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Find the equation of the line passing through (5, –3) and parallel to x – 3y = 4. - Mathematics

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Question

Find the equation of the line passing through (5, –3) and parallel to x – 3y = 4.

Sum

Solution

x − 3y = 4

`=>` 3y = x – 4

`=> y = 1/3x - 4/3`

Slope of this line =`1/3`

Slope of a line parallel to this line = `1/3`

Required equation of the line passing through (5, −3) is

y − y1 = m(x − x1)

`y - (-3) = 1/3 (x - 5)`

`y + 3 = 1/3 (x -5)`

3y + 9 = x – 5

3y = x – 5 – 9

3y = x – 14

x − 3y − 14 = 0

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Conditions for Two Lines to Be Parallel Or Perpendicular
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Chapter 14: Equation of a Line - Exercise 14 (D) [Page 201]

APPEARS IN

Selina Mathematics [English] Class 10 ICSE
Chapter 14 Equation of a Line
Exercise 14 (D) | Q 13.1 | Page 201
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