A ray of light passing through the point P(2,3) reflects on the x-axis at point A and the …

Question

A ray of light passing through the point $P(2,3)$ reflects on the $x$-axis at point $A$ and the reflected ray passes through the point $Q(5,4)$. Let $R$ be the point that divides the line segment $AQ$ internally into the ratio $2:1$. Let the co-ordinates of the foot of the perpendicular $M$ from $R$ on the bisector of the angle $PAQ$ be $(\alpha,\beta)$. Then, the value of $7\alpha+3\beta$ is equal to:

Answer: 31

Accepted Answer

Image of $P(2,3)$ in the $x$-axis is $(2,-3)$. Line through $(2,-3)$ and $Q(5,4)$ gives point $A$ where it crosses the $x$-axis:
$$\frac{4-(-3)}{5-2}=\frac73, \quad y+3=\frac73(x-2), \text{ set } y=0: 3=\frac73(x-2)\implies x=2+\frac97=\frac{23}7$$

So $A=\left(\dfrac{23}7,0\right)$.

$$R = \frac{2Q+A}3 = \frac{2(5,4)+\left(\frac{23}7,0\right)}3 = \left(\frac{10+\frac{23}7}3,\frac83\right) = \left(\frac{31}7\cdot\frac{7}{7}\Big/3\cdots\right)$$

Working directly: $R_x = \dfrac{10+\frac{23}7}3 = \dfrac{\frac{93}7}3=\dfrac{31}7$, $R_y=\dfrac83$.

The bisector of $\angle PAQ$ is vertical, $x=\dfrac{23}7$ (since $P$ and its reflection are symmetric about the $x$-axis, the angle bisector at $A$ between $AP$ and $AQ$ is the vertical line through $A$). So the foot of perpendicular from $R$ onto this bisector is $M=\left(\dfrac{23}7,\dfrac83\right)$.

$$7\alpha+3\beta = 7\cdot\frac{23}7+3\cdot\frac83 = 23+8=31$$

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