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
11 and 12 Grade Math
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