AIEEE 2012 Solved Paper

Equation of a line passing through $(x_1, y_1)$ having slope $m$ is given by $y-y_1=m (x-x_1)$Since the line $P Q$ is passing through $(1,2)$ therefore its equation is$(y-2)=m(x-1)$ where $m$ is the slope of the line $P Q$. Now, point $P(x, 0)$ will also satisfy the equation of $P Q$$\; \therefore y-2=m(x-1)$$\; \Rightarrow 0-2=m(x-1)$$\; \Rightarrow -2=m(x-1)$$\; \Rightarrow x-1=\frac{-2}{m}$$\; \Rightarrow x=\frac{-2}{m}+1$ Also, $O P=\sqrt{(x-0)^2+(0-0)^2}=x=\frac{-2}{m}+1$Similarly, point $Q(0,y)$ will satisfy equation of $P Q$$\; \therefore y-2=m(x-1)$$\; \Rightarrow y-2=m(-1) \Rightarrow y=2-m b$and $O Q=y=2-m$ Area of $\triangle P O Q=\frac{1}{2}(O P)(O Q)=\frac{1}{2} \left(1-\frac{2}{m}\right) (2-m)$$\; \; \text{(As Area of } \; \Delta = \frac{1}{2} \times \; \text{ base } \; \times \; \text{ height)} \;$$\; =\frac{1}{2} \left[2-m-\frac{4}{m}+2\right] =\frac{1}{2} \left[4- \left(m+\frac{4}{m}\right) \right]$$\; =2-\frac{m}{2}-\frac{2}{m}$Let Area $=f(m)=2-\frac{m}{2}-\frac{2}{m}$Now, $f'(m)=\frac{-1}{2}+\frac{2}{m^2}$Put $f'(m)=0$$\Rightarrow m^2=4 \Rightarrow m=\pm 2$Now, $f''(m)=\frac{-4}{m^3}$$\left. f''(m) \right|_{m=2} =-\frac{1}{2}< 0$$\left. f''(m) \right|_{m=-2} =\frac{1}{2}>0$ Area will be least at $m=-2$Hence, slope of $P Q$ is $-2$.