Question number 5 bhi..

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the solution is way too big tbh.

note that area of parallelogram A= 2*area of triangle APB= 2k (say). also, the two given lines are parallel with slopes ¾.

so we need to minimise k, the area of triangle APB, which can be written as ½ * PB * perpendicular distance b/w the given lines. since the perpendicular distance b/w the given parallel lines is a constant, so we need to minimise PB.

let angle APB= x then you can drop perpendicular from A onto the other line and write

PB= dcotx + dtanx= d(T+1/T), where d is perpendicular distance b/w the given lines and noting that angle b/w AP and AB is 90. for PB= d(T+1/T) to be minimum, T needs to be 1 by AM-GM. or tanx= 1 so x= 45. if the foot of perpendcular from A onto other line is C, then AC= CP= CB as x is 45 which implies 90 – x is also 45. note that AC= d which you can easily calculate using the formula for the perpendicular distance b/w 2 parallel lines. so you can find B and P (by first determining coordinates of C and then using dist formula for CB= d)

further, P and B will be interchangeable (due to the possiblity of a symmetric pllogram), i.e, B will not b unique. hence there can be 2 eqns of AB. similarly, there will be 2 sets of coordinates possible.