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Joint equation of pair of lines through `(3,-2)` and parallel to `x^2-4xy+3y^2=0` is

A

` x^(2)+3y^(2)-4xy-14x+24y+45=0`

B

` x^(2)+3y^(2)+4xy-14x+24y+45=0`

C

` x^(2)+3y^(2)+4xy-14x+24y-45=0`

D

` x^(2)+3y^(2)+4xy-14x-24y-45=0`

Text Solution

Verified by Experts

The correct Answer is:
A

Given equation of line is ` x^(2)-4xy+3y^(2)=0`
`:. m_(1)+m_(2)=(4)/(3) and m_(1)m_(2)=(1)/(3)` .
On solving these equations , we get `m_(1)=1,m_(2)=(1)/(3)`
Let the line parallel to given line are `y=m_(1)x+c_(1) and y=m_(2)x+c_(2) `
`:. y=(1)/(3) x+ c_(1) and y=x+c_(2)`
Also, these lines passes through the point (3,-2)
`:. -2=(1)/(3)xx 3 +c_(1)`
`implies c_(1)=-3 `
and `-2=1 xx 3 +c_(2)`
`implies c_(2)=-5 `
`:.` Required equation of pair lines is
(3y-x+9)(y-x+5)=0
`implies x^(2)+3y^(2)-4xy-14x+24y+45=0`
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