Distance between Two Parallel Lines Examples

Distance between 2 parallel lines is the perpendicular distance from any point to one of the lines. In this article, you will learn parallel lines definition, how to find the distance between them, and solved examples. 

The graph of the function, a pair of railroad tracks, or the opposite sides of a parallelogram or keys of a piano can be a few examples of parallel lines. A common property that can be found in the above examples is that the two railroad tracks never meet, the opposite sides of a parallelogram will never intersect or the piano keys are parallel to each other. This article explains the method of finding the distance between two parallel lines with appropriate examples.

Parallel Lines Definition

Parallel lines are those lines that never meet each other. When the distance between a pair of lines is the same throughout, it can be called parallel lines. It is denoted by “||”. The main criteria for any two lines to be parallel is that they have to be drawn on the same plane. They are always equidistant from each other.

The lines can be extended till infinity. The slopes of two parallel lines are equal.

How to Find the Distance Between Two Parallel Lines

The method for calculating the distance between two parallel lines is as follows:

  • Ensure whether the equations of the given parallel lines are in slope-intercept form (y=mx+c).
  • The intercepts (c1 and c2) and slope value which is common for both the lines has to be determined.
  • After obtaining the above values, substitute them in the slope-intercept equation to find y.
  • Finally, put all the above values in the distance formula to find the distance between two parallel lines.

The two parallel lines can be taken in the form

y = mx + c1 … (i)

and y = mx + c2 … (ii)

The line (ii) will intersect the x-axis at point A (–c1/m, 0) as shown in the figure.

Problems on Distance between Two Parallel Lines

 

The length of the perpendicular from point A to the line (i) is of the same length as the distance between two lines.

Therefore, the distance between the lines (i) and (ii) is

|(–m)(–c1/m) + (–c2)|/√(1 + m2) or d = |c1–c2|/√(1+m2).

Distance d between two parallel lines y = mx + c1 and y = mx + c2 is given by

d = |C1–C2|/√(A2 + B)

Distance between a point (x1, y1) and a line is d= |a1x1+b1y1+c1| / √(a12+b12) ,

where a1 and b1 are the coefficients of variable x and y in the equation of the line. 

The equation of line is

a1x+b1y+c1 = 0

Considering the following equations of 2 parallel lines, we can calculate the distance between those lines using the distance formula

ax+by+c = 0

ax+by+c1 = 0

Using the above 2 equations we can conclude that 

Distance between 2 parallel lines, d = |c-c1| / √(a2+b2)

Also read

Height and distance problems

Trigonometric equations

Solved Examples

Example 1: Find the distance of the point (4, –6) from the line 2x – 7y – 24 = 0.

Solution:

Given line is 2x – 7y – 24 = 0. …… (1)

Comparing (1) with general equation of line Ax + By + C = 0, we get

A = 2, B = –7 and C = –24.

Given point is (x1, y1) = (4, –6).

The distance of the given point from given line is d = |Ax1 + By1 + C|/√A2+B2 = 26/7.2

= 3.6

Example 2: Estimate the distance between the two parallel lines y=2x+7 and y=2x+5.

Solution:

The distance between two parallel lines is given by |c1-c2|/√(a2+b2).

Here, the equations of parallel lines are y = 2x + 7 and y = 2x + 5.

Slopes are same m1 = m2 = 2 and c1 = 7 ,c2 = 5.

here a = 2, b = -1

So, the distance between two parallel lines is given by

=> |c1-c2| = |7-5|/√(22+12) = 2/√5

Example 3: Calculate the distance between the parallel lines 3x+4y+7=0 and 3x+4y−5=0 .

Solution:

The distance between two parallel lines is given by d = |c1-c2|/√(a2+b2).

Here c= 7, c2 = -5, a = 3, b = 4

so d = |7 – (-5)|/√(32+42)

= 12/5

Example 4: Find the distance from the line 6x – 4y + 36 = 0 to point (0, 0).

