How do you change the form of a linear function from standard to point-slope and vice versa?

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The standard form of a linear equation is Ax+By+C=0.

The point slope form of a linear equation: given a point `(x_1,y_1)` and a slope `m` the equation is `y-y_1=m(x-x_1)`

The slopeintercept form is `y=mx+b` where m is the slope and b the y-intercept.

(1) If you are given the standard form, if you solve for y you will have the slope-intercept form.

`Ax+By+C=0`

`By=-Ax-C`

`y=-A/Bx-C/B` where `m=-A/B` and the y-intercept `b=-C/B`

If you are asked to write in point-slope form (and this shouldn't happen) then you can find teh slope (`m=-A/B` ), and you will have to find a point that lies on the line. (My choice would be to substitute x=0 to get the y-intercept) then use `y-y_1=m(x-x_1)` where the point you found was `(x_1,y_1)` .

(2) Given point-slope form:

Again solve for y; this is easier with an example. Given (1,2) and m=3 you have `y-2=3(x-1)`

Solve for y to get `y-2=3x-3==>y=3x-1`

This is slope-intercept form.

To put into standard form we need everything on one side:

`3x-y-1=0`

** Note that there is more than 1 correct "standard form"

Given 3x-y-1=0 these are the same line:

-3x+y+1=0

6x-2y-2=0

etc...

**Sources:**

Standard form is: Ax + By + C where A,B, and C are non-zero integer coefficients.

Point slope form is: y - y1 = m(x - x1) where m is the slope and (x1, y1) is a point.

Lets give an easy example to show how to go between the two forms:

-10x + 2y = 20

-5x + y = 10

y = 5x + 10 (This is slope-intercept form.)

y = 5(x + 2) (Pull out the slope)

And there you have point-slope form!

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