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=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#

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"#