A line L is drawn from P(4,3) to meet the lines L1 and L2 given by 3x+4y+5=0 and 3x+4y+15=0 at points
A and B , respectively. From A¯ , a line perpendicular to L is drawn meeting the line L2 at A1. Similarly, from point
B , a line perpendicular to L is drawn meeting the line L1 at B1 . Thus, a parallelogram AA1BB1 is formed. Then
the the equation of L so that the area of the parallelogram AA1BB1 is the least is
x−7y+17=0
7x+y+31=0
x−7y−17=0
x+7y−31=0
The given lines L1 and L2 are parallel and the distance between them (BC or AD) is (15−5)5=2 units. Let
∠BAC=θ . So, AB=BCcosecθ=2cosecθ and AA1=ADsecθ=2secθ . Now, the area of parallelogram AA1BB1 is
Δ=AB×AA1=4secθcosecθ =8sin2θ
Clearly, Δ is the least for θ=π/4 . Let the slope of AB be m .
then,
1=m+341−(3m4)
or 4m+3=±(4−3m) or m=17 or −7
Hence, the equation of L is
x−7y+17=0 or 7x+y−31=0