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.
`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:
** Note that there is more than 1 correct "standard form"
Given 3x-y-1=0 these are the same line:
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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