The equations of sides AB and AC of a triangle ABC are (λ + 1)x + =...

Given,

The equations of sides $AB$ and $AC$ of a triangle $ABC$ and have orthocentre at \left(1, 2\right)$</p><p>So,$AB ≡ \left(\lambda + 1\right)x + \lambday = 4 . . . . . . . . \left(1\right)$</p><p>And$AC ≡ \lambdax + \left(1 - \lambda\right)y + \lambda = 0 . . . . . . \left(2\right)$</p><p>Also given vertex$A$is on the$y -$axis</p><p>So, putting the value$x = 0$in equation \left(1\right) \text{ and }$\left(2\right)$we get,</p><p>$y = \dfrac{4}{\lambda} \text{ and } y = \dfrac{\lambda}{\lambda - 1}$</p><p>$⇒ \dfrac{4}{\lambda} = \dfrac{\lambda}{\lambda - 1}$</p><p>$⇒ \lambda = 2$</p><p>Now putting the value of$\lambda$in both equation we get,</p><p>$AB ≡ 3x + 2y = 4 . . . . . . . . $\left(3\right)$

$AC ≡ 2x - y + 2 = 0 . . . . . . \left(4\right)$

Now solving above equation we get, $A\left(0, 2\right)$

Now let $C\left(\alpha, 2\alpha + 2\right)$ {using equation \left(4\right)}</p><p>Now we know that slopes of perpendicular lines is- 1$,</p><p>So, from diagram slope of altitude through$C \times \left(\text{slope of}AB\right)$we get,</p><p>$\dfrac{2\alpha}{\alpha - 1} \times \left(\dfrac{- 3}{2}\right) = - 1$</p><p>$⇒ \alpha = \dfrac{- 1}{2}$</p><p>Hence, point$C$will be$C\left(\dfrac{- 1}{2}, 1\right)$</p><p>Now let the equation of tangent be$y = mx + \dfrac{3}{2m}$which passes through point$C\left(\dfrac{- 1}{2}, 1\right)$</p><p>So,$1 = m\left(\dfrac{- 1}{2}\right) + \dfrac{3}{2m}$</p><p>$⇒ {m}^{2} + 2m - 3 = 0$</p><p>$⇒ m = 1, - 3$</p><p>Also we know that point of contact of tangent to parabola is given by \left(\dfrac{a}{{m}^{2}}, \dfrac{2a}{m}\right) ≡ \left(\dfrac{3}{2}, 3\right)$

Hence, $CT = \sqrt{{\left(\dfrac{3}{2} + \dfrac{1}{2}\right)}^{2} + {\left(3 - 1\right)}^{2}} = \sqrt{8} = 2\sqrt{2}$