Question

How do you write an equation of a line that has an x-intercept of 3 and a y-intercept of 4?

Answer 1

One quick way to do this is to use the fact that `3\cdot 4=12` to say an equation for this line is `4x+3y=12` (since, when `y=0`, it follows that `x=3` and when `x=0` it follows that `y=4`).

You could also find the slope of the line containing the points `(x,y)=(3,0)` and `(x,y)=(0,4)` as `\frac{\Delta y}{\Delta x}=\frac{4-0}{0-3}=-4/3` so that the equation has the form `y=-4/3x+b`, where `b` is the `y`-intercept so the equation becomes `y=-4/3x+4`. This is equivalent to the last answer because you can multiply everything by 3 and rearrange to get `4x+3y=12`.

Answer 2

The x-intercept is the value of x when y=0. The y-intercept is the value when x=0.

The x-intercept of 3 corresponds to y=0, and the point on the line is (3,0).
The y-intercept of 4 corresponds to x=0, and the point on the line is (0,4).

Since we have two points on the line, we can find the slope using the formula `m=(Deltay)/(Deltax)=(y_2-y_1)/(x_2-x_1)`

For the line running through the points `(3,0)` and `(0,4)`, `y_2=4 and y_1=0`, and `x_2=0 and x_1=3`.

`m=(y_2-y_1)/(x_2-x_1)=(0-4)/(0-3)=-4/3`

Now that we have the slope, we can write the equation for the line in point-slope form.

`y-y_1=m(x-x_1)` =

`y-0=-4/3(x-3)`