

Find the distance between point $($\frac{1}{2}$,$\frac{1}{2}$)$ to the line $x=-$\frac{1}{2}$$ .

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Distance cannot be a negative number. The function $x=-$\frac{1}{2}$$ is a vertical line. Subtract the value of the line to the x-value of the given point to find the distance.

$D=\left | (-$\frac{1}{2}$)-$\frac{1}{2}$ \right | = \left | -1\right |=1$

Find the distance between point to line .

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The line is vertical covering the first and fourth quadrant on the coordinate plane.

The x-value of is negative one.

Find the perpendicular distance from the point to the line by subtracting the values of the line and the x-value of the point.

Distance cannot be negative.

$\left | 1-(-1)\right |=\left |2 \right |=2$

How far apart are the line $y = $\frac{ 1}{2 }$ x - 3$ and the point ?

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To find the distance, use the formula $distance = $$\frac{|ax_{0}$$$+by_{0}$+c $|}{$\sqrt{a^2$ + $b^2$ }$}$ where the point is and the line is

First, we'll re-write the equation $y = $\frac{1}{2}$ x - 3$ in this form to identify a, b, and c:

$y = $\frac{1}{2}$ x - 3$ subtract half x and add 3 to both sides

$y - $\frac{1}{2}$x + 3 = 0$ multiply both sides by 2

Now we see that . Plugging these plus into the formula, we get:

$distance = $\frac{|-1(2)+2(7)+6 $|}{\sqrt{(-1)^2$ +( $2)^2$ }$} = $\frac{|-2 + 14+6|}{\sqrt{1+4}$ } = $\frac{18}{\sqrt5}$$

Find the distance between and

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To find the distance, use the formula $distance = $$\frac{|ax_{0}$$$+by_{0}$+c $|}{$\sqrt{a^2$ + $b^2$ }$}$ where the point is and the line is

First, we'll re-write the equation in this form to identify , , and :

subtract and from both sides

Now we see that . Plugging these plus into the formula, we get:

$distance = $\frac{|-4(2)+1(-3)-1 $|}{\sqrt{(1)^2$ $+(-4)^2$ }$} = $\frac{|-8 -3-1|}{\sqrt{1+16}$ } = $\frac{12}{\sqrt{17}$}$

How far apart are the line $y = -$\frac{2}{3}$x + 8$ and the point ?

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To find the distance, use the formula $distance = $$\frac{|ax_{0}$$$+by_{0}$+c $|}{$\sqrt{a^2$ + $b^2$ }$}$ where the point is and the line is

First, we'll re-write the equation $y = -$\frac{2}{3}$ x +8$ in this form to identify a, b, and c:

$y = -$\frac{2}{3}$ x+8$ add $$\frac{2}{3}$x$ to and subtract 8 from both sides

$y + $\frac{2}{3}$x -8 = 0$ multiply both sides by 3

Now we see that . Plugging these plus into the formula, we get:

$distance = $\frac{|2(1)+3(3)-24 $|}{\sqrt{(3)^2$ +( $2)^2$ }$} = $\frac{|2 + 9-24|}{\sqrt{9+4}$ } = $\frac{13}{\sqrt{13}$}$

Find the distance between $y = -$\frac{4}{5}$x - 2$ and .

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To find the distance, use the formula $distance = $$\frac{|ax_{0}$$$+by_{0}$+c $|}{$\sqrt{a^2$ + $b^2$ }$}$ where the point is and the line is

First, we'll re-write the equation $y = -$\frac{4}{5}$ x -2$ in this form to identify , , and :

$y = -$\frac{4}{5}$ x-2$ add $$\frac{4}{5}$x$ and to both sides

$y + $\frac{4}{5}$x +2 = 0$ multiply both sides by

Now we see that . Plugging these plus into the formula, we get:

$distance = $\frac{|4(-3)+5(4)+10 $|}{\sqrt{(4)^2$ $+(5)^2$ }$} = $\frac{|-12+20+10|}{\sqrt{16+25}$ } = $\frac{18}{\sqrt{41}$}$

Find the distance between $y = $\frac{7}{2}$ x - 14$ and

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To find the distance, use the formula $distance = $$\frac{|ax_{0}$$$+by_{0}$+c $|}{$\sqrt{a^2$ + $b^2$ }$}$ where the point is and the line is

First, we'll re-write the equation $y = $\frac{7}{2}$ x -14$ in this form to identify , , and :

