Question

How do you find the equation of the line that goes through `(1,- 3)` and `( 0,8)`?

Answer 1

`y=-11x+8`

Explanation:

First, we find the slope using `(y^2-y^1)/(x^2-x^1)`

Therefore, `m=(8-(-3))/(0-1)` which is equal to `m=-11`

To get the line, the equation would be `y=mx+b`

`y=11x+b`

We will put in `(1, -3)` as `x=1` and `y=-3`
`-3=(-11)1+b->`
`-3=(-11)*1+b->`
`-3=-11*1+b->`
`-3=-11+b->`
`8=b` (Add `11` to both sides of the equation)

`8=b`

Therefore, the line will be `y=mx+b->`

`y=-11x+8` graph{y=-11x+8 [-10, 10, -5, 5]}

Answer 2

`11x+1y=8`

Explanation:

For a straight line the slope between any two points is the same for all pairs of points on the line.

By slope we mean the ratio of the difference between the `y` coordinates divided by the difference between the `x` coordinates.

`color(white)("xxxx")"slope"=(delta y)/(delta x)=(y_2-y_1)/(x_2-x_1)`
`color(white)("xxxxxx")`for two points `(x_1,y_1)` and `(x_2,y_2)`

For a general point `(x,y)` and the point `(1,-3)` (this is one of the given points,
we have
`color(white)("xxxx")"slope"=(y-(-3))/(x-1)=(y+3)/(x-1)`
and
for the two given points `(0,8)` and `(1,-3)`
we have
`color(white)("xxxx")"slope" = (8-(-3))/(0-1)=11/(-1)=-11`

Since the slope must be the same for all pairs of points on a straight line
`color(white)("XXX")(y+3)/(x-1)=-11`

While is is a valid equation that answers the given question, it would be normal to convert this into standard `Ax+By=C` form:

`color(white)("XXX")(y+3)=-11(x-1)`

`color(white)("XXX")y+3=-11x+11`

`color(white)("XXX")11x+y+3=11`

`color(white)("XXX")11x+1y=8`

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When doing this kind of problem it is always a good idea to verify that the derived equation satisfies the given points:

For `(x,y)=(1,-3)`
`color(white)("XXX")11 * (1)+1 * (-3)=11-3=8` ... as required
and
for `(x,y)=(0,8)`
`color(white)("XXX")11 * (0) + 1 * (8)= 0+8=8` ...again, as required.