Two lines draw through the point P(4, 0) divide the area bounded by the curves $y = \sqrt{2} \sin \frac{\pi x}{4}$ and x-axis, between the line ax = 2, and x = 4, in to three equal parts. Sum of the slopes of the drawn lines is equal to

$-\frac{2 \sqrt{2}}{\pi}$

Area bounded by $y=\sqrt{2} . \sin \frac{\pi x}{4}$ and x-axis between the lines x = 2 and x= 4,

$\Delta=\sqrt{2} \int_2^4 \sin \frac{\pi x}{4} d x=-\left.\frac{4 \sqrt{2}}{\pi} . \cos \frac{\pi x}{4}\right|_2 ^4 $

$= \frac{4 \sqrt{2}}{\pi}$ sq. units.

Let the drawn lines are $L_1: y-m_1(x-4)=0$ and $L_2: y-m_2(x-4)=0$, meeting the line x = 2 at the points A and B respectively Clearly $A=\left(2,-2 m_1\right) ; B=\left(2,-2 m_2\right)$

Now $\triangle_{A C D}=\frac{\Delta}{3} \Rightarrow \frac{4 \sqrt{2}}{3 \pi}=\frac{1}{2} . 2 .-2 m_1$

$\Rightarrow m_1=-\frac{2 \sqrt{2}}{3 \pi}$ Also $\Delta_{B C D}=\frac{2 \Delta}{3}$

$\Rightarrow \frac{8 \sqrt{2}}{3 \pi}=\frac{1}{2} . 2-2 m_2$

$\Rightarrow m_2=\frac{-4 \sqrt{2}}{3 \pi}$

Required sum= $-\frac{2 \sqrt{2}}{\pi}$