The maximum value of p for which the lines 3x-4y=2, 3x-4y=12, 12x+5y=7 and 12x+5y=p constitute the sides of a rhombus is

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MARKS Content Team · Accepted Answer

Correct answer: (A) 33. If the parallelogram is a rhombus, the distance between the pair of parallel sides are equal Hence, 12-2√(3^2+4^2)=±p-7√(12^2+5^2)⇒p-7=±26 ⇒p=33 or -19 So the maximum value of p is 33

The maximum value of $p$ for which the lines $3 x - 4 y = 2, 3 x - 4 y = 12, 12 x + 5 y = 7$ and $12 x + 5 y = p$ constitute the sides of a rhombus is

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