How do you write the equation of a line in slope intercept, point slope and standard form given (1,-2), and (3,-8)?

The slope intercept form of equation of a line is `y=mx+c`, where `m` is the slope of the line and `c` is its intercept on `y`-axis.

Point slope form of equation of a line is `y-y_1=m(x-x_1)`, where `m` is the slope of the line and `(x_1,y_1)` are the coordinates of the point through which the line passes.

The general form of equation of a line is `ax+by+c=0`.

For the equation of a line passing through two points `(x_1,y_1)` and `(x_2,y_2)`, the formula is `(y-y_1)/(y_2-y_1)=(x-x_1)/(x_2-x_1)`. However, we first find the slope of line `(x_1,y_1)` and `(x_2,y_2)`, which is `(y_2-y_1)/(x_2-x_1)`. For points `(1,-2)` and `(3,-8)` is `(-8-(-2))/(3-1)=-6/2=-3`.

Hence selecting the point slope form of equation of desired line (choosing point `(1,-2)`) is `y-(-2)=(-3)(x-1)` or `y+2=-3x+3`,

which in general form is `3x+y-1=0`

Note that had we chosen `(3,-8)`, the point slope form of equation could also have been `y-(-8)=(-3)(x-3)` or `y+8=-3x+9` which again gives us `3x+y-1=0`.

We can also write the equation as `y=-3x+1` in slope intercept form which shows slope as `-3` and `y`-intercept as `1`.

graph{-3x+1 [-10, 10, -5, 5]}