QJEEM16Ma07fb: JEE Main 2016 Mathematics - Class 11 JEEM16Ma07fb Solution: If a variable line drawn through the intersection of the lines…

L₁ : 4x + 3y $-$ 12 = 0

L₂ : 3x + 4y $-$ 12 = 0

Equation of line passing through the intersection of these two lines L₁ and L₂ is

L₁ + $\lambda$ L₂ = 0

$\Rightarrow$ $\,\,\,$ (4x + 3y $-$ 12) + $\lambda$ (3x + 4y $-$ 12) = 0

$\Rightarrow$ $\,\,\,$ x(4 + 3 $\lambda$ ) + y(3 + 4 $\lambda$ ) $-$ 12(1 + $\lambda$ ) = 0

this line meets x coordinate at point A and y coordinate at point B.

$\therefore\,\,\,$ Point A = $\left( {{{12\left( {1 + \lambda } \right)} \over {4 + 3\lambda }},0} \right)$

and Point B = $\left( {0,\,\,{{12\left( {1 + \lambda } \right)} \over {3 + 4\lambda }}} \right)$

Let coordinate of midpoint of line AB is (h, k).

$\therefore\,\,\,$ h = ${{6\left( {1 + \lambda } \right)} \over {4 + 3\lambda }}$ . . . . . (1)

and k = ${{6\left( {1 + \lambda } \right)} \over {3 + 4\lambda }}$ . . . . (2)

Eliminate $\lambda$ from (1) and (2), then we get

6(h + k) = 7 hk

$\therefore\,\,\,$ Locus of midpoint of line AB is ,

6(x + y) = 7xy