Question 223927
How do I graph a line using the slope of m=1/3 and point (3,-1)?

Step 1.  The slope m is given as


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


Step 2.  Let (x1,y1)=(3,-1) or x1=2 and y1=1 .  Let other point be ((x2,y2)=(x,y) or x2=x and y2=y.


Step 3.  Now we're given {{{m=2}}}.  Substituting above values and variables in the slope equation m yields the following steps:


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


{{{1/3=(y-(-1))/(x-3)}}}


Step 4.  Multiply x-3 to both sides to get rid of denominator on right side of equation.


{{{(x-3)/3=y+1}}} 


{{{x/3-3/3=y+1}}} 


{{{x/3-1=y+1}}}


Step 5.  Now add -1 to both sides of equation to solve for y.


{{{x/3-1+(-1)=y+1+(-1)}}}


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


The slope m=1/3 and y-intercept is -2 at point (0,-2). 



Now you have two points  (0,-2) and   (3,-1) to draw a line.


Note:  the above equation can be rewritten as 

{{{x/3-y=2}}}  Note in this form: when y=0 then x=6 at point (6,0).  


And the graph is shown below which is consistent with the above steps.


*[invoke describe_linear_equation 1/3, -1, 2 ]



I hope the above steps were helpful.  Good luck in your studies!


Respectfully,
Dr J


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