JEE Main 2016 (7)

Let the line through the origin have the equation $y = mx.$ This line will intersect the line $x + y = 1 \Longrightarrow y = -x + 1$ when

$mx = -x + 1 \Longrightarrow x(m + 1) = 1 \Longrightarrow x = \dfrac{1}{m + 1} \Longrightarrow y = 1 - x = \dfrac{m}{m + 1}, m \ne -1.$

(Note that if $m \lt 0$ then $|OA|$ would exceed $\dfrac{1}{2}$ and $|OB|$ would exceed $1$ , so since we require that $|OA||OB| = \dfrac{1}{3}$ we know that $m \ge 0$ and specifically that $m \ne -1$ .)

Thus $|OA| = \sqrt{\left(\dfrac{1}{m + 1}\right)^{2} + \left(\dfrac{m}{m + 1}\right)^{2}} = \dfrac{\sqrt{1 + m^{2}}}{m + 1}.$

Next, the lines $y = mx$ and $x + 2y = 1 \Longrightarrow y = -\dfrac{x}{2} + \dfrac{1}{2}$ intersect when

$mx = -\dfrac{x}{2} + \dfrac{1}{2} \Longrightarrow x(2m + 1) = 1 \Longrightarrow x = \dfrac{1}{2m + 1}$

$\Longrightarrow 2y = 1 - x = \dfrac{2m}{2m + 1} \Longrightarrow y = \dfrac{m}{2m + 1}, m \ne -\dfrac{1}{2}.$

Thus $|OB| = \sqrt{\left(\dfrac{1}{2m + 1}\right)^{2} + \left(\dfrac{m}{2m + 1}\right)^{2}} = \dfrac{\sqrt{1 + m^{2}}}{2m + 1}.$

We then need to find $m$ such that $|OA||OB| = \dfrac{1}{3}$

$\Longrightarrow \dfrac{\sqrt{1 + m^{2}}}{m + 1} \times \dfrac{\sqrt{1 + m^{2}}}{2m + 1} = \dfrac{1 + m^{2}}{(m + 1)(2m + 1)} = \dfrac{1}{3}$

$\Longrightarrow 3 + 3m^{2} = (m + 1)(2m + 1) = 2m^{2} + 3m + 1$

$\Longrightarrow m^{2} - 3m + 2 = 0 \Longrightarrow (m - 1)(m - 2) = 0.$

Thus the possible values for $m$ are $m = 1$ and $m = 2$ , which sum to $\boxed{3}$ .