Question 213249: How do I find the slope of the line containing the points?
p1(0,2) and p2(4,-2)
Answer by drj(1380)   (Show Source):
You can put this solution on YOUR website! How do I find the slope of the line containing the points?
p1(0,2) and p2(4,-2)
However, here are the steps showing you how you can check your work with one of the points.
Step 1. The slope of the line m is given as
where for our example is x1=0, y1=2, x2=4 and y2=-2 (think of  ). You can
choose the points the other way around but be consistent with the x and y coordinates. You will
get the same result.
Step 2. Substituting the above values in the slope equation gives
Step 3. The slope is calculated as -1 or m=-1
Step 4. Now use the slope equation of step 1 and choose one of the given points. I'll choose point (0,2). Letting y=y2 and x=x2 and substituting m=-1 in the slope equation given as,
Step 5. Multiply both sides of equation by x to get rid of denomination found on the right side of the equation
Step 6. Now simplify and put the above equation into slope-intercept form.
 ANSWER in slope-intercept form. m=-1 and y-intercept=2
Step 7. See if the other point (4,-2) or x=4 and y=-2 satisfies this equation
 so it's ok...
In other words, you can use the other point to check your work.
Note; above equation can be also be transform into standard form as
See graph below to check 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. 2/1 = 2.
- Y intercept is found by setting x to 0: the equation becomes by=c, and therefore y = c/b. Y intercept is 2/1 = 2.
- Slope is -1/1 = -1.
- Equation in slope-intercept form: y=-1*x+2.
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I hope the above steps were helpful.
And good luck in your studies!
For free Step-By-Step Videos on Introduction to Algebra, please visit
http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit
http://www.FreedomUniversity.TV/courses/Trigonometry.
Respectfully,
Dr J
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