SOLUTION: Please help!!! Thank you!! A line passes through the point (x,y) =(-2,-3) and has a slope of -4 . Write an equation for this line.

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Question 215168: Please help!!! Thank you!!
A line passes through the point (x,y) =(-2,-3) and has a slope of -4 . Write an equation for this line.
Found 2 solutions by stanbon, drj:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
A line passes through the point (x,y) =(-2,-3) and has a slope of -4 . Write an equation for this line.
-----------------------
Form is y = mx +b ; solve for "b":
-3 = -4*-2 + b
-3 = 8 + b
b = -11
--------------
Equation:
y = -4x - 11
=======================
Cheers,
Stan H.
Answer by drj(1380)   (Show Source): You can put this solution on YOUR website!
A line passes through the point (x,y) =(-2,-3) and has a slope of -4 . Write an equation for this line.

Step 1. The slope m is given as



Step 2. Let (x1,y1)=(-2,-3) 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 . Substituting above values and variables in the slope equation m yields the following steps:







Step 4. Multiply x+2 to both sides to get rid of denominator on right side of equation.





Step 5. Now subtract 3 to both sides of equation to solve for y.





Step 6. ANSWER. y=-4x-11

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. -11/4 = -2.75.
  • Y intercept is found by setting x to 0: the equation becomes by=c, and therefore y = c/b. Y intercept is -11/1 = -11.
  • Slope is -4/1 = -4.
  • Equation in slope-intercept form: y=-4*x+-11.




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

Respectfully,
Dr J

For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra.

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