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

Q: Wait, why do we use the image $P'$? Can't we just use the law of reflection directly?

You absolutely can, but it is much harder! Using the law of reflection directly involves calculating the slopes of $AP$ and $AQ$, finding the angle between them, and then finding the bisector line. By using the image $P'$, we leverage the geometric property that the path $P \to A \to Q$ is equivalent to the straight line $P' \to A \to Q$. It is a classic, elegant shortcut in optics and coordinate geometry.

Q: Why is the normal at $A$ the angle bisector of $\angle PAQ$?

This is a fundamental consequence of the Law of Reflection: the angle of incidence equals the angle of reflection. If you draw the normal at the point of incidence, it splits the angle between the incident ray and the reflected ray into two equal parts. Thus, the normal is, by definition, the angle bisector.

Q: How do I find the foot of the perpendicular $M$ to a vertical line?

Think about the geometry. A vertical line has the equation $x = k$. Any line perpendicular to it must be a horizontal line, which has the equation $y = c$. If you drop a perpendicular from a point $R(x_R, y_R)$ to the line $x = k$, the foot of the perpendicular $M$ will have the same $x$-coordinate as the line ($k$) and the same $y$-coordinate as the point $R$ ($y_R$). So, $M$ is simply $(k, y_R)$.

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 _______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem involves a light ray starting at $P(2, 3)$, reflecting off the $x$-axis at point $A$, and passing through $Q(5, 4)$. The path of light is governed by the laws of reflection, which can be complex to solve using trigonometry alone.

### The Mirror Trick

To find point $A$ efficiently, we employ the **Reflection Principle**. By reflecting point $P$ across the $x$-axis (the mirror), we obtain the image point $P'(2, -3)$.

The reflected ray $AQ$ is a continuation of the straight line segment $P'A$. Consequently, the points $P'$, $A$, and $Q$ are collinear. This geometric insight simplifies the problem significantly.

### The Path of the Ray

Since $P'$, $A$, and $Q$ lie on the same line, we calculate the slope $m$ using $P'(2, -3)$ and $Q(5, 4)$:

$$m = \frac{4 - (-3)}{5 - 2} = \frac{7}{3}$$

Using the point-slope form, the equation of the line $P'Q$ is:

$$y - 4 = \frac{7}{3}(x - 5)$$

This simplifies to the linear equation $7x - 3y = 23$. Point $A$ is the intersection of this ray with the $x$-axis, found by setting $y = 0$:

$$7x = 23 \Rightarrow x = \frac{23}{7}$$

Thus, the point of incidence is $A(\frac{23}{7}, 0)$.

### The Normal and the Bisector

According to the **Law of Reflection**, the normal at the point of incidence $A$ bisects the angle $\angle PAQ$. Because the reflecting surface is the horizontal $x$-axis, the normal is a vertical line passing through $A$.

Therefore, the equation of the angle bisector is simply:

$$x = \frac{23}{7}$$

### The Final Projection

We are given a point $R$ that divides the segment $AQ$ in the ratio $2:1$. Using the section formula with $A(\frac{23}{7}, 0)$ and $Q(5, 4)$, we find the coordinates of $R$:

$$x_R = \frac{2(5) + 1(\frac{23}{7})}{2+1} = \frac{10 + \frac{23}{7}}{3} = \frac{93/7}{3} = \frac{31}{7}$$

$$y_R = \frac{2(4) + 1(0)}{2+1} = \frac{8}{3}$$

Thus, $R$ is $(\frac{31}{7}, \frac{8}{3})$. We now find the foot of the perpendicular $M(\alpha, \beta)$ from $R$ to the bisector $x = \frac{23}{7}$.

For a vertical line, the foot of the perpendicular shares the $x$-coordinate of the line and the $y$-coordinate of the point. Therefore, $\alpha = \frac{23}{7}$ and $\beta = \frac{8}{3}$.

The final step is to evaluate $7\alpha + 3\beta$:

$$7\left(\frac{23}{7}\right) + 3\left(\frac{8}{3}\right) = 23 + 8 = 31$$

The final answer is **31**.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Reflection Principle

Image of Point $P$

Path of the Reflected Ray

Equation of Line $P^{'} Q$

Locating Point $A$

The Angle Bisector

Equation of the Bisector

Section Formula for $R$

Setting up Coordinates of $R$

Calculating Coordinates of $R$

Foot of the Perpendicular $M$

Coordinates of $M$

Setting up the Final Expression

Final Calculation

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Mirror Trick

The Path of the Ray

The Normal and the Bisector

The Final Projection

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