The sum of distances of the point P lying inside the triangle formed by lines y = 0; 4x – 3y = 0 and 3x + 4y – 9 = 0 is 3. Then the locus of P is

7x + 6y – 24 = 0

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x – 2y + 24 = 0

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7x – 4x + 6 = 0

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x – 2y – 6 = 0

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Solution

The correct option is D
x – 2y – 6 = 0

$∴ | k | + \frac{| 3 h + 4 k - 9 |}{5} + \frac{| 4 h - 3 k |}{5} = 3 ∴ k + \frac{9 - 3 h - 4 k}{5} + \frac{4 h - 3 k}{5} = 3 h - 2 k = 6$
$∴$ Required locus is x – 2y = 6

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{(\frac{15}{A B})}^{2} + {(\frac{10}{A C})}^{2} = {(\frac{6}{A D})}^{2}

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