If ( a, b ) be the orthocentre of the triangle whose vertices are (...

Equation of $CE$ is given by,

$⇒ \left(y - 1\right) = \dfrac{- 1}{\dfrac{3 - 2}{2 - 1}}\left(x - 3\right)$

$⇒ \left(y - 1\right) = - ( x - 3 )$

$⇒ x + y = 4 . . . \left(i\right)$

Similarly, equation of $AD$ is given by,

$⇒ \left(y - 2\right) = \dfrac{- 1}{\dfrac{3 - 1}{2 - 3}}\left(x - 1\right)$

$⇒ \left(y - 2\right) = \dfrac{1}{2}\left(x - 1\right)$

$⇒ 2y - 4 = x - 1$

$⇒ x - 2y + 3 = 0$

Using \left(i\right)$</p><p>$⇒ 4 - y - 2y + 3 = 0$</p><p>$⇒ y = \dfrac{7}{3}$</p><p>$⇒ x = 4 - \dfrac{7}{3}$</p><p>$⇒ x = \dfrac{5}{3}$</p><p>So, Orthocentre is$H ≡ \left(a, b\right) = \left(\dfrac{5}{3}, \dfrac{7}{3}\right)$.</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/0fa5deeb-0b17-4bc6-bc8e-7dd1df3d7619.png" style="width: 275px; height: 200px;"/></p><p>Orthocentre lies on the line$x + y = 4$</p><p>So,$a + b = 4$</p><p>Now,${I}_{1} = {∫}_{a}^{b}xsin ( x ( 4 - x ) ) dx . . . ( 1 )$</p><p>$⇒ {I}_{1} = {∫}_{a}^{b} ( a + b - x ) \sin ( \left(a + b - x\right) ( 4 - \left(a + b - x\right) ) ) dx$</p><p>$⇒ {I}_{1} = {∫}_{a}^{b} ( 4 - x ) \sin ( x ( 4 - x ) ) dx . . . ( 2 )$</p><p>Using,$( 1 ) + ( 2 )$</p><p>$⇒ 2{I}_{1} = {∫}_{a}^{b}4\sin ( x ( 4 - x ) ) dx$</p><p>$⇒ 2{I}_{1} = 4{I}_{2}$</p><p>$⇒ {I}_{1} = 2{I}_{2}$</p><p>$⇒ \dfrac{{I}_{1}}{{I}_{2}} = 2$</p><p>$⇒ \dfrac{36{I}_{1}}{{I}_{2}} = 72