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.

Answer 1

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