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 \[7xy=6(x+y)\]You need to login to perform this action.
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