$y = $\frac{7}{2}$ x-14$ subtract $$\frac{7}{2}$x$ from and add to both sides

$y -$\frac{7}{2}$x +14 = 0$ multiply both sides by

Now we see that . Plugging these plus into the formula, we get:

$distance = $\frac{|-7(-6)+2(2)+28 $|}{\sqrt{(-7)^2$ +( $2)^2$ }$} = $\frac{|13 + 4+28|}{\sqrt{49+4}$ } = $\frac{45}{\sqrt{53}$}$

Find the distance between $y = $\frac{12}{5}$ x - 3$ and the point

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To find the distance, use the formula $distance = $$\frac{|ax_{0}$$$+by_{0}$+c $|}{$\sqrt{a^2$ + $b^2$ }$}$ where the point is and the line is

First, we'll re-write the equation $y = $\frac{12}{5}$ x-3$ in this form to identify , , and :

$y = $\frac{12}{5}$ x-3$ subtract $$\frac{12}{5}$x$ from and add to both sides

$y -$\frac{12}{5}$x +3 = 0$ multiply both sides by

Now we see that . Plugging these plus into the formula, we get:

$distance = $\frac{|-12(3)+5(-4)+15 $|}{\sqrt{(5)^2$ +( $12)^2$ }$} = $\frac{|-36 + -20+15|}{\sqrt{25+144}$ } = $\frac{41}{\sqrt{169}$} = $\frac{41}{13}$$

Find the minimum distance between the point and the following line:

$f(x)=-$\frac{1}{3}$x+2$

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The minimum distance from the point to the line would be found by drawing a segment perpendicular to the line directly to the point. Our first step is to find the equation of the new line that connects the point to the line given in the problem. Because we know this new line is perpendicular to the line we're finding the distance to, we know its slope will be the negative inverse of the line its perpendicular to. So if the line we're finding the distance to is:

$f(x)=-$\frac{1}{3}$x+2$

Then its slope is -1/3, so the slope of a line perpendicular to it would be 3. Now that we know the slope of the line that will give the shortest distance from the point to the given line, we can plug the coordinates of our point into the formula for a line to get the full equation of the new line:

$-4=3(-2)+b\rightarrow b=2$

Now that we know the equation of our perpendicular line, our next step is to find the point where it intersects the line given in the problem:

$3x+2=-$\frac{1}{3}$x+2$

$$\frac{10}{3}$x=0\rightarrow x=0$

So if the lines intersect at x=0, we plug that value into either equation to find the y coordinate of the point where the lines intersect, which is the point on the line closest to the point given in the problem and therefore tells us the location of the minimum distance from the point to the line:

So we now know we want to find the distance between the following two points:

and

Using the following formula for the distance between two points, which we can see is just an application of the Pythagorean Theorem, we can plug in the values of our two points and calculate the shortest distance between the point and line given in the problem:

$d=$\sqrt{(x_2-x_$1)^2$+(y_2-y_$1)^2$}$$

Which we can then simplify by factoring the radical:

$\sqrt{40}$=$\sqrt{4\cdot 10}$=2$\sqrt{10}$

What is the shortest distance between the line and the origin?

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The shortest distance from a point to a line is always going to be along a path perpendicular to that line. To be perpendicular to our line, we need a slope of $-$\frac{1}{3}$$ .

To find the equation of our line, we can simply use point-slope form, using the origin, giving us

$y-0=\left(-$\frac{1}{3}\right)(x-0)$ which simplifies to $y=-$\frac{x}{3}$$ .

Now we want to know where this line intersects with our given line. We simply set them equal to each other, giving us $-$\frac{x}{3}$=3x-4$ .

If we multiply each side by , we get .

We can then add to each side, giving us .

Finally we divide by , giving us $x=\frac{6}{5}$$ .

This is the x-coordinate of their intersection. To find the y-coordinate, we plug $\frac{6}{5}$ into $y=-$\frac{x}{3}$$ , giving us $y=-$\frac{2}{5}$$ .

Therefore, our point of intersection must be $\left($\frac{6}{5}$,-$\frac{2}{5}\right)$ .

We then use the distance formula using $\left($\frac{6}{5}$,-$\frac{2}{5}\right)$ and the origin.

This give us $d=$\sqrt{\left(\frac{36}{25}\right)+\left($\frac{4}{25}\right)}=$\sqrt{\frac{40}{25}$}=\frac{2\sqrt{10}$}{5}$ .

Find the distance from point $\left ( 1,-$\frac{1}{2}\right )$ to the line .

