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Analytical Geometry

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28 Oct 2007 17:58:11 IST

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please help me to solve this question.

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Q. a line through A(-5,-4) meets the lines x+3y+2=0, 2x+y+4=0 and x-y-5=0 at the points, B, C, D respectively. if (15/AB)²+(10/AC)²=(6/AD)²find the equation of the line.

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sowjanya gudipati

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28 Oct 2007 21:38:04 IST

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the equation of line is 2x+3y+22=0

solution:

let the equation of line be y+4=m(x+5)

solving x+3y+2=0 and y+4=m(x+5)

we get B=(10-15m/3m+1,3m-4/3m+1)

solving 2x+y+4=0 and y=4=m(x+5)

we get C=(-5m/2+m,6m-8/2+m)

solving x-y-5=0 and y+4=m(x+5)

we get D=(5m+1/1-m,10m-4/1-m)

given condition is (15/AB)²+(10/AC)²=(6/AD)²

sowjanya gudipati

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28 Oct 2007 21:45:26 IST

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hey i am continuing here plz have a look even at this

(15/AB)²=(3m+1)²/m²+1

(10/AC)²=(2+m)²/1+m²

(6/AD)²=(1-m)²/1+m²

using these values u'll get the value of m=-2/3

therefore equation of line is y+4=m(x+5)

2x+3y+22=0

Bipin Dubey

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Joined: 23 Jan 2007

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29 Oct 2007 12:12:37 IST

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Well done sowjanya_reddygudipati.

shohitaa choudhry

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29 Oct 2007 23:04:42 IST

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the question can also be done using parametric form
let the point on the line be (-5+rcos theta, -4+rsin theta)
then let for each eqn put these points and solve for r
which will be AB, AC , AD for the 3 eqns
now put the value of ab , ac , ad in the given relation and solve

the condition is satisfied

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