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

Question

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

Infinity Learn · Accepted Answer

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=BCcosec$$⁡θ$$=2cosec$$⁡θ and $$AA1=ADsec$$⁡θ$$=2sec$$⁡θ . Now, the area of parallelogram $$AA1BB1$$ is Δ$$=AB$$×$$AA1=4sec$$⁡θcosec⁡θ  $$=8sin$$⁡$$2$$θ Clearly, Δ is the least for θ$$=$$π$$/4$$ . Let the slope of AB be m . then,$$1=m+341$$−$$(3m4) or  $$4m+3=$$±(4$$−$$3m) or $$m=17$$ or −$$7Hence$$, the equation of L is x−$$7y+17=0$$ or  $$7x+y$$−$$31=0$$

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