If the line y = mx bisects the area enclosed by the lines x = 0, y...

Given, the equation of line, $y = mx$

And the line $x = 0, y = 0, x = \dfrac{3}{2}$

The intersection point \left(\dfrac{3}{2}, 0\right)$</p><p>Area of triangle is half the total area </p><p>$\dfrac{1}{2}{∫}_{0}^{3 / 2}\left(1 + 4x - {x}^{2}\right)dx = {∫}_{0}^{3 / 2}mxdx$</p><p>$\dfrac{1}{2}{\left[x + 2{x}^{2} - \dfrac{{x}^{3}}{3}\right]}_{0}^{3 / 2} = \dfrac{m}{2}{\left[{x}^{2}\right]}_{0}^{3 / 2}$</p><p>\left(\dfrac{3}{2} + \dfrac{9}{2} - \dfrac{9}{8}\right) = \dfrac{9m}{4}$

$\dfrac{39}{8} = \dfrac{9m}{4}$

$m = \dfrac{39}{18}$

Then, $12m$

$12m = \dfrac{39}{18} \times 12$

$12m = 26$