Find the Distance Between Two Parallel Lines - Pre-Calculus

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Find the distance between the two lines.

$\ y_1=2.65x-13 \ y_2=2.65x+4$

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Since the slope of the two lines are equivalent, we know that the lines are parallel. Therefore, they are separated by a constant distance. We can then find the distance between the two lines by using the formula for the distance from a point to a nonvertical line:

$d=\frac{|Ax_1 +By_$1+C|}{\sqrt{A^{2}$$$+B^{2}$}}$

First, we need to take one of the line and convert it to standard form.

where $A=2.65,B=-1, \textup{and }C=-13$

Now we can substitute A, B, and C into our distance equation along with a point, $\left( x_1,y_1\right)$ , from the other line. We can pick any point we want, as long as it is on line . Just plug in a number for x, and solve for y. I will use the y-intercept, where x = 0, because it is easy to calculate:

Now we have a point, , that is on the line . So let's plug our values for $x_1,y_1,A,B,\textup{and }C}$ :

$d=\frac{|(2.65)\cdot (0) +(-1)\cdot $(4)+(-13)|}{\sqrt{(2.65)^{2}$$$+(-1)^{2}$}}=\frac{|0-4-13|}{\sqrt{(7.0225+1)}$}=\frac{|-17|}{\sqrt{8.0225}$}$

$=\frac{17}{2.83}$=6.002\approx6$

Find the distance between and

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To find the distance, choose any point on one of the lines. Plugging in 2 into the first equation can generate our first point:

this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

subtract the whole right side from both sides

now we see that

We can plug the coefficients and the point into the formula

$distance = $\frac{|ax + $by+c|}{\sqrt{a^2$ + $b^2$ }$}$ where represents the point.

$$\frac{|-2(2) + 1(1)-1| }{ $\sqrt{1^2$ + $2^2$ }$} = $\frac{|-4+1-1|}{\sqrt{1+4}$} = $\frac{4}{\sqrt{5}$}$

Find the distance between $y = $\frac{1}$5 x +4$ and $y = $\frac{1}{5}$x - 3$

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To find the distance, choose any point on one of the lines. Plugging in into the second equation can generate our first point:

$y = $\frac{1}{5}$(5) - 3 = 1 -3 = -2$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y = $\frac{1}{5}$x+4$ subtract the whole right side from both sides

$- $\frac{1}{5}$x + y - 4 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = $\frac{|ax + $by+c|}{\sqrt{a^2$ + $b^2$ }$}$ where represents the point.

$$\frac{|-1(5) + 5(-2)-20| }{ $\sqrt{1^2$ + $5^2$ }$} = $\frac{|-5-10-20|}{\sqrt{1+25}$} = $\frac{35}{\sqrt{26}$}$

How far apart are the lines $y = -$\frac{3}{2}$x -3$ and $y = -$\frac{3}{2}$x + 6$ ?

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To find the distance, choose any point on one of the lines. Plugging in into the first equation can generate our first point:

$y = -$\frac{3}{2}$(2) - 3 = -3 - 3 =-6$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y =- $\frac{3}{2}$x+6$ subtract the whole right side from both sides

$$\frac{3}{2}$x + y - 6 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = $\frac{|ax + $by+c|}{\sqrt{a^2$ + $b^2$ }$}$ where represents the point.

$$\frac{|3(2) + 2(-6)-12| }{ $\sqrt{3^2$ + $2^2$ }$} = $\frac{|6-12-12|}{\sqrt{9+4}$} = $\frac{18}{\sqrt{13}$}$

Find the distance between $y = $\frac{4}{3}$x +12$ and $y = $\frac{4}{3}$x + 2$

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To find the distance, choose any point on one of the lines. Plugging in into the second equation can generate our first point:

$y = $\frac{4}{3}$(3) + 2 = 4 + 2 = 6$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y = $\frac{4}{3}$x + 12$ subtract the whole right side from both sides

$- $\frac{4}{3}$x + y - 12 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = $\frac{|ax + $by+c|}{\sqrt{a^2$ + $b^2$ }$}$ where represents the point.

$$\frac{|-4(3) +3(6)-36| }{ $\sqrt{(-4)^2$ + $3^2$ }$} = $\frac{|-12+18-36|}{\sqrt{16+9}$} = $\frac{30}{\sqrt{25}$} = $\frac{30}{5}$ = 6$

