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Intern
Joined: 14 Oct 2009
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Can anyone help me with this problem? Find the equation of a line passing through (2,5) and the sum of whose intercepts on the axes is 14.
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Manager
Joined: 04 Nov 2009
Posts: 66
Schools: London Business School (int)
WE 1: Research
WE 2: Corporate Strat
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Re: Geometry Lines [#permalink]
04 Nov 2009, 10:41
I get 2 possible answers, after a rather drawn out calculation.
Assume line is y=mx+c so for (2,5): 5=2m+c ---------- (A)
Sum of intercepts = 14 i.e. y-intercept (when x=0) + x-intercept (when y=0) = 14 Put x=0 to get y-intercept = c Put y=0 to get x-intecept = -(c/m)
So c - c/m = 14 ----------------(B)
So we have 2 eqns A and B for 2 variables - we can solve for each
We can rearrange eqn B to get c = 14m/(m-1)
Put that for c in A to get: 5 = 2m +14m(m-1) 2m^2 + 7m + 5 = 0 I used the formula for quadratic roots to get the solution for m = -5/2 or m = -1
We know 5=2m+c For m = -5/2, c=10. so the line is y= -5x/2 + 10 Check: sum of intercepts = 10 + 4 = 14
For m = -1, c = 7 the line is y = -x + 7 Check: sum of intercepts= 7 + 7 = 14.
So two possible solutions. Is there a faster way to do this?
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Re: Geometry Lines
[#permalink]
04 Nov 2009, 10:41
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