Question 218680: Find the slope-intercept form of the equation of the line that passes through the given points.
(2, 2)
(6, -2/3)
Answer by drj(1380) (Show Source):
You can put this solution on YOUR website! Find the slope-intercept form of the equation of the line that passes through the given points.
(2, 2)
(6, -2/3)
Step 1. The slope of the line m is given as
where for our example is x1=2, y1=2, x2=6 and y2=-2/3 (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 -2/3 or m=-2/3
Step 4. Now use the slope equation of step 1 and choose one of the given points. I'll choose point (2,2). Letting y=y2 and x=x2 and substituting m=-3 in the slope equation given as,
Step 5. Multiply both sides of equation by x-2 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.
Add 2 from both sides of the equation
ANSWER in slope-intercept form. m=-2/3 and y-intercept b=10/3
Step 7. See if the other point (6,-2/3) or x=2 and y=0 satisfies this equation
So the point (6,-2/3) satisfies the equation and is on the line. 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. 10/2 = 5.
- Y intercept is found by setting x to 0: the equation becomes by=c, and therefore y = c/b. Y intercept is 10/3 = 3.33333333333333.
- Slope is -2/3 = -0.666666666666667.
- Equation in slope-intercept form: y=-0.666666666666667*x+3.33333333333333.
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I hope the above steps 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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