Solving for Line L1: Finding equation of Line L1

[ritwik06]

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:

\frac{15^{2}}{AB^{2}}+\frac{10^{2}}{AC^{2}}=\frac{6^{2}}{AD^{2}}

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
\frac{x+5}{cos \theta}=\frac{y+4}{sin \theta}=r

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
\frac{15^{2}}{AB^{2}}+\frac{10^{2}}{AC^{2}}=\frac{6^{2}}{AD^{2}}

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

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

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.

 

[ritwik06]

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:

\frac{15^{2}}{AB^{2}}+\frac{10^{2}}{AC^{2}}=\frac{6^{2}}{AD^{2}}

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
\frac{x+5}{cos \theta}=\frac{y+4}{sin \theta}=r

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
\frac{15^{2}}{AB^{2}}+\frac{10^{2}}{AC^{2}}=\frac{6^{2}}{AD^{2}}

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

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

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.

Are my equations in m correct?

Just give it a thought to my second method. Will converting the coordinates of the intersecting point to parametric forms an then applying the distance along a line from a point formula help??

Distance(r) of a point(x,y) from A along the line L1
\frac{x+5}{cos \theta}=\frac{y+4}{sin \theta}=r

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

 

[Mark44]

I replied before I understood what you were trying to do. Let me think about it further.

 

[Mark44]

Your equation for L1 is incorrect. It's given that A(-5, -4) is on line L1, so its equation is y - (-4) = m(x - (-5), or
y + 4 = mx + 5m, or
y = mx + 5m - 4

You have

L1: y=mx+4m-5

 

[ritwik06]

Your equation for L1 is incorrect. It's given that A(-5, -4) is on line L1, so its equation is y - (-4) = m(x - (-5), or
y + 4 = mx + 5m, or
y = mx + 5m - 4

You have

Oh, I am sorry. My calculations were wrong. but still I would get a nasty equation in m. Wouldnt I?
Isnt there any simple method?

 

[Mark44]

I don't see any simpler approach.