Equation of line : y=mx + c
Gradient = (-5-0)/(-3-3)
= 5/6
Thus y=(5/6)x + c
Sub. in Point (3, 0)
0= (5/6)(3) + c
c= (-5/2)
thus the line is y = (5/6)x - (5/2)
2006-09-04 20:54:11
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answer #1
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answered by maczh2002 2
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First, we need the slope. This is given by m = dy/dx. dy = 0 - -5, or 5; dx = 3 - -3 or 6. So, in the straight line equation y = mx + c, we have m = 5/6. It is convenient to use the second point to determine c: 0 = (5/6)3 + c, so c is obviously -2.5. The complete equation is then
y = (5/6)x - 2.5. Checking, we have:
-5 = (5/6) -3 - 2.5 -- OK;
0 = (5/6) 3 - 2.5 -- OK. We're done.
2006-09-05 04:01:03
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answer #2
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answered by Anonymous
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Well... let it be a*x+b=y the equation of the line, where a, b real numbers. Then, if the line passes through the spots (-3,-5) and (3,0), this equation should be true for these pairs of (x,y). So, you must solve the sustem:
a*(-3)+b=-5
a*3+b=0
That is equal to:
b=a*(-3)
a*(-3)+a*(-3)=-5
That means:
a*(-6)=-5
b=a*(-3)
And the solution is:
a=5/6
b=-3*5/6=-5/2
So nthe equation of the line passing through the spots (-3,-5) and (3,0) is: y=(5/6)*x-5/2.
2006-09-05 04:02:38
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answer #3
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answered by Leonidas Nikolakis 3
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-5-0
m= -------- =5/6 so the line would be:
-3-3
y-0 = 5/6 (x - 3 ) so :
y = 5/6 x -5/2
2006-09-05 05:50:17
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answer #4
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answered by Sarah 2
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It is (x+3)/(y+5)=(-3-3)/(-5), which simplifies to:
5x-6y-15=0
2006-09-05 03:56:05
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answer #5
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answered by Anonymous
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y=5/6x+-3
or
y=5/6x-3
I didn't get it that well on the y-intercept.
2006-09-09 01:25:10
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answer #6
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answered by _anonymous_ 4
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yup..thts guy's method is correct though i didnt check the calculations..
2006-09-05 03:57:02
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answer #7
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answered by Anonymous
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