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Draw a line that connects the point and intersects the line at a perpendicular angle.

The vertical distance from the point $\left ( 1,-$\frac{1}{2}\right)$ to the line will be the difference of the 2 y-values.

The distance can never be negative.

$5-\left(-$\frac{1}{2}\right) = $\frac{10}{2}$+\frac{1}{2}$=\frac{11}{2}$$

Find the distance between point $($\frac{1}{2}$,$\frac{1}{2}$)$ to the line $x=-$\frac{1}{2}$$ .

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Distance cannot be a negative number. The function $x=-$\frac{1}{2}$$ is a vertical line. Subtract the value of the line to the x-value of the given point to find the distance.

$D=\left | (-$\frac{1}{2}$)-$\frac{1}{2}$ \right | = \left | -1\right |=1$

Find the distance between point to line .

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The line is vertical covering the first and fourth quadrant on the coordinate plane.

The x-value of is negative one.

Find the perpendicular distance from the point to the line by subtracting the values of the line and the x-value of the point.

Distance cannot be negative.

$\left | 1-(-1)\right |=\left |2 \right |=2$

How far apart are the line $y = $\frac{ 1}{2 }$ x - 3$ and the point ?

Tap to see back →

To find the distance, use the formula $distance = $$\frac{|ax_{0}$$$+by_{0}$+c $|}{$\sqrt{a^2$ + $b^2$ }$}$ where the point is and the line is

First, we'll re-write the equation $y = $\frac{1}{2}$ x - 3$ in this form to identify a, b, and c:

$y = $\frac{1}{2}$ x - 3$ subtract half x and add 3 to both sides

$y - $\frac{1}{2}$x + 3 = 0$ multiply both sides by 2

Now we see that . Plugging these plus into the formula, we get:

$distance = $\frac{|-1(2)+2(7)+6 $|}{\sqrt{(-1)^2$ +( $2)^2$ }$} = $\frac{|-2 + 14+6|}{\sqrt{1+4}$ } = $\frac{18}{\sqrt5}$$

Find the distance between and

Tap to see back →

To find the distance, use the formula $distance = $$\frac{|ax_{0}$$$+by_{0}$+c $|}{$\sqrt{a^2$ + $b^2$ }$}$ where the point is and the line is

First, we'll re-write the equation in this form to identify , , and :

subtract and from both sides

Now we see that . Plugging these plus into the formula, we get:

$distance = $\frac{|-4(2)+1(-3)-1 $|}{\sqrt{(1)^2$ $+(-4)^2$ }$} = $\frac{|-8 -3-1|}{\sqrt{1+16}$ } = $\frac{12}{\sqrt{17}$}$

How far apart are the line $y = -$\frac{2}{3}$x + 8$ and the point ?

Tap to see back →

To find the distance, use the formula $distance = $$\frac{|ax_{0}$$$+by_{0}$+c $|}{$\sqrt{a^2$ + $b^2$ }$}$ where the point is and the line is

First, we'll re-write the equation $y = -$\frac{2}{3}$ x +8$ in this form to identify a, b, and c:

$y = -$\frac{2}{3}$ x+8$ add $$\frac{2}{3}$x$ to and subtract 8 from both sides

$y + $\frac{2}{3}$x -8 = 0$ multiply both sides by 3

Now we see that . Plugging these plus into the formula, we get:

$distance = $\frac{|2(1)+3(3)-24 $|}{\sqrt{(3)^2$ +( $2)^2$ }$} = $\frac{|2 + 9-24|}{\sqrt{9+4}$ } = $\frac{13}{\sqrt{13}$}$

Find the distance between $y = -$\frac{4}{5}$x - 2$ and .

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To find the distance, use the formula $distance = $$\frac{|ax_{0}$$$+by_{0}$+c $|}{$\sqrt{a^2$ + $b^2$ }$}$ where the point is and the line is

First, we'll re-write the equation $y = -$\frac{4}{5}$ x -2$ in this form to identify , , and :

$y = -$\frac{4}{5}$ x-2$ add $$\frac{4}{5}$x$ and to both sides

$y + $\frac{4}{5}$x +2 = 0$ multiply both sides by

Now we see that . Plugging these plus into the formula, we get:

$distance = $\frac{|4(-3)+5(4)+10 $|}{\sqrt{(4)^2$ $+(5)^2$ }$} = $\frac{|-12+20+10|}{\sqrt{16+25}$ } = $\frac{18}{\sqrt{41}$}$