Question

How do you write an equation of a line with point (3,-3) slope 3?

Answer 1

`y = 3x -12`

Explanation:

We can simply apply the formula; `y - y_1 = m(x-x_1)`
Where `x_1` and `y_1` are points on the line, and m is the gradiant

Hence `y-(-3) = 3(x-3)`
`therefore` `y + 3 = 3x - 9`

Hence when rearanging to yields; `y = 3x - 12`

Answer 2

`y=3x-12`

Explanation:

`"the equation of a line in "color(blue)"slope-intercept form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=mx+b)color(white)(2/2)|)))`

`"where m is the slope and b the y-intercept"`

`"here "m=3`

`rArry=3x+blarrcolor(blue)"is the partial equation"`

`"to find b substitute "(3,-3)" into the partial equation"`

`-3=9+brArrb=-12`

`rArry=3x-12larrcolor(red)"in slope-intercept form"`

Answer 3

The answer is `y=3x-12`.

Explanation:

I am assuming that you need this in slope-intercept form. You already know that the slope is `3`. Since the slope-intercept form is

`y=mx+b`

you can fill in `m` with `3`. Now you have:

`y=3x+b`

Now, all that you need to do is solve for `b`. There are two ways to do this. The way that I personally prefer is changing the slope into a fraction.

So first you change `3` into `3/1`. The next step is to find the reciprocal of `3/1`. This is `1/3`. Now you just subtract `1` from `3` until you get zero. Then you take the number of times you had to do this `(3)` and multiply it by your denominator, `3`, and add it to `y` `(-3)`. Now you have your `y`-intercept `(b)`. It is `(0,-12)`.

Another way to do this is by plugging in `x` and `y`. Here's how to do it:

`y=3x+b`

`-3=3(3)+b`

`-3=9+b`

`-12=b`

Now just plug in `b` into your equation and you have:

`y=3x-12`