Pre-Calculus - Find the Distance Between Two Parallel Lines

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

$\frac{2}{\sqrt{5}}$

$\frac{18}{\sqrt{34}}$

$\frac{42}{\sqrt{34}}$

$\frac{9}{\sqrt{34}}$

$\frac{9}{\sqrt{5}}$

Explanation

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.

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

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

$\frac{18}{\sqrt{13}}$

$\frac{30}{\sqrt{13}}$

$\frac{30}{\sqrt{40}}$

$\frac{26}{\sqrt{13}}$

$\frac{6}{\sqrt{13}}$

Explanation

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

$\frac{24}{\sqrt5}$

$\frac{30}{\sqrt5}$

$\frac{24}{\sqrt{52}}$

Explanation

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 $y = \frac{1}5 x +4$ and $y = \frac{1}{5}x - 3$

$\frac{35}{\sqrt{26}}$

$\frac{5}{\sqrt{26}}$

$\frac{5}{\sqrt{29}}$

$\frac{7}{\sqrt{29}}$

$\frac{7}{\sqrt{26}}$

Explanation

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.

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

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

$\frac{51}{5}$

$\frac{51}{\sqrt{45}}$

$\frac{30}{\sqrt{45}}$

Explanation

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 and

$\frac{4}{\sqrt5}$

$\frac{6}{\sqrt5}$

Explanation

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 the two lines.

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

Explanation

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: