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Question
A line 5x + 3y + 15 = 0 meets y-axis at point P. Find the co-ordinates of points P. Find the equation of a line through P and perpendicular to x – 3y + 4 = 0.
Solution
P lies on y-axis and let the co-ordinates of P be (0, y)
∵ P lies also on the line 5x + 3y + 15 = 0
∴ It will satisfy it.
∴ 5 × 0 + 3y + 15 = 0 `\implies` 3y = –15
∴ y = –5
∴ Co-ordinates of P are (0, –5)
Now, writing the line x – 3y + 4 = 0 is form of y = mx + c
–3y = –x – 4
`\implies` 3y = x + 4
`\implies y = 1/3 x + 4/3`
∴ Slope of given line = `1/3`
∴ Slope of the line which is perpendicular to it
= `-(3/1)`
= –3 ...(∵ Product of slopes = –1)
∴ Equation of the line passing through P(0, –5)
y – y1 = m(x – x1)
`\implies` y + 5 = –3(x – 0)
`\implies` y + 5 = –3x
`\implies` 3x + y + 5 = 0
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