Question 212773
need to find the slope of line containing points

  p1(2,0) and p2(3,-2)


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=2, y1=0, x2=3 and y2=-2 (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=(-2-0)/(3-2)}}}


{{{m=-2/1}}}


{{{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 (3,-2).   Letting y=y2 and x=x2 and substituting m=-3 in the slope equation given as,


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



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


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


Step 5.  Multiply both sides of equation by x-3 to get rid of denomination found on the right side of the equation



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



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



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


{{{-2x+6=y+2}}}


Subtract 2 from both sides of the equation


{{{-2x+6-2=y+2-2}}}


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


{{{y=-2x+4}}}   ANSWER in slope-intercept form.  m=-2 and y-intercept=4


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


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


{{{0=-2*2+4}}}


{{{0=0}}}  So the point (2,0) 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=4}}}


See graph below to check the above steps.


*[invoke describe_linear_equation 2, 1,  4]


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