SOLUTION: Find the slope-intercept form of the equation of the line that passes through the given points. (2, 2) (6, -2/3)

Question 217337: 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. We need to find an equation in slope intercept form given as y=mx+b where m is the slope and b is the y-intercept at point (0,b).

Step 2. 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 3. Substituting the above values in the slope equation gives

Step 4. The slope is calculated as or .

Step 5. 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=60 in the slope equation given as,

Step 6. Multiply both sides of equation by 3(x-2) to get rid of denomination found on the right side of the equation

Step 7. Now simplify and put the above equation into slope-intercept form.

Add 6 to both sides of the equation

Divide by 3 to both sides of the equation

ANSWER in slope-intercept form is where slope m=-2/3 and y-intercept=10/3

Step 8. See if the other point (6,-2/3) or x=5 and y=-2/3 satisfies this equation

So the other point satisfies this equation and lies 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. Note the slope and y-intercept as well as the x-intercept.

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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