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August 9th, 2008, 08:21 AM   #1
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straight line Q

the point (2,3) lies on the line ax + by = 4 which has a gradient of 2.
find the values of the constants a and b and hence, in its general form, the equation of the straight line.

i know the equation of the straight line is y = 2x - 1, and in its general form [ax + by + c = 0] is 2x - y - 1 = 0.

but i don't quite get how to work out the constants. could someone explain this please also because its not explained in my textbook

thanks!
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August 9th, 2008, 09:24 AM   #2
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Re: straight line Q

Quote:
i know the equation of the straight line is y = 2x - 1....
but i don't quite get how to work out the constants.
How did you then find that the equation is y=2x-1 without first finding the values of constants ...which is really the essence of the problem. Can you show what you have done so far?

The text must contain some information. If you re-arrange the equation ax + by = 4 to "slope-intercept form", what do you get? If you substitute the point (2,3) into the equation what do you get? Both of these will give you two equations in a and b. Solve them for the solution.
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August 9th, 2008, 09:29 AM   #3
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Re: straight line Q

sorry, i didn't work it out, should've mentioned this is the answer from the back of the book.

the equation is 2x - y = 1

or y = 2x - 1 [when rearranged]
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