A line 5x + 3y + 15 = 0 meets y-axis at point P. Find the co-ordinates of point P. Find the equation of a line through P and perpendicular to x - 3y + 4 = 0.

As the point P lies on y-axis, the x co-ordinate of point P will be = 0.

Putting x = 0 in the equation 5x + 3y + 15 = 0, we get

⇒ 5(0) + 3y + 15 = 0

⇒ 3y + 15 = 0

⇒ 3y = -15

⇒ y = -5.

Hence, the co-ordinates of the point P are (0, -5).

Given line equation,

x - 3y + 4 = 0

3y = x + 4

y = $\dfrac{1}{3}x + \dfrac{4}{3}$

Comparing above equation with y = mx + c we get,

Slope of this line (m) = $\dfrac{1}{3}$

Product of slope of perpendicular lines = -1.

Let slope of line perpendicular to x - 3y + 4 = 0 be m₁.

Then,

⇒ m₁ × m = -1

⇒ m₁ $\times \dfrac{1}{3} = -1$

⇒ m₁ = -3.

Slope of the required line = -3.

By point-slope form,

Equation of line through P and slope = -3 is,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-5) = -3(x - 0)

⇒ y + 5 = -3x

⇒ 3x + y + 5 = 0.

Hence, P = (0, -5) and equation of required line is 3x + y + 5 = 0.