Question

How do you write an equation of the line that passes through (-3,4) and (1,0)?

Answer 1

`y = -x+1`

Explanation:

The line equation is of the form
`y = ax + b`

As the line passes through this two points
`(x_0,y_0) = (-3,4)`
`(x_1,y_1) = (1,0)`

they both obey the equation
`y_1 = ax_1 + b => 0 = a+b => a = -b`
`y_0 = ax_0 + b => 4 = -3a+b`

Therefore:
`4 = -3a - a => 4 = -4a => a = -1 and b = 1`

So, the equation of the line that passes through those points is
`y = -x+1`

Answer 2

We will write the equation in slope-intercept form, `y=mx+c`. I have calculated below that the slope of the line is -1 and its y-intercept is 1.

The equation of the line can be written: `y=-x+1`

(we don't write the number '1' in front of the `x`, since `1x=x`)

Explanation:

First step: find the slope (gradient) of the line:

`m=(y_2-y_1)/(x_2-x_1)`

It doesn't matter which of the two given points we decided is 'Point 1' `(x_1,y_1)`, but let's choose the point `(1,0)`, so `x_1=1` and `y_1=0`.

Similarly for the other point, `(-3,4)`, so `x_2=-3` and `y_2=4`.

So `m=(y_2-y_1)/(x_2-x_1)=(4-0)/(-3-1) =4/(-4)=-1`

Now that we know the slope, we can use it and the value of one of the points we were given to find the y-intercept of the line. This is the point where the line crosses the y-axis. The y-axis is the line `x=0`, so if we substitute `m=-1` and, for example, `x_1=1` and `y_1=0`, into the equation:

`y=mx+c`

`0=-1(1)+c`

Rearranging, `c=1`.