Question

How do you write an equation of a line that goes through (6,-2); m= -4/3 in standard form?

Answer 1

Standard form is `y = mx + c` where `m` is the slope and `c` is the `y` coordinate of the intersection of the line with the `y` axis.

We are given `m`, so it remains to determine `c`.

Subtracting `mx` from both sides of the formula yields `c = y - mx`.

Given that the line passes through (6, -2) all we need do is substitute `x = 6` and `y = -2` to get

`c = y - mx`

`= -2 - (-4/3)*6 = -2 + 4*6/3 = -2 + 8 = 6`

So the standard form is `y = -(4/3)x + 6`