Find the distance between $y = -$\frac{1}{2}$x + 7$ and $y = -$\frac{1}{2}$x + 4$

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To find the distance, choose any point on one of the lines. Plugging in into the first equation can generate our first point:

$y = -$\frac{1}{2}$(4)+7 = -2 + 7 = 5$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y =-$\frac{1}{2}$x + 4$ subtract the whole right side from both sides

$- $\frac{1}{2}$x + y - 4 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = $\frac{|ax + $by+c|}{\sqrt{a^2$ + $b^2$ }$}$ where represents the point.

$$\frac{|1(4) + -2(5)+8| }{ $\sqrt{1^2$ + $(-2)^2$ }$} = $\frac{|4-10+8|}{\sqrt{5}$} = $\frac{2}{\sqrt{5}$}$

Find the distance between the lines $y = $\frac{1}{2}$ x - 11$ and $y = $\frac{1}{2}$ x + 1$

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To find the distance, choose any point on one of the lines. Plugging in into the first equation can generate our first point:

$y = $\frac{1}{2}$(6) +1 = 3+1= 4$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y = $\frac{1}{2}$ x - 11$ subtract the whole right side from both sides

$-$\frac{1}{2}$ x + y + 11 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = $\frac{|ax + $by+c|}{\sqrt{a^2$ + $b^2$ }$}$ where represents the point.

$$\frac{|-1(6) + 2(4)+22| }{ $\sqrt{(-1)^2$ + $2^2$ }$} = $\frac{|-6+8+22|}{\sqrt{1+4}$} = $\frac{24}{\sqrt{5}$}$

Find the distance between the two lines.

$\ y_1=2.65x-13 \ y_2=2.65x+4$

Tap to see back →

Since the slope of the two lines are equivalent, we know that the lines are parallel. Therefore, they are separated by a constant distance. We can then find the distance between the two lines by using the formula for the distance from a point to a nonvertical line:

$d=\frac{|Ax_1 +By_$1+C|}{\sqrt{A^{2}$$$+B^{2}$}}$

First, we need to take one of the line and convert it to standard form.

where $A=2.65,B=-1, \textup{and }C=-13$

Now we can substitute A, B, and C into our distance equation along with a point, $\left( x_1,y_1\right)$ , from the other line. We can pick any point we want, as long as it is on line . Just plug in a number for x, and solve for y. I will use the y-intercept, where x = 0, because it is easy to calculate:

Now we have a point, , that is on the line . So let's plug our values for $x_1,y_1,A,B,\textup{and }C}$ :

$d=\frac{|(2.65)\cdot (0) +(-1)\cdot $(4)+(-13)|}{\sqrt{(2.65)^{2}$$$+(-1)^{2}$}}=\frac{|0-4-13|}{\sqrt{(7.0225+1)}$}=\frac{|-17|}{\sqrt{8.0225}$}$

$=\frac{17}{2.83}$=6.002\approx6$

Find the distance between and

Tap to see back →

To find the distance, choose any point on one of the lines. Plugging in 2 into the first equation can generate our first point:

this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

subtract the whole right side from both sides

now we see that

We can plug the coefficients and the point into the formula

$distance = $\frac{|ax + $by+c|}{\sqrt{a^2$ + $b^2$ }$}$ where represents the point.

$$\frac{|-2(2) + 1(1)-1| }{ $\sqrt{1^2$ + $2^2$ }$} = $\frac{|-4+1-1|}{\sqrt{1+4}$} = $\frac{4}{\sqrt{5}$}$

Find the distance between $y = $\frac{1}$5 x +4$ and $y = $\frac{1}{5}$x - 3$

Tap to see back →

To find the distance, choose any point on one of the lines. Plugging in into the second equation can generate our first point:

$y = $\frac{1}{5}$(5) - 3 = 1 -3 = -2$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y = $\frac{1}{5}$x+4$ subtract the whole right side from both sides

$- $\frac{1}{5}$x + y - 4 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = $\frac{|ax + $by+c|}{\sqrt{a^2$ + $b^2$ }$}$ where represents the point.

$$\frac{|-1(5) + 5(-2)-20| }{ $\sqrt{1^2$ + $5^2$ }$} = $\frac{|-5-10-20|}{\sqrt{1+25}$} = $\frac{35}{\sqrt{26}$}$

How far apart are the lines $y = -$\frac{3}{2}$x -3$ and $y = -$\frac{3}{2}$x + 6$ ?

