Question

What is the equation of the line perpendicular to `y=-3x` that passes through `(5,8)`?

Answer 1

Equation of the line perpendicular to `y=-3x` and passing trough `(5,8)` is `x-3y+19=0`.

Explanation:

The equation is equivalent to `3x+y=0` and hence equation of a line perpendicular to it will be `x-3y=k`.

This is so because for two line to be perpendicular, product of their slopes should be `-1`.

Using this it is easy to deduce that lines `Ax+By=C_1` and `Bx-Ay=C_2` (i.e.just reverse the coefficients of `x` and `y` and change sign of one of them) are perpendicular to each other.

Putting the values `(5,8)` in `x-3y=k`, we get `k=5-3*8=5-24=-19`

Hence, equation of the line perpendicular to `y=-3x` is `x-3y=-19` or `x-3y+19=0`.