Given two points that make up a linear equation, how can we create the point-slope form of that equation and then use it to determine the slope-intercept and standard forms of that line?

a) Use the two points to compute the slope, then put that and one of the points into the point-slope form

b) Eliminate parentheses and solve for y to get the equation in slope-intercept form

c) From slope-intercept form, subtract the x-term, then multiply by a common denominator of there are any fractions. Multiply by -1 if necessary to make the x-coefficient positive.

Step-by-step explanation:

a) The slope (m) is computed from two points by ...

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

That value and one of the points goes into the point-slope form ...

... y -y1 = m(x -x1)

b) Putting the above equation into slope-intercept form is a matter of consolidating all of the constants.

... y = mx +(-m·x1 +y1)

c) Rearranging to standard form puts the x- and y-terms on the same side of the equal sign, preferably with mutually prime integer coefficients. This may require that the equation be multiplied by an appropriate number. The x-coefficient should be positive.

Example:

y -3 = 1/2(x +7) . . . . . . line with slope 1/2 through (-7, 3)

-1/2x + y = 7/2 + 3

x -2y = -13 . . . . . . . . . multiply by -2 to get standard form

michael scott & dwight schrute

step-by-step explanation:

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