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.

**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**

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