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November 28, 2014

November 28, 2014

Posted by **Bob** on Wednesday, January 14, 2009 at 11:40am.

With a slope of 2/3 that goes through the point (4, 5)

With a slope of (-1/3) that goes through the point (2, -4)

With a slope of 3 that goes through the point (-5, -1)

- algebra II -
**DrBob222**, Wednesday, January 14, 2009 at 12:19pmI would do the first one this way.

The equation for a straight line is

y = mx + b where m is the slope and b is the y intercept. If you want the line to go through (4,5) and slope = m = 2/3, then

set y = 5, m = 2/3, and x = 4

5=(2/3)*4 + b

Solve for b.

Multiply both sides by 3 to clear the fraction.

15 = 8 + 3b

3b = 7 and b = 7/3 so th equation is

y = (2/3)x + (7/3) or to get rid of the fractions, it would be

3y = 2x + 7

CHECK:

plug in 4 for x and 5 for y

3*5=2*4 + 7

15 = 8 + 7

15 = 15.

The others are done the same way.

Check my work.

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