Question 212931
How do I find the equation of the line which passes through the points
(-3,2) and (0,4)


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 given (-3,2) and (0,4) then x1=-3, y1=2, x2=0 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=(4-2)/(0-(-3))}}}


{{{m=2/3}}}


Step 3.  The slope is calculated as {{{2/3}}}  or {{{m=2/3}}}


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=2/3 in the slope equation given as,


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



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


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


Step 5.  Multiply both sides of equation by {{{x+3}}} to get rid of denominators on right side of equation.


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


Then multiply by 3 to get rid of denominator on left side


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


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


{{{2x+6=3y-6}}}


Add 6 to both sides of the equation to isolate 3y on right side


{{{2x+6+6=3y-6+6}}}


{{{2x+12=3y}}}


{{{3y=2x+12}}} 
  

Divide 3 from both sides of equation to solve for y


{{{3y/3=(2x+12)/3}}} 


{{{y=(2x/3)+4}}}    ANSWER in slope-intercept form.  {{{m=2/3}}} and y-intercept=4


Step 7.  See if the other point (0,4) or x=0 and y=4 satisfies this equation


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


{{{4=(2*0/3)+4}}}


{{{4=4}}}

So the point (0,4) 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


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


OR


{{{-2x+3y=-12}}}


See graph below to check the above steps.


*[invoke describe_linear_equation -2, 3, 12]


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