Question

How do you write an equation in standard form for a line passing through (3, -5) and is perpendicular to the line y = 3x - 6?

Answer 1

You may use the general relationship:
`y-y_0=m(x-x_0)`
Where (`x_0,y_0`) are the coordinates of your point (`3,-5`) and `m` is the slope.
Now, for your line to be perpendicular to the given one it needs to have a slope: `m=-1/m_1` where `m_1` is the slope of your given line (i.e., `m_1=3`, the coefficient of `x` in `y=3x-6`), and `m=-1/3`.
So finally:
`y-(-5)=-1/3(x-3)`