JEE Mains · Maths · STD 11 - 9. straight line
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
- A \(7xy=6 (x+y)\)
- B \(4 (x + y)^2 - 28 (x+y) + 49 = 0\)
- C \(6xy=7 (x + y)\)
- D \(14 (x+y)^2 - 97 (x+y)+ 168 = 0\)
Answer & Solution
Correct Answer
(A) \(7xy=6 (x+y)\)
Step-by-step Solution
Detailed explanation
\({L_1}:4x + 3y - 12 = 0\) \({L_2}:3x + 4y - 12 = 0\) \({L_1} + \lambda {L_2} = 0\) \(\left( {4x + 3y - 12} \right) + \lambda \left( {3x + 4y - 12} \right) = 0\) \(x\left( {4 + 3\lambda } \right) + y\left( {3 + 4\lambda } \right) - 12\left( {1 + \lambda } \right) = 0\) Point…
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