Tap to see back →

To find the distance, choose any point on one of the lines. Plugging in into the first equation can generate our first point:

$y = -$\frac{3}{2}$(2) - 3 = -3 - 3 =-6$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y =- $\frac{3}{2}$x+6$ subtract the whole right side from both sides

$$\frac{3}{2}$x + y - 6 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = $\frac{|ax + $by+c|}{\sqrt{a^2$ + $b^2$ }$}$ where represents the point.

$$\frac{|3(2) + 2(-6)-12| }{ $\sqrt{3^2$ + $2^2$ }$} = $\frac{|6-12-12|}{\sqrt{9+4}$} = $\frac{18}{\sqrt{13}$}$

Find the distance between $y = $\frac{4}{3}$x +12$ and $y = $\frac{4}{3}$x + 2$

Tap to see back →

To find the distance, choose any point on one of the lines. Plugging in into the second equation can generate our first point:

$y = $\frac{4}{3}$(3) + 2 = 4 + 2 = 6$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y = $\frac{4}{3}$x + 12$ subtract the whole right side from both sides

$- $\frac{4}{3}$x + y - 12 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = $\frac{|ax + $by+c|}{\sqrt{a^2$ + $b^2$ }$}$ where represents the point.

$$\frac{|-4(3) +3(6)-36| }{ $\sqrt{(-4)^2$ + $3^2$ }$} = $\frac{|-12+18-36|}{\sqrt{16+9}$} = $\frac{30}{\sqrt{25}$} = $\frac{30}{5}$ = 6$

Find the distance between $y = -$\frac{1}{2}$x + 7$ and $y = -$\frac{1}{2}$x + 4$

Tap to see back →

To find the distance, choose any point on one of the lines. Plugging in into the first equation can generate our first point:

$y = -$\frac{1}{2}$(4)+7 = -2 + 7 = 5$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y =-$\frac{1}{2}$x + 4$ subtract the whole right side from both sides

$- $\frac{1}{2}$x + y - 4 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = $\frac{|ax + $by+c|}{\sqrt{a^2$ + $b^2$ }$}$ where represents the point.

$$\frac{|1(4) + -2(5)+8| }{ $\sqrt{1^2$ + $(-2)^2$ }$} = $\frac{|4-10+8|}{\sqrt{5}$} = $\frac{2}{\sqrt{5}$}$

Find the distance between the lines $y = $\frac{1}{2}$ x - 11$ and $y = $\frac{1}{2}$ x + 1$

Tap to see back →

To find the distance, choose any point on one of the lines. Plugging in into the first equation can generate our first point:

$y = $\frac{1}{2}$(6) +1 = 3+1= 4$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y = $\frac{1}{2}$ x - 11$ subtract the whole right side from both sides

$-$\frac{1}{2}$ x + y + 11 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = $\frac{|ax + $by+c|}{\sqrt{a^2$ + $b^2$ }$}$ where represents the point.

$$\frac{|-1(6) + 2(4)+22| }{ $\sqrt{(-1)^2$ + $2^2$ }$} = $\frac{|-6+8+22|}{\sqrt{1+4}$} = $\frac{24}{\sqrt{5}$}$

Find the distance between the two lines.

$\ y_1=2.65x-13 \ y_2=2.65x+4$

Tap to see back →

Since the slope of the two lines are equivalent, we know that the lines are parallel. Therefore, they are separated by a constant distance. We can then find the distance between the two lines by using the formula for the distance from a point to a nonvertical line:

$d=\frac{|Ax_1 +By_$1+C|}{\sqrt{A^{2}$$$+B^{2}$}}$

First, we need to take one of the line and convert it to standard form.

where $A=2.65,B=-1, \textup{and }C=-13$

Now we can substitute A, B, and C into our distance equation along with a point, $\left( x_1,y_1\right)$ , from the other line. We can pick any point we want, as long as it is on line . Just plug in a number for x, and solve for y. I will use the y-intercept, where x = 0, because it is easy to calculate:

Now we have a point, , that is on the line . So let's plug our values for $x_1,y_1,A,B,\textup{and }C}$ :

