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The line L is drawn through the point (1, 1) to intersect $L_{1} : x + 2 y = 1$ and $L_{2} : x + 2 y = 3$ at P and Q respectively. Line perpendicular to L from P intersects $L_{2}$ at R and line perpendicular to L from Q intersects $L_{1}$ at S. If the area of parallelogram PRQS is least, then which of the following is/are correct?

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the equation of line L can be y + 3x = 4

the equation of the L can be 3y – x = 2

the least area of PQRS is $\frac{8}{5}$ square unit

the equation of line L can be 3y – x = 4

answer is A, B, C.

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Detailed Solution

For area of parallelogram PQRS to be least line L makes angle $\frac{π}{4}$ with $L_{1}$ and $L_{2}$

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