A ray of light passing through the point P(2, 3) reflects on the X...

By observation we know the point A $\left(\alpha, 0\right)$ and $\beta = y$-coordinate of $R= \dfrac{2 \times 4 + 1 \times 0}{2 + 1} = \dfrac{8}{3} ⋯ \left(1\right)</p><p>Now${P}^{'} is image of $P in y = 0$ which will be ${P}^{'}\left(2, - 3\right)$

$∴$ Equation of ${P}^{'}Q$ is $\left(y + 3\right) = \dfrac{4 + 3}{5 - 2}$\left(x - 2\right)

i.e. $3y + 9 = 7x - 14$

Now ${P}^{'}Q$ meets $x -$axis at A$</p><p>so A ≡ \left(\dfrac{23}{7}, 0\right)$ by solving with $y = 0$

$∴ \alpha = \dfrac{23}{7} ⋯ \left(2\right)</p><p>By \left(1\right), \left(2\right), we get</p><p>$7\alpha + 3\beta = 23 + 8 = 31$</p>$