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# Geometry Lines

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Joined: 14 Oct 2009
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04 Nov 2009, 08:24
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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Joined: 04 Nov 2009
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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?
Re: Geometry Lines   [#permalink] 04 Nov 2009, 10:41
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