Question 218680
Find the slope-intercept form of the equation of the line that passes through the given points. 


(2, 2)
(6, -2/3)


Step 1.  The slope of the line m is given as


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


where for our example is x1=2, y1=2, x2=6 and y2=-2/3 (think of {{{slope=rise/run}}}).  You can choose the points the other way around but be consistent with the x and y coordinates.  You will get the same result.


Step 2.  Substituting the above values in the slope equation gives


{{{m=(-2/3-2)/(6-2)}}}


{{{m=-(8/3)/4}}}


{{{m=-8/12=-2/3}}}


Step 3.  The slope is calculated as -2/3 or m=-2/3


Step 4.  Now use the slope equation of step 1 and choose one of the given points.  I'll choose point (2,2).   Letting y=y2 and x=x2 and substituting m=-3 in the slope equation given as,


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



{{{ -2/3=(y-2)/(x-2)}}}


{{{ -2/3=(y-2)/(x-2)}}}


Step 5.  Multiply both sides of equation by x-2 to get rid of denomination found on the right side of the equation



{{{ -2(x-2)/3=(x-2)(y-2)/(x-2)}}}



{{{ -2(x-2)/3=y-2}}}



Step 6.  Now simplify and put the above equation into slope-intercept form.


{{{-2x/3+4/3=y-2}}}


Add 2 from both sides of the equation


{{{-2x/3+4/3+2=y-2+2}}}


{{{-2x/3+10/3=y}}}


{{{y=-2x/3+10/3}}}   ANSWER in slope-intercept form.  m=-2/3 and y-intercept b=10/3


Step 7.  See if the other point (6,-2/3) or x=2 and y=0 satisfies this equation


{{{y=-2x/3+10/3}}}


{{{-2/3=-2*6/3+2/3}}}


{{{0=0}}}  So the point (6,-2/3) satisfies the equation and is on the line.  In other words, you can use the other point to check your work.


Note;  above equation can be also be transform into standard form as


{{{2x+3y=10}}}


See graph below to check the above steps.


*[invoke describe_linear_equation 2, 3,  10]

I hope the above steps were helpful.


For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.


And good luck in your studies!


Respectfully,
Dr J

http://www.FreedomUniversity.TV