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slope 3

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vannu
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slope 3 [#permalink] New post   14 Jul 2009, 14:09
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The x-intercept of a line passing through points (5,-2) and (6,9) is _________________________
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57/11


please explain.



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Re: slope 3 [#permalink] New post   14 Jul 2009, 14:16
You have to use a system of 2 equations. Plug in your given points into the formula for a line: y=mx+b

A: -2=m(5)+b
B: 9=m(6)+b

Subtract the bottom from the top:
-11 = -m
m = 11

Plug back into a to find the full equation:
A: -2=11(5)+b
b=-57

x-int. is where the line crosses the x-axis which means y=0. Plug this in:
y=11x-57
0=11x-57
x=57/11
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Re: slope 3 [#permalink] New post   15 Jul 2009, 08:07
given 2 points there is a direct eqn for a line

\(y-y2 = y2-y1/x2-x1 (x-x1)\)

using this , you get the eqn y+2 = 11(x-5)

for x-interecept you put y=0 and you get x = 57/11
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Re: slope 3 [#permalink] New post   15 Jul 2009, 09:42
skpMatcha wrote:
given 2 points there is a direct eqn for a line

\(y-y2 = y2-y1/x2-x1 (x-x1)\)

using this , you get the eqn y+2 = 11(x-5)

for x-interecept you put y=0 and you get x = 57/11


I think your equation is wrong. I tried plugging in numbers and didn't get the correct equation. I believe the left side of the equation should be y - y1, not y - y2. Please confirm.
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Re: slope 3 [#permalink] New post   15 Jul 2009, 10:00
you are correct, you can use either of the points here but you need to stick with one point and use x & y coordinates of that point !

thanks for the correction!
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Re: slope 3 [#permalink] New post   15 Jul 2009, 10:07
\(y-y2= y2-y1/x2-x1(x-x2)\)

or

\(y-y1= y2-y1/x2-x1(x-x1)\)
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Re: slope 3 [#permalink] New post   15 Jul 2009, 18:50
Thanks for the replies.



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Re: slope 3 [#permalink]
15 Jul 2009, 18:50
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