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

3 Answers
May 11, 2018

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

May 11, 2018

4x+y=104x+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)P1=(x1,y1) and P_2 = (x_2,y_2)P2=(x2,y2), the slope mm 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

May 11, 2018

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"