IIT JEE , AIEEE Forum - Mathematics , Algebra

raulrag009

Q2

Here

$6x^{2}+17xy+12y^{2}=0\\\\ (3x+4y)(2x+3y)=0\\\\$

Thus

$(3x+4y)=0\\\\(2x+3y)=0\\\\$

are the two lines.

eqn of perpendicular lines is given by

$4x-3y+\alpha=0\\\\ 3x-2y+\mu=0\\\\ where \;\alpha\;and\;\mu\;are\;constants$

since they pass through (2,1)

(2,1) should surely satisfy them

$\alpha=-5\\\\ \mu=-4$

thus eqn of pair of lines is

$(4x-3y-5)(3x-2y-4)=0\\\\ 12x^{2}-9xy-31x+14y+6y^{2}+20=0$

forgive me for calculation mistakes