what is the point slope form for (0,-1) (-2,-2)

Question

i also want to find the slope -intercept form, and the standard form for it. 

Answer 1

Let's say you have an equation of a line

 

y = mx + b

 

Consider a point that's on that line: (x_o, y_o)

 

Then this must satisfy this equation:

 

y_o = mx_o + b

 

Whenever I have an equation, I can subtract the same thing from both sides:

 

y - y_o = mx + b - y_o

 

but we have that y_o = mx_o + b, so this line above can become

 

y - y_o = mx + b - (mx_o + b)

 

y - y_o = mx + b - mx_o - b

 

y - y_o = mx - mx_o

 

y - y_o = m(x - x_o)

 

and that's where point slope form comes from.

 

Given the two points you're given, you can find the slope, m, and you can use the point-slope form to write an equation describing the line passing through them.

 

Let's see first the rise: when I go from -1 to -2, I'm going down one, so the numerator of the slope will be -1.

 

Now the run: when I go from 0 to -2, I'm going down two, so the denominator of the slope will be -2.

 

That is m = -1/-2 which simplifies to m = 1/2.

 

Now that equation in point slope form, y - y_o = m(x - x_o), can be rewritten as y - y_o = (1/2)(x - x_o).

 

To find x_o and y_o, pick your favorite of those two points and call the x coordinate x_o and the y coordinate y_o.

 

It doesn't matter which point you pick, they're both valid.

 

One way you could write this equation in point-slope form is

 

x_o = 0 and y_o = -1             so

 

y - (-1) = (1/2)(x - 0)

 

Another way you could write it is choosing the other point:

 

x_o = -2 and y_o = -2           so

 

y - (-2) = (1/2)(x - (-2))

 

 

YOU DON'T HAVE TO FIND BOTH UNLESS YOU WANT TO.

 

 

 

That's point slope form.

 

To get to slope-intercept form, pick one of those equations above and solve for y:

 

y - (-1) = (1/2)(x - 0)                        (simplify)

 

y + 1 = (1/2)x                                  (bring the 1 over to the right)

 

y = (1/2)x - 1                                    SLOPE INTERCEPT FORM looks like y = mx + b

 

To get to standard form, pick one of those equations and get x and y both on one side (and I think you generally like to present it with integral coefficients...): 

 

y - (-1) = (1/2)(x - 0)                    (simplify)

 

y + 1 = (1/2)x                              (bring the x term to the left and the one to the right)

 

y - (1/2)x = -1                              (multiply everything by 2)

 

2y - x = -2                                    STANDARD FORM looks like Ax + By = C

 

 

Answer 2

When given two points and your need to find the equation of the line, the starting point is the point-slope form.

 

y-y₁ = m(x-x₁)  where (x₁,y₁) is one of the points and m is the slope

 

The formula for the slope is 

         y₂-y₁

m =  -------

         x₂-x₁

 

Calculate the slope:

        -2-(-1)        -1        1

m = --------- = ------ = ----

         -2-0          -2         2

 

Now enter the point and slope into the equation for the line

 

y - (-1) = (1/2) (x - 0)

 

And simplify

 

y + 1 = (1/2) x       

 

I used the first point given, but you would get the same result if you used the 2nd point.  Try it.

 

The point slope version of the equation of a line is y = mx + b

We can get this by rearranging the terms in the point-slope equation.

Subtract 1 from both sides

y = (1/2) x - 1

 

Finally, the standard form of the equation has the x and y on the same side and whole number coefficients.  So working with the the above equation

multiply by 2

2y = x - 2

Then subtract x from both sides

2y - x = -2

Answer 3

m = (y2 - y1)/(x2 - x1)

 

m = (-2 - (-1))/(-2 - 0)

 

m = (-1)/(-2) = 1/2 or .5

 

y - y₁ = m(x -x₁)

y - (-1)  (1/2)(x - 0)    or  y - (-2) = (1/2)(x - (-2))

 

y + 1 = (1/2)x              or  y + 2 = (1/2)x + 1   Point-slope intercept form options

 

y = (1/2)x - 1    Slope intercept form: results from either of the point-slope options

 

x - 2y = 2    Standard form  Ax + By = C; A, B, C are integers.