Question

# If point $$P(h,k)$$ lies on the line $$2x + 3y = 5$$ such that $$|PA .PB|$$ is maximum where $$A(2,3)$$ and $$B(1,2)$$ then the value of $$(3h + 2k)$$ is

A
3
B
2
C
6
D
4

Solution

## The correct option is C $$4$$$$| PA - PB| < AB$$$$|PA PB|_{max} = AB$$In this case P, A & B are collinearSo 'P' is intersecting point of these two lines as shown in figureEquation of AB: $$\left.\begin{matrix}y-x & =1\\ 2x+3y & =5\end{matrix}\right\}$$$$\Rightarrow$$ Intersection point $$P\left (\dfrac {2}{5}, \dfrac {7}{5}\right )$$$$\therefore 3h+2k=4$$Mathematics

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