Question 217928
Find the slope if it exists of the line containing the pair of points: 

(4,5) and (8,-3)


However, here are the steps showing you how you can check your work with one of the points.


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


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


where for our example is x1=4, y1=5, x2=8 and y2=-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=(-3-5)/(8-4)}}}


{{{m=-8/4}}}


{{{m=-2}}}


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


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


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



{{{ -2=(y-5)/(x-4)}}}


{{{ -2=(y-5)/(x-4)}}}


Step 5.  Multiply both sides of equation by x-4 to get rid of denomination found on the right side of the equation



{{{ -2(x-4)=(x-4)(y-5)/(x-4)}}}



{{{ -2(x-4)=y-5}}}



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


{{{-2x+8=y-5}}}


Add 5 to both sides of the equation


{{{-2x+8+5=y-5+5}}}


{{{-2x+13=y}}}


{{{y=-2x+13}}}   ANSWER in slope-intercept form.  m=-2 and y-intercept=13


Step 7.  See if the other point (8,-3) or x=8 and y=-3 satisfies this equation


{{{y=-2x+13}}}


{{{-3=-2*8+13}}}  which is a true statement


{{{-3=-3}}}  So the point (8,-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+y=13}}}


See graph below to check the above steps and note the slope and the y-intercept at x=0


{{{graph(400,400, -5,15,-5,15, -2x+13)}}} 


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