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## Homework Statement

There is a point A(-5,-4) through which th line L1 passes.

Straight lines

1)x+3y+2=0

2)2x+y+4=0

3)x-y-5=0

intersect L1 at B,C,D respectively! It is known that:

[tex]\frac{15^{2}}{AB^{2}}+\frac{10^{2}}{AC^{2}}=\frac{6^{2}}{AD^{2}}[/tex]

Find equation of Line L1.

## Homework Equations

General equation of straight line

y=mx+c

Distance(r) of a point(x,y) from A along the line L1

[tex]\frac{x+5}{cos \theta}=\frac{y+4}{sin \theta}=r[/tex]

where theta is the angle made with positive x axis by L1

## The Attempt at a Solution

Let L1: y=mx+c

(-5,-4) lies on the line

therefore:

L1: y=mx+4m-5

where m is slope

I solved this equation simultaneously with the three given above to get coordinates of intersecting points in terms of 'm'.

I used the distance along line formula to get AB,AC,AD in terms of m. I plug it back to

[tex]\frac{15^{2}}{AB^{2}}+\frac{10^{2}}{AC^{2}}=\frac{6^{2}}{AD^{2}}[/tex]

what I get is the following beautiful equation in 'm'

[tex]\frac{225(3m+1)^{2}}{(2m-5)^2}+\frac{100(1-m)^{2}}{(9m-5)^2}=\frac{9(m+2)^{2}}{(2m-5)^2}[/tex]

First of all, I cannot solve for 'm'! Second thing is that the method I adopted was a tedious one. Someone please help me with a more clever and short technique.