$d=\frac{|(2.65)\cdot (0) +(-1)\cdot $(4)+(-13)|}{\sqrt{(2.65)^{2}$$$+(-1)^{2}$}}=\frac{|0-4-13|}{\sqrt{(7.0225+1)}$}=\frac{|-17|}{\sqrt{8.0225}$}$

$=\frac{17}{2.83}$=6.002\approx6$

Find the distance between and

Tap to see back →

To find the distance, choose any point on one of the lines. Plugging in 2 into the first equation can generate our first point:

this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

subtract the whole right side from both sides

now we see that

We can plug the coefficients and the point into the formula

$distance = $\frac{|ax + $by+c|}{\sqrt{a^2$ + $b^2$ }$}$ where represents the point.

$$\frac{|-2(2) + 1(1)-1| }{ $\sqrt{1^2$ + $2^2$ }$} = $\frac{|-4+1-1|}{\sqrt{1+4}$} = $\frac{4}{\sqrt{5}$}$

Find the distance between $y = $\frac{1}$5 x +4$ and $y = $\frac{1}{5}$x - 3$

Tap to see back →

To find the distance, choose any point on one of the lines. Plugging in into the second equation can generate our first point:

$y = $\frac{1}{5}$(5) - 3 = 1 -3 = -2$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y = $\frac{1}{5}$x+4$ subtract the whole right side from both sides

$- $\frac{1}{5}$x + y - 4 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = $\frac{|ax + $by+c|}{\sqrt{a^2$ + $b^2$ }$}$ where represents the point.

$$\frac{|-1(5) + 5(-2)-20| }{ $\sqrt{1^2$ + $5^2$ }$} = $\frac{|-5-10-20|}{\sqrt{1+25}$} = $\frac{35}{\sqrt{26}$}$

How far apart are the lines $y = -$\frac{3}{2}$x -3$ and $y = -$\frac{3}{2}$x + 6$ ?

Tap to see back →

To find the distance, choose any point on one of the lines. Plugging in into the first equation can generate our first point:

$y = -$\frac{3}{2}$(2) - 3 = -3 - 3 =-6$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y =- $\frac{3}{2}$x+6$ subtract the whole right side from both sides

$$\frac{3}{2}$x + y - 6 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = $\frac{|ax + $by+c|}{\sqrt{a^2$ + $b^2$ }$}$ where represents the point.

$$\frac{|3(2) + 2(-6)-12| }{ $\sqrt{3^2$ + $2^2$ }$} = $\frac{|6-12-12|}{\sqrt{9+4}$} = $\frac{18}{\sqrt{13}$}$

Find the distance between $y = $\frac{4}{3}$x +12$ and $y = $\frac{4}{3}$x + 2$

Tap to see back →

To find the distance, choose any point on one of the lines. Plugging in into the second equation can generate our first point:

$y = $\frac{4}{3}$(3) + 2 = 4 + 2 = 6$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y = $\frac{4}{3}$x + 12$ subtract the whole right side from both sides

$- $\frac{4}{3}$x + y - 12 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = $\frac{|ax + $by+c|}{\sqrt{a^2$ + $b^2$ }$}$ where represents the point.

$$\frac{|-4(3) +3(6)-36| }{ $\sqrt{(-4)^2$ + $3^2$ }$} = $\frac{|-12+18-36|}{\sqrt{16+9}$} = $\frac{30}{\sqrt{25}$} = $\frac{30}{5}$ = 6$

Find the distance between $y = -$\frac{1}{2}$x + 7$ and $y = -$\frac{1}{2}$x + 4$

Tap to see back →

To find the distance, choose any point on one of the lines. Plugging in into the first equation can generate our first point:

$y = -$\frac{1}{2}$(4)+7 = -2 + 7 = 5$ this gives us the point

We can find the distance between this point and the other line by putting the second line into the form :

$y =-$\frac{1}{2}$x + 4$ subtract the whole right side from both sides

$- $\frac{1}{2}$x + y - 4 = 0$ multiply both sides by

now we see that

We can plug the coefficients and the point into the formula

$distance = $\frac{|ax + $by+c|}{\sqrt{a^2$ + $b^2$ }$}$ where represents the point.

$$\frac{|1(4) + -2(5)+8| }{ $\sqrt{1^2$ + $(-2)^2$ }$} = $\frac{|4-10+8|}{\sqrt{5}$} = $\frac{2}{\sqrt{5}$}$