Parallel lines are two or more lines that lie in the same plane but never intersect, no matter how far they are extended. The symbol ‖ is used to indicate parallel lines.
In contrast, perpendicular lines are two lines that intersect at a 90-degree angle. They are represented by the symbol ⊥.
Parallel Lines
Two lines are parallel if they have the same slope but different y-intercepts.
Mathematically, if two lines are parallel, then
m1 = m2, here m1 and m2 are the slopes of the two lines
The lines y = 2x + 3 and y = 2x – 5 are parallel because they have the same slope, i.e., 2.
On a Graph
To graph the above equations, we will use the slope and the y-intercept.
Finding the Equation of a Parallel Line
Now, let us find the equation of a parallel line corresponding to y = 3x + 5 and passing through (2, 4).
Step 1: Identifying the Slope
Here, y = 3x + 5 …..(i)
As we know, the general form of a line equation is y = mx + b …..(ii)
Now, comparing equations (i) and (ii), we get
m = 3
Thus, the given line has a slope of 3
Since parallel lines have the same slope, the new line must have a slope of 3.
Step 2: Using the Point-Slope Form
Now, using the point-slope form, we will find the equation of the parallel line that passes through the point (2, 4) and has a slope of 3.
The point-slope form of a line is:
y – y1 = m(x – x1),
Here,
(x1, y1) = (2, 4) and m = 3
Substituting the values, we get
⇒ y – 4 = 3(x – 2)
Step 3: Solving for y
⇒ y – 4 = 3x – 6
⇒ y = 3x – 6 + 4
⇒ y = 3x – 2
Thus, the required equation is y = 3x – 2
Perpendicular Lines
Two lines are perpendicular if they intersect at a 90-degree angle. Their slopes are negative reciprocals of each other.
Mathematically, if two lines are perpendicular, then
m1 × m2 = -1, here m1 and m2 are the slopes of the two lines
This means that if the slope of one line is m1, the slope of the perpendicular line is: ${m_{2}=-\dfrac{1}{m_{1}}}$
Example:
The lines y = 2x + 3 and y = ${-\dfrac{1}{2}x+5}$ are perpendicular because they have the negative reciprocal slopes, i.e., 2 and ${-\dfrac{1}{2}}$
On a Graph
To graph the above equations, we will use the slope and the y-intercept.
Finding the Equation of a Perpendicular Line
Now, let us find the equation of a line perpendicular to y = 3x + 5 and passing through the point (2, 4)
Step 1: Identifying the Slope
Here, y = 3x + 5 …..(i)
As we know, the general form of a line equation is y = mx + b …..(ii)
Now, comparing equations (i) and (ii), we get
m = 3
The given line has a slope of 3
Step 2: Finding the Negative Reciprocal
Since the perpendicular slope is the negative reciprocal of the original line:
${m_{1}=-\dfrac{1}{m}}$
m1 = ${-\dfrac{1}{3}}$
Step 3: Using the Point-Slope Form
Using the point-slope form, we can find the equation of the perpendicular line that passes through the point (2, 4) and has a slope of 3.
