If the line segment joining the points (5, 2) and (2, a) subtends an angle π/4 at the origin, then t

Q: Wait, why do we have two values for $a$? Shouldn't there be only one?

That is a fantastic question! Think of it geometrically: the line $OA$ is fixed. To make a $45^{\circ}$ angle with $OA$, you could rotate $45^{\circ}$ clockwise or $45^{\circ}$ counter-clockwise. Since point $B$ lies on the vertical line $x=2$, there are two distinct points on that line that satisfy the angle condition—one above the x-axis and one below. That is why we get two valid values for $a$.

Q: Can I use the dot product formula instead of the slope formula?

Absolutely! Using the dot product $\vec{OA} \cdot \vec{OB} = |\vec{OA}| |\vec{OB}| \cos \theta$ is a perfectly valid and elegant approach. It will lead you to the same result. However, for problems involving lines and angles in the Cartesian plane, the slope formula is often more direct. Both methods show your deep understanding of the subject!

Q: Why do we take the absolute value at the end?

The question specifically asks for the 'absolute value of the product of all possible values of $a$'. This is a common way to ensure a unique, positive answer in competitive exams. Even if your product turns out to be negative, the absolute value operator $|x|$ ensures we report the magnitude, which is the standard convention for such problems.

Question

If the line segment joining the points $(5, 2)$ and $(2, a)$ subtends an angle $\frac{\pi}{4}$ at the origin, then the absolute value of the product of all possible values of $a$ is :

Options:

(A) 6

(B) 8

(C) 2

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the origin of a coordinate plane with a fixed point