Equation of a Line Parallel to a Line

We will learn how to find the equation of a line parallel to a line.

Prove that the equation of a line parallel to a given line ax + by + λ = 0, where λ is a constant.

Let, ax + by + c = 0 (b ≠ 0) be the equation of the given straight line.

Now, convert the equation ax + by + c = 0 to its slope-intercept form.

ax + by+ c = 0

⇒ by = - ax - c

Dividing both sides by b, [b ≠ 0] we get,      

y =  -\(\frac{a}{b}\) x - \(\frac{c}{b}\), which is the slope-intercept form.

Now comparing the above equation to slope-intercept form (y = mx + b) we get,

The slope of the line ax + by + c = 0 is (- \(\frac{a}{b}\)).

Since the required line is parallel to the given line, the slope of the required line is also (- \(\frac{a}{b}\)).

Let k (an arbitrary constant) be the intercept of the required straight line. Then the equation of the straight line is

y = - \(\frac{a}{b}\) x + k

⇒ by = - ax + bk        

⇒ ax +  by = λ, Where λ = bk = another arbitrary constant.

Note: (i) Assigning different values to λ in ax + by = λ we shall get different straight lines each of which is parallel to the line ax + by + c = 0. Thus, we can have a family of straight lines parallel to a given line.

(ii) To write a line parallel to a given line we keep the expression containing x and y same and simply replace the given constant by a new constant λ. The value of λ can be determined by some given condition.

To get it more clear let us compare the equation ax + by = λ with equation ax + by + c = 0. It follows that to write the equation of a line parallel to a given straight line we simply need to replace the given constant by an arbitrary constant, the terms with x and y remain unaltered. For example, the equation of a straight line parallel to the straight line 7x - 5y + 9 = 0 is 7x - 5y + λ = 0 where λ is an arbitrary constant.

Solved examples to find the equations of straight lines parallel to a given line:

1.  Find the equation of the straight line which is parallel to 5x - 7y = 0 and passing through the point (2, - 3).

Solution:    

The equation of any straight line parallel to the line 5x - 7y = 0 is 5x - 7y + λ = 0 …………… (i)  [Where λ is an arbitrary constant].

If the line (i) passes through the point (2, - 3) then we shall have,

5 ∙ 2 - 7 ∙ (-3) + λ = 0

⇒ 10 + 21 + λ = 0

⇒ 31 + λ = 0

⇒ λ = -31

Therefore, the equation of the required straight line is 5x - 7y - 31 = 0.

2. Find the equation of the straight line passing through the point (5, - 6) and parallel to the straight line 3x - 2y + 10 = 0.

Solution:

The equation of any straight line parallel to the line 3x - 2y + 10 = 0 is 3x - 2y + k = 0 …………… (i) [Where k is an arbitrary constant].

According to the problem, the line (i) passes through the point (5, - 6) then we shall have,

3 ∙ 5 - 2 ∙ (-6) + k = 0

⇒ 15 + 21 + k = 0

⇒ 36 + k = 0

⇒ k = -36

Therefore, the equation of the required straight line is 3x - 2y - 36 = 0.

● The Straight Line

• Straight Line
• Slope of a Straight Line
• Slope of a Line through Two Given Points
• Collinearity of Three Points
• Equation of a Line Parallel to x-axis
• Equation of a Line Parallel to y-axis
• Slope-intercept Form
• Point-slope Form
• Straight line in Two-point Form
• Straight Line in Intercept Form
• Straight Line in Normal Form
• General Form into Slope-intercept Form
• General Form into Intercept Form
• General Form into Normal Form
  
• Point of Intersection of Two Lines
  
• Concurrency of Three Lines
• Angle between Two Straight Lines
• Condition of Parallelism of Lines
• Equation of a Line Parallel to a Line
• Condition of Perpendicularity of Two Lines
• Equation of a Line Perpendicular to a Line
• Identical Straight Lines
• Position of a Point Relative to a Line
• Distance of a Point from a Straight Line
• Equations of the Bisectors of the Angles between Two Straight Lines
• Bisector of the Angle which Contains the Origin
• Straight Line Formulae
• Problems on Straight Lines
• Word Problems on Straight Lines
• Problems on Slope and Intercept

11 and 12 Grade Math 

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