The locus of the middle points of the segment of a line passing through the point of intersection of the lines $2 x - y + 1 = 0$ and $3 x + 2 y - 1 = 0$ and intercepted between the axes, is:

14 x y = 5 x - y

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14 x y = 5 x + y

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14 x y = x - 5 y

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14 x y = x + 5 y

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Solution

The correct option is A $14 x y = 5 x - y$
Given
$L_{1} = 2 x - y + 1 = 0$ - $(i)$
$L_{2} = 3 x + 2 y - 1 = 0$ - $(i i)$
$4 x - 2 y + 2 = 0$
$3 x + 2 y - 1 = 0$
________________
$7 x + 1 = 0$
$∴ x = \frac{- 1}{7}$
$∴ y = \frac{5}{7}$
$y \frac{- 5}{7} = m (x + \frac{1}{7})$
$∴ C (\frac{- 5 - m}{14 m}, \frac{m + 5}{14})$
$14 k - 5 = m$
$λ = \frac{- 5 - (14 k - 5)}{14 (14 k - 5)} \Rightarrow λ = \frac{- k}{14 k - 5}$
$\Rightarrow 14 λ k - 5 λ = - k$
$∴ 14 x y = 5 x - y$

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The locus of the middle points of the segment of a line passing through the point of intersection of the lines 2x−y+1=0 and 3x+2y−1=0 and intercepted between the axes, is:

The correct option is A 14xy=5x−yGiven L1=2x−y+1=0 - (i)L2=3x+2y−1=0 - (ii)4x−2y+2=03x+2y−1=0________________7x+1=0∴x=−17∴y=57y−57=m(x+17)∴C(−5−m14m,m+514)14k−5=mλ=−5−(14k−5)14(14k−5)⇒λ=−k14k−5⇒14λk−5λ=−k∴14xy=5x−y

The locus of the middle points of the segment of a line passing through the point of intersection of the lines $2 x - y + 1 = 0$ and $3 x + 2 y - 1 = 0$ and intercepted between the axes, is: