SOLUTION: Write the slope intercept form of the equation of the line described. through:(-2,4), parallel to y=-5/2x+5

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Question 224557: Write the slope intercept form of the equation of the line described.
through:(-2,4), parallel to y=-5/2x+5
Answer by drj(1380)   (Show Source): You can put this solution on YOUR website!
Write the slope intercept form of the equation of the line described.
through:(-2,4), parallel to y=-5x/2+5

Step 1. We can find the slope by recognizing that parallel lines have the same slope. Since is in slope-intercept form given as y=mx+b where the slope m=-5/2 and the y-intercept b=5 when x=0 or at point (0,b) or (0,5).

Step 2. Now we have to find the line with slope m=-5/2 going through point (-2,4).

Step 3. Given two points (x1,y1) and (x2,y2), then the slope m is given as



Step 4. Let (x1,y1)=(-2,4) or x1=-2 and y1=4. Let other point be (x2,y2)=(x,y) or x2=x and y2=y.

Step 5. Now we're given . Substituting above values and variables in the slope equation m yields the following steps:





Step 6. Multiply x+2 to both sides to get rid of denominator on right side of equation.





Step 7. Add 4 to both sides of the equation





Step 7. ANSWER: The equation in slope-intercept form is


Note: the above equation can be rewritten as



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 describes a sloping line. For any
equation ax+by+c = 0, slope is .
  • X intercept is found by setting y to 0: ax+by=c becomes ax=c. that means that x = c/a. -2/5 = -0.4.
  • Y intercept is found by setting x to 0: the equation becomes by=c, and therefore y = c/b. Y intercept is -2/2 = -1.
  • Slope is -5/2 = -2.5.
  • Equation in slope-intercept form: y=-2.5*x+-1.



I hope the above steps and explanation were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

And good luck in your studies!

Respectfully,
Dr J
http://www.FreedomUniversity.TV
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