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