Question 212929
How do I find the equation of the line which passes through the points


   (-1,2) and (-2,0)


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  (-1,2) and (-2,0)  x1=-1, y1=2, x2=-2 and y2=0 (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=(0-2)/(-2-(-1))}}}


{{{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 (-1,2).   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-(-1))}}}


{{{2=(y-2)/(x+1)}}}


Step 5.  Multiply both sides of equation by x+1 to get rid of denomination found on the right side of the equation



{{{ 2(x+1)=(x+1)(y-2)/(x+1)}}}



{{{2(x+1)=y-2}}}



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


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


Add 2 to both sides of the equation


{{{2x+2+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