Question

Let $\left(\alpha_1, \beta_1\right),\left(\alpha_2, \beta_2\right),\left(\alpha_3, \beta_3\right)$ be vertices of a $\triangle ABC$ with $\alpha_1, \alpha_2, \alpha_3, \beta_1, \beta_2, \beta_3$ are prime values of $k$ in increasing order for which one root of equation $(k-5) x^2-2 k x+k-4=0$ is smaller than 1 and other exceed 2 . If $P ( p , q )$ is a point inside the triangle such that area of $\triangle PAC =$ area of $\triangle PAB =$ area of $\triangle PBC$, then find the value of $\left(\frac{ p + q }{10}\right)$.

Answer 1

$\Theta 1$ and 2 lies between roots,
$\therefore( k -5) f (1)<0 $
$\Rightarrow( k -5)( k -5-2 k + k -4)<0$
$\Rightarrow( k -5)(-9)<0 \Rightarrow k -5>0 \Rightarrow k >5 $
$\text { and }( k -5) f (2)<0 $
$\Rightarrow( k -5)(4( k -5)-4 k + k -4)<0$
$\Rightarrow( k -5)( k -24)<0 \Rightarrow 5< k <24 $
$\therefore k \in(5,24)$
$\text { prime numbers }=7,11,13,17,19,23$
$\therefore \text { points will be }=(7,17)(11,19) \text { and }(13,23) $
$\text { Clearly, Pis centroid }$
$\therefore P =\left(\frac{31}{3}, \frac{59}{3}\right) $
$ \frac{ p + q }{10}=\frac{\frac{90}{3}}{10}=3 $