SOLUTION: need help to find the equation of the line which contains the point (5,-2)and has slope negative two fifths.

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




Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
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=%28y2-y1%29%2F%28x2-x1%29

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%2F5. Substituting above values and variables in the slope equation m yields the following steps:

m=%28y2-y1%29%2F%28x2-x1%29

-2%2F5=%28y-%28-2%29%29%2F%28x-5%29

-2%2F5=%28y%2B2%29%2F%28x-5%29

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

%28-2%2F5%29%28x-5%29=y%2B2

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

-2%28x-5%29=5%28y%2B2%29

Step 6. Now multiply everything out and solve for y

-2x%2B10=5y%2B10

Subtract 10 from both sides to isolate y

-2x%2B10-10=5y%2B10-10

-2x=5y

Divide by 5 to get y

5y%2F5=-2x%2F5

y=-2x%2F5

Note: the above equation can be rewritten as
5y%2B2x=0
And the graph is shown below which is consistent with the above steps.

Solved by pluggable solver: DESCRIBE a linear EQUATION: slope, intercepts, etc
Equation 2+x+%2B+5+y+=+0 describes a sloping line. For any
equation ax+by+c = 0, slope is -a%2Fb+=+-2%2F5.
  • X intercept is found by setting y to 0: ax+by=c becomes ax=c. that means that x = c/a. 0/2 = 0.
  • Y intercept is found by setting x to 0: the equation becomes by=c, and therefore y = c/b. Y intercept is 0/5 = 0.
  • Slope is -2/5 = -0.4.
  • Equation in slope-intercept form: y=-0.4*x+0.
graph%28+500%2C+500%2C+0-8%2C+0%2B8%2C+0-8%2C+0%2B8%2C+-0.4%2Ax%2B0+%29+




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

Respectfully,
Dr J

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