question_answer

If a variable line drawn through the intersection of the lines\[\frac{x}{3}+\frac{y}{4}=1\]and\[\frac{x}{4}+\frac{y}{3}=1,\] meets the coordinate axes at A and B, \[(A\ne B),\] then the locus of the midpoint of AB is   JEE Main Online Paper (Held On 09 April 2016)

A) \[7xy=6(x+y)\]

B) \[6xy=7(x+y)\]

C) \[4{{(x+y)}^{2}}-28(x+y)+49=0\]

D) \[14{{(x+y)}^{2}}-97(x+y)+168=0\]

Correct Answer: A

Solution :

                4x + 3y = 12                                                         ....(1) 3x + 4y = 12                                                         ....(2) equation of lines passing through the intersection of the lines \[4x+3y-12+\lambda (3x+4y-12)=0\] \[A=C\left( \frac{12(1+\lambda )}{4+3\lambda },0 \right)\]\[B=\left( 0,\frac{12(1+\lambda )}{3+4\lambda } \right)\] \[\ell n=\frac{6(1+\lambda )}{4+3\lambda }\]                                     ?(3) \[k=\frac{6(1+\lambda )}{3+4\lambda }\]                                             ?(4) from (3) & (4) \[\lambda =\frac{3k-4h}{3h-4k}\]put in (1) 7hk = 6(h + k) hence locus is \[7xy=6(x+y)\]