How To Find The Distance Between 2 Lines

In this article, we volition discuss how to find the distance between 2 parallel and skew lines.

The shortest distance between ii lines refer to how far away two lines are located from each other. In other words, we tin can say that the shortest distance between two lines in a plane is the minimum distance between whatsoever two points that are present on both the lines. The shortest distance is the measure of the length of a perpendicular line between two lines.

In geometry, we come across different lines such as parallel lines and skew lines. Nosotros have formulas to summate the distance betwixt 2 parallel and skew lines. In this commodity, we will focus on using formulas to find the minimum distance between these types of lines.

Distance Between Parallel Lines

What are parallel lines?

Two lines that are parallel never meet each other and they take the same slopes.

How to find the distance between ii parallel lines?

Before proceeding to discuss the method to observe the distance between 2 parallel lines, let us first run across what is meant by the distance.

It is the length of the perpendicular line that stretches from any point on 1 line to the other one.

To detect the distance between two parallel lines, you should follow the following method:

The formula for distance between two parallel lines is given below if the lines are in the slope intercept class. Remember that the slope intercept form of the line is $y = mx + c$ .

$d = \frac {|c_2 - c_1|} {\sqrt{1 + m^2}}$

Hither, $c_1$ is the constant of line ane and $c_2$ is the constant of line 2. $m$ represents the gradient of

If the equations of the parallel lines are given in the $ax + by + d$ , then the formula is slightly unlike than the first ane:

$d = \frac {|d_2 - d_1|} {\sqrt{a^2 + b^2}}$

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Let's become

Example i

Observe the distance between the parallel lines having the following equations:

Line 1 : $y = 5x + 5$

Line 2 : $y = 5x - 7$

Solution

These lines have the same slopes, so we volition apply the post-obit formula to compute the distance between them:

$d = \frac {|c_2 - c_1|} {\sqrt{1 + m^2}}$

Here, $c_1 = 5$ , $c_2 = -7$ , and $m = 5$ . Substitute these values in the to a higher place formula:

$d = \frac {|-7 - 5|} {\sqrt{1 + 5^2}}$

$d = \frac {|-12|} {\sqrt{26}}$

$d = \frac {|-12|} {5.09} = 2.35$ units

Hence, the distance between two lines is two.35 units.

Example 2

Find the altitude between the parallel lines having the following equations:

Line one : $y = 3x + 2$

Line 2 : $y = 3x - 1$

Solution

These lines accept the aforementioned slopes, so nosotros will utilise the post-obit formula to compute the distance between them:

$d = \frac {|c_2 - c_1|} {\sqrt{1 + m^2}}$

Here, $c_1 = 2$ , $c_2 = -1$ , and $m = 3$ . Substitute these values in the above formula:

$d = \frac {|-1 - 2|} {\sqrt{1 + 3^2}}$

$d = \frac {|-3|} {\sqrt{10}}$

$d = \frac {3} {3.16} = 0.94$ units

Hence, the altitude between 2 lines is 0.94 units.

Example iii

Find the distance between the following two parallel lines:

Line one : $4x + 2y + 5$

Line ii : $4x + 2y + 9$

Solution

Since the equations are given in the class $ax + by + d$ , hence we will employ the following formula to compute the distance:

$d = \frac {|d_2 - d_1|} {\sqrt{a^2 + b^2}}$

Hither, $d_1 = 5$ , $d_2 = 9$ , $a = 4$ and $b = 2$ . Substitute these values in the higher up formula to get the distance:

$d = \frac {|9 - 5|} {\sqrt{4^2 + 2^2}}$

$d = \frac {|4|} {\sqrt{20}}$

$d = \frac {|4|} {4.47}$

$d = 0.89$ units

Hence, the distance between these lines is 0.89 units.

Case 4

Observe the distance between the following two parallel lines:

Line 1 : $5x + 2y + 1$

Line two : $5x + 2y + 7$

Solution

Since the equations are given in the class $ax + by + d$ , hence we will utilise the following formula to compute the distance:

$d = \frac {|d_2 - d_1|} {\sqrt{a^2 + b^2}}$

Here, $d_1 = 1$ , $d_2 = 7$ , $a = 5$ and $b = 2$ . Substitute these values in the above formula to get the altitude:

$d = \frac {|7 - 1|} {\sqrt{4^2 + 2^2}}$

$d = \frac {|6|} {\sqrt{20}}$

$d = \frac {|6|} {4.47}$

$d = 1.34$ units

Hence, the distance betwixt these lines is 1.34 units.

Example v

Find the distance between the post-obit two parallel lines:

Line 1 : $3x + 2y + 8$

Line two : $3x + 2y + 9$

Solution

Since the equations are given in the course $ax + by + d$ , hence nosotros volition employ the post-obit formula to compute the distance:

$d = \frac {|d_2 - d_1|} {\sqrt{a^2 + b^2}}$

Hither, $d_1 = 8$ , $d_2 = 9$ , $a = 3$ and $b = 2$ . Substitute these values in the higher up formula to get the altitude:

$d = \frac {|9 - 8|} {\sqrt{3^2 + 2^2}}$

$d = \frac {|1|} {\sqrt{13}}$

$d = \frac {|1|} {3.60}$

$d = 0.27$ units

Hence, the distance between these lines is 0.27 units.

Example 6

Notice the distance betwixt the following two parallel lines:

Line 1 : $7x + y + 8$

Line two : $7x + y + 9$

Solution

Since the equations are given in the form $ax + by + d$ , hence we will utilise the following formula to compute the altitude:

$d = \frac {|d_2 - d_1|} {\sqrt{a^2 + b^2}}$

Here, $d_1 = 8$ , $d_2 = 9$ , $a = 7$ and $b = 1$ . Substitute these values in the in a higher place formula to go the distance:

$d = \frac {|9 - 8|} {\sqrt{7^2 + 1^2}}$

$d = \frac {|1|} {\sqrt{50}}$

$d = \frac {|1|} {7.07}$

$d = 0.14$ units

Hence, the distance between these lines is 0.fourteen units.

Altitude Betwixt Skew Lines

These lines do non intersect each other at any signal, just they are also non parallel to each other. These lines are as well called agonic lines and most of the time these lines be in three or more dimensions. In the effigy beneath, the lines r and southward are skew lines.

Distance between skew lines

The general formula to calculate the distance between two skew lines of the grade $l_1 =a_1 +tb_1$ and $l_2 = a_2 + tb_2$ is given beneath:

$d = \frac{ (\overrightarrow {a_2} - \overrightarrow {a_1}) (\overrightarrow{b_2} - \overrightarrow {b_1})} {(\overrightarrow {b_1} \times \overrightarrow {b_2})}$

Example

Notice the altitude between the following lines:

$r = 2i - j + k + \lambda (2 + j + 2k)$ and $s = 3i + 2j + 3k + \mu (2i + j + 2k)$

Solution

Here, $a_1 = 2i -j + k, a_2 = 3i +2 j + 3k$ and values of $b_1$ and $b_2$ are same, i.east., $b_1 = b_2$ . Nosotros will use the following formula to compute the distance: