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The area of parallelogram whose sides are along the straight lines $y = 3 x + 5; y = 3 x + 2; y = 5 x + 4; y = 5 x - 1$ is

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$\frac{15}{2} s q . u n i t s$

15 sq.units

$\frac{15}{\sqrt{10} \sqrt{26}} s q . u n i t s$

$\sqrt{10} \sqrt{26} s q . u n i t s$

answer is A.

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Detailed Solution

$y = m x + a; y = m x + b; y = n x + c; y = n x + d,$ are the sides of a parallelogram

$∴ a r e a = |\frac{(a - b) (c - d)}{(m - n)}|$ $= |\frac{(5 - 2) (4 + 1)}{3 - 5}| = \frac{15}{2}$

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