Question

A line is drawn through the point (1,2) to meet the coordinate axes at P and Q such that it forms a triangle OPQ, where O is the origin. If the area of the triangle OPQ is least, then the slope of the line PQ is :

• A. $- \frac{1}{4}$
• B. -4
• C. -2
• D. $- \frac{1}{2}$

Answer 1

The Correct Option is C

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 PQ is passing through (1,2) therefore its equation is $(y - 2) = m (x - 1) $ where m is the slope of the line P Now, point P (x,0) will also satisfy the equation of PQ $\therefore \, y - 2 = m (x -1) \, \Rightarrow \, 0 - 2 = m (x - 1)$ $ \Rightarrow \, - 2 = m (x -1) \Rightarrow \, x - 1 = \frac{-2}{m}$ $ \Rightarrow \, x = \frac{-2}{m} + 1 $ Also , $OP = \sqrt{(x - 0)^2 + (0 -0)^2} = x $ $ = \frac{ - 2}{m} + 1 $ Similarly, point Q (0,y) will satisfy equation of PQ $\therefore y -2 = m \left(x-1\right) $ $ \Rightarrow y - 2 = m \left(-1\right)$ $ \Rightarrow y = 2 - m$ and $OQ = y = 2 -m$ Area of $ \Delta POQ = \frac{1}{2}\left(OP\right) \left(OQ\right) $ $= \frac{1}{2}\left(1- \frac{2}{m}\right) \left(2-m\right) $ $\left(\because \, \text{Area} \, \text{of} \Delta = \frac{1}{2} \times\text{base} \times\text{height}\right) $ $= \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\left(m\right) = 2 - \frac{m}{2} - \frac{2}{m}$ Now, $ f'\left(m\right) = \frac{-1}{2} + \frac{2}{m^{2}}$ Put $ f'\left(m\right) = 0$ $ \Rightarrow m^{2} = 4 \Rightarrow m=\pm2$ Now, $ f'' \left(m\right) = \frac{-4}{m^{3}}$ $ f''\left(m\right)|_{m=2} = - \frac{1}{2} < 0 $ $ f''\left(m\right)|_{m= - 2} =\frac{1}{2} < 0 $ Area will be least at m = -2 Hence, slope of PQ is -2.