Question 226744
Find an equaion of the line containing the given pair of points.

(2,3) and (6,4)


What is the equation of the line?
y = ?


Step 1.  We will put the equation of the line in slope-intercept form given as y=mx+b where m is the slope and b is the y-intercept when x=0 or 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=2, y1=3, x2=6 and y2=4 (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=(4-3)/(6-2)}}}


{{{m=1/4}}}


Step 4.  The slope is calculated as {{{1/4}}} or {{{m=1/4}}}


Step 5.  Now use the slope equation of Step 2 and choose one of the given points.  I'll choose point (2,3).   Letting y=y2 and x=x2 and substituting {{{m=1/4}}} in the slope equation given as,


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



{{{ 1/4=(y-3)/(x-2)}}}


Step 6.  Multiply both sides of equation by x-2 to get rid of denomination found on the right side of the equation



{{{ (x-2)/4=(x-2)(y-3)/(x-2)}}}



{{{ (x-2)/4=y-3}}}



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


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


{{{x/4-1/2=y-3}}}


Add 3 from both sides of the equation


{{{x/4-1/2+3=y-3+3}}}


{{{x/4+5/2=y}}}


{{{y=x/4+5/2}}}   This is in slope-intercept form where the slope m=1/4 and y-intercept b=0


Step 8.  See if the other point (6,4) or x=6 and y=4 satisfies this equation


{{{y=x/4+5/2}}}


{{{4=6/4+5/2}}}


{{{4=3/2+5/2}}}  


{{{4=4}}} So the point (6,4) satisfies the equation and is on the line.  In other words, you can use the other point to check your work.


Step 9.  ANSWER:  The equation of the line is {{{y=x/4+5/2}}}


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


{{{-x+4y=10}}}


See graph below to check the above steps.


{{{graph(800,600,-15,15,-10,10, x/4+5/2)}}}


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.


Good luck in your studies!


Respectfully,
Dr J