Question

How do you write in standard form an equation of the line with the slope -4 through the given point (2,2)?

Answer 1

y `=` -4x `+` 10

Explanation:

First off you have to know the standard form formula which is:
y `=` mx `+` b

Plug in the slope and the points (x,y) to get b

y `=` mx `+` b
2 `=` -4(2) `+` b
2 `=` -8 `+` b

Next, you add 8 to both sides to get b alone:
10 `=` b

Plug your slope and b value into the standard formula

Answer 2

`4x+y=10`

Explanation:

Let's start with the very definition of the slope of a line: take two points `P_1 = (x_1,y_1)` and `P_2 = (x_2,y_2)`, the slope `m` is defined as

`m = \frac{\Delta y}{\Delta x} = \frac{y_2-y_1}{x_2-x_1}`

From here, we have

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

To have the generic expression of the line, let's change this equation a little bit. Instead of having two fixed points `P_1 = (x_1,y_1)` and `P_2 = (x_2,y_2)`, assume we have a fixed point `P_0 = (x_0,y_0)` and any other generic point on the line `P = (x,y)`. The equation becomes

`y-y_0 = m(x-x_0)`

Plug your values: `P_0=(2,2)`, and `m = -4`, to get

`y-2 = -4(x-2)`

From here, with a bit of algebra you get

`y = -4x +10`

EDIT:

as pointed out, we're not in the standard form yet. To achieve it, we must separate variables from "pure" numbers. Just move `-4x` to the left hand side to obtain

`4x+y=10`

Answer 3

`4x+y=10`

Explanation:

`"the equation of a line in "color(blue)"standard form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(Ax+By=C)color(white)(2/2)|)))`

`"where A is a positive integer and B, C are integers"`

`"obtain the equation in "color(blue)"point-slope form ""and"`
`"rearrange into standard form"`

`•color(white)(x)y-y_1=m(x-x_1)`

`"where m is the slope and "(x_1,y_1)" a point on the line"`

`"here "m=-4" and "(x_1,y_1)=(2,2)`

`rArry-2=-4(x-2)larrcolor(blue)"in point-slope form"`

`rArry-2=-4x+8`

`"add "4x" to both sides"`

`rArr4x+y-2=8`

`"add 2 to both sides"`

`rArr4x+y=10larrcolor(red)"in standard form"`