Question 220921
Find an equation of the line containg the given pair of points (-2,-9) and (-6,-4)


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=-9, 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 2.  Substituting the above values in the slope equation gives


{{{m=(-4-(-9))/(-6-(-2))}}}


{{{m=-5/4}}}


Step 3.  The slope is calculated as -5/4 or m=-5/4


Step 4.  Now use the slope equation of step 1 and choose one of the given points.  I'll choose point (-2, -9).   Letting y=y2 and x=x2 and substituting m=-3 in the slope equation given as,


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



{{{ -5/4=(y-(-9))/(x-(-2))}}}


{{{ -5/4=(y+9)/(x+2)}}}


Step 5.  Multiply both sides of equation by x+2 to get rid of denomination found on the right side of the equation



{{{ (-5/4)(x+2)=(x+2)(y+9)/(x+2)}}}



{{{ -(5/4)(x+2)=y+9}}}



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


{{{-5x/4-5/2=y+9}}}


Subtract 9 from both sides of the equation


{{{-5x/4-5/2-9=y+9-9}}}


{{{-5x/4-23/2=y}}}


{{{y=-5x/4-23/2}}}   ANSWER in slope-intercept form.  m=-5/4 and y-intercept=-23/2


Step 7.  See if the other point (-6,-4) or x=-6 and y=-4 satisfies this equation


{{{y=-5*(-6)/4-23/2}}}


{{{-4=15/2-23/2}}}


{{{-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.


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


{{{5x+4y=-46}}}


See graph below to check the above steps.


{{{graph(400,400, -15, 5, -15, 5, -5x/4-23/2)}}}


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