Question 212520
Need help to find the equation of the line which contains the point (5,-2)and has slope  negative two fifths.


Step 1.  The slope m is given as


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


Step 2.  Let (x1,y1)=(5,-2) or x1=5 and y1=-2 .  Let other point be ((x2,y2)=(x,y) or x2=x and y2=y.


Step 3.  Now we're given {{{m=-2/5}}}.  Substituting above values and variables in the slope equation m yields the following steps:


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


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


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


Step 4.  Multiply x-5 to both sides to get rid of denominator on right side of equation.


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


Step 5.  Now multiply 5 to both sides of equation to get rid of denominator on left side of equation.


{{{-2(x-5)=5(y+2)}}} 


Step 6.  Now multiply everything out and solve for y


{{{-2x+10=5y+10}}}


Subtract 10 from both sides to isolate y


{{{-2x+10-10=5y+10-10}}}


{{{-2x=5y}}}


Divide by 5 to get y


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


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


Note:  the above equation can be rewritten as 

{{{5y+2x=0}}}

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


*[invoke describe_linear_equation 2, 5, 0 ]



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


Respectfully,
Dr J


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