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JEE MainMathematicsToughStraight Lines

If a variable line, drawn through the point of intersection of
the straight lines 3x + 2y - 6 = 0 and 2x + 3y - 6 = 0, meets the
x-axis and y-axis in A and B respectively, then the locus of point P
such that PB: PA = 3: 1

Show answer and solution
Let the straight line be 3x + 2y - 6 + λ\lambda (2x + 3y -
6) = 0
∴\therefore Coordinates of A and B are
(6+6λ3+2λ,0)\left( \frac{6 + 6\lambda}{3 + 2\lambda},0 \right) and (0,6+6λ2+3λ)\left( 0,\frac{6 + 6\lambda}{2 + 3\lambda} \right)
∴\therefore Coordinates of P are
(34(6+6λ3+2λ),14(6+6λ2+3λ))\left( \frac{3}{4}\left( \frac{6 + 6\lambda}{3 + 2\lambda} \right),\frac{1}{4}\left( \frac{6 + 6\lambda}{2 + 3\lambda} \right) \right)
∴\therefore Locus P is given by eliminating λ\lambda between
6+6λ3+2λ=4x3\frac{6 + 6\lambda}{3 + 2\lambda} = \frac{4x}{3} and 6+6λ2+3λ=4y\frac{6 + 6\lambda}{2 + 3\lambda} = 4y
 i.e. 12x−1818−8x=8y−66−12y\frac{12x - 18}{18 - 8x} = \frac{8y - 6}{6 - 12y}
i.e. 72x - 108 - 144xy + 216y = 144y - 64xy - 108 + 48x
i.e. 24x + 72y - 80xy = 0
3x + 9y - 10xy = 0

Source: Grass-G / M-11 Mathematics Paper-2 1Nov.docx

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