Question 217410
Write an equation in slope-intercept form that passes through (4, 9) and (5, 12)


Step 1.  We need to find an equation in slope intercept form given as y=mx+b where m is the slope and b is the y-intercept at point (0,b).


Step 2.  The slope of the line m is given as


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


where for our example is x1=4, y1=9, x2=5 and y2=12 (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 3.  Substituting the above values in the slope equation gives


{{{m=(12-9)/(5-4)}}}


{{{m=3/1=3}}}


Step 4.  The slope is calculated as {{{3}}} or {{{m=3}}}.


Step 5.  Now use the slope equation of step 1 and choose one of the given points.  I'll choose point (4,9).   Letting y=y2 and x=x2 and substituting m=60 in the slope equation given as,


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



{{{ 3=(y-9)/(x-4)}}}


Step 6.  Multiply both sides of equation by (x-4) to get rid of denomination found on the right side of the equation



{{{ 3(x-4)=(x-4)(y-9)/(x-4)}}}



{{{ 3x-12=(y-9)}}}



Step 7.  Now simplify and put the above equation into slope-intercept form.


{{{3x-12=y-9}}}


Add 9 to both sides of the equation


{{{3x-12+9=y-9+9}}}


{{{3x-3=y}}}


ANSWER in slope-intercept form is {{{y=3x-3}}} where slope m=3 band y-intercept=-3


Step 8.  See if the other point (5,12) or x=5 and y=12 satisfies this equation


{{{y=3x-3}}}


{{{12=3*5-3=12}}}  So the other point satisfies this equation and lies 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


{{{3x-y=3}}}


See graph below to check the above steps.  Note the slope and y-intercept as well as the x-intercept.


{{{graph(600,600,-15,15,-10,10,3x-3)}}}


I hope the above steps were helpful. 

 
And good luck in your studies!


For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.


Respectfully,
Dr J