The point-slope form of a line is:
y – y1 = m(x – x1),
Here,
(x1, y1) = (2, 4) and m = m1 = ${-\dfrac{1}{3}}$
Substituting the values, we get
⇒ ${y-4=-\dfrac{1}{3}\left( x-2\right)}$
Step 4: Solving for y
⇒ ${3\left( y-4\right) =-\left( x-2\right)}$
⇒ 3y – 12 = -x + 2
⇒ 3y = -x + 2 + 12
⇒ 3y = -x + 14
⇒ ${y=-\dfrac{x}{3}+\dfrac{14}{3}}$
Thus, the required equation is ${y=-\dfrac{x}{3}+\dfrac{14}{3}}$
Summary
Basis
Parallel
Perpendicular
Slope
m1 = m2 (same slope)
m1 × m2 = -1 (negative reciprocals)
Angle Between Lines
0° (never intersect)
90° (intersect perpendicularly)
Solved Examples
Find the equation of a line that is parallel to 2x – 3y = 6 and passes through the point (-1, 4)
Solution:
Given, 2x – 3y = 6 ⇒ -3y = -2x + 6 ⇒ 3y = 2x – 6 ⇒ y = ${\dfrac{2}{3}x-\dfrac{6}{3}}$ ⇒ y = ${\dfrac{2}{3}x-2}$ As we know, the general form of a line is y = mx + b Comparing the equations, m = ${\dfrac{2}{3}}$ Since the new line is parallel, it has the same slope m = ${\dfrac{2}{3}}$ Here, the equation of the new line is: y – y1 = m(x – x1) with (x1, y1) = (-1, 4) ⇒ y – 4 = ${\dfrac{2}{3}}$(x + 1) ⇒ 3(y – 4) = 2(x + 1) ⇒ 3y – 12 = 2x + 2 ⇒ 3y = 2x + 2 + 12 ⇒ 3y = 2x + 14 ⇒ y = ${\dfrac{2}{3}x+\dfrac{14}{3}}$ Thus, the required line is y = ${\dfrac{2}{3}x+\dfrac{14}{3}}$
Find the equation of a line perpendicular to 4x + 5y = 10 and passing through the point (3, -2)
Solution:
Given, 4x + 5y = 10 ⇒ 5y = -4x + 10 ⇒ y = ${-\dfrac{4}{5}x+10}$ As we know, the general form of a line is y = mx + b Comparing the equations, m = ${-\dfrac{4}{5}}$ Since the new line is perpendicular, it has the negative reciprocal slope of m = ${-\dfrac{4}{5}}$ The slope of the new line is m1 = ${\dfrac{5}{4}}$ Here, the equation of the new line is: y – y1 = m(x – x1) with (x1, y1) = (3, -2) ⇒ y + 2 = ${\dfrac{5}{4}}$(x – 3) ⇒ 4(y + 2) = 5(x – 3) ⇒ 4y + 8 = 5x – 15 ⇒ 4y = 5x – 15 – 8 ⇒ 4y = 5x – 23 ⇒ y = ${\dfrac{5}{4}x-\dfrac{23}{4}}$ Thus, the required line is y = ${\dfrac{5}{4}x-\dfrac{23}{4}}$
Are the lines given by 3x – 6y = 12 and ${y=\dfrac{1}{2}x-5}$ parallel?
Solution:
Given lines are 3x – 6y = 12 …..(i) ${y=\dfrac{1}{2}x-5}$ …..(ii) From equation (i), 3x – 6y = 12 ⇒ -6y = 12 – 3x ⇒ 6y = 3x + 12 ⇒ y = ${\dfrac{3}{6}x+\dfrac{12}{6}}$ ⇒ y = ${\dfrac{1}{2}x+2}$ …..(iii) Now, comparing equations (ii) and (iii), m1 = m2 = ${\dfrac{1}{2}}$ (same slope) Thus, the lines are parallel.
Verify whether the lines 2x + 3y = 6 and 3x – 2y = 8 are perpendicular.
Solution:
Given lines are 2x + 3y = 6 …..(i) 3x – 2y = 8 …..(ii) From equation (i), 2x + 3y = 6 ⇒ 3y = 6 – 2x ⇒ y = ${-\dfrac{2}{3}x+\dfrac{6}{3}}$ ⇒ y = ${-\dfrac{2}{3}x+2}$ …..(iii) From equation (ii), 3x – 2y = 8 ⇒ -2y = 8 – 3x ⇒ 2y = 3x – 8 ⇒ y = ${\dfrac{3}{2}x-\dfrac{8}{2}}$ ⇒ y = ${\dfrac{3}{2}x-4}$ …..(iv) Now, comparing equations (iii) and (iv), m1 = ${-\dfrac{2}{3}}$ and m2 = ${\dfrac{3}{2}}$ (negative reciprocals) Thus, the lines are perpendicular.
