A line 'L' is drawn from (4, 3) to meet the lines L 1 : 3x + 4y + 5 = 0 and L 2 : 3x + 4y + 15 = 0 at points A and B respectively. From 'A' a line, perpendicular to L is drawn meeting the line L 2 at A 1 . Similarly, from point 'B' a line, perpendicular to L is drawn meeting the line L 1 at B 1 . Thus parallelogram AA 1 BB 1 is formed. Least value of area of parallelogram AA 1 BB 1 is.....
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Sol.
L 1 and L 2 are parallel lines.
Line L divides this parallelogram in two triangles of equal area. Altitudes of these triangle's is fixed = h = 
= 2
base length of each triangle is = h tan θ + h cot θ
= h(tan θ + cot θ ) = 
= 
for area to be least, this base length must be least, so sin2 θ = 1. So θ = 45º
So least area = 2.(1/2.2h.h) = 2h 2 = 8 sq. units
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