Solution:

Distance between a point (x1, y1) and a line a1x+b1y+c1 = 0 is

d = |a1x1+b1y1+c1| / √(a12+b12)

Here a1 = 6, b1 = -4, c1 = 36, x1 = 0, y1 = 0

so d = 36/√(36 + 16)

= 36/√52

Example 5: Let PS be the median of the triangle with vertices P(2, 2), Q(6,  1)P(2,\ 2),\ Q(6,\ -\ 1) and R(7, 3)R(7,\ 3). The equation of the line passing through (1, -1) and parallel to PS is

A) 2x − 9y − 7 = 0

B) 2x − 9y − 11 = 0

C) 2x + 9y − 11 = 0

D) 2x + 9y + 7 = 0

Solution:

S = midpoint of QR=(6+72,1+32)=(132,1)QR=\left( \frac{6+7}{2},\,\frac{-1+3}{2} \right)=\left( \frac{13}{2},\,1 \right)\\

Slope of PS=212132=29PS=\frac{2-1}{2-\frac{13}{2}}=-\frac{2}{9}\\

Equation of line passing through (x1, y1) having slope m is y – y1 = m(x – x1)

The required equation is y+1=29(x1)y+1=\frac{-2}{9}(x-1) that is 2x+9y+7=0.2x+9y+7=0..

Example 6: The graph of the function cosx cos(x+2)cos2(x+1)\cos x\ \cos (x+2)-{{\cos }^{2}}(x+1) is

A) A straight line passing through (0,−sin2 1) with slope 0

B) A straight line passing through (0, 0)

C) A parabola with vertex 75 degrees

D) A straight line passing through the point (π/ 2, −sin2 1) and parallel to the x-axis

Solution:

y=cos(x+11)cos(x+1+1)cos2(x+1)=cos2(x+1)sin21cos2(x+1)=sin21,y=\cos (x+1-1)\cos (x+1+1)-{{\cos }^{2}}(x+1) \\={{\cos }^{2}}(x+1)-{{\sin }^{2}}1-{{\cos }^{2}}(x+1)=-{{\sin }^{2}}1,\\ which represents a straight line parallel to x-axis, it passes through  (0, −sin2 1).

Here slope = 0

Hence option A is the answer.

Example 7: The line 3x + 2y = 24 meets y-axis at A and x-axis at B. The perpendicular bisector of AB meets the line through (0,−1) parallel to the x-axis at C. The area of the triangle ABC is

A) 182sq.units

B) 91sq.units

C) 48sq.units

D) None of these

Solution:

The coordinates of A and B are (0, 12) and (8, 0) respectively.

Slope of given line  3x + 2y = 24 is m = -3/2

Slope of perpendicular bisector = 2/3

Midpoint of AB = (4, 6)

The equation of the perpendicular bisector of AB is  y6=23(x4) or 2x3y+10=0..(i)y-6=\frac{2}{3}(x-4) \text \ or \ 2x-3y+10=0 …..(i)\\

Equation of a line passing through (0, 1) and parallel to the x-axis is y=−1.

(substitute y = -1 in 2x-3y+10=0)

This meets (i) at C, Therefore the coordinates of C are (132,1)\left( -\frac{13}{2},-1 \right).

Hence, the area of the triangle ABC is Δ=12012180113211=91 sq. units.\Delta =\frac{1}{2}\left| \begin{matrix} 0 & 12 & 1 \\ 8 & 0 & 1 \\ -\frac{13}{2} & -1 & 1 \\ \end{matrix}\, \right|=91 \text \ sq. \ units.

Example 8: The diagonals of a parallelogram PQRS are along the lines x + 3y = 4 and 6x − 2y = 7. Then PQRS must be a

A) Rectangle

B) Square

C) Cyclic quadrilateral

D) Rhombus

Solution:

Given x + 3y = 4  ..(i)

6x − 2y = 7  ..(ii)

m1 = −1/3 and m2 = 3.

here m1m2 = -1

Hence, lines x + 3y = 4 and 6x − 2y = 7 are perpendicular to each other.

Therefore, the parallelogram is a rhombus.

Example 9:

Find the distance between the lines  4x +3y+6= 0 and  4x+3y-3= 0. 

Solution:

Here A = 4, B = 3,  C1 = 6 and C2 = -3

So the distance =| (C1-C2) | / √(A2+B2

=| (6–3) |  / √(16+9) = 9/√25 = 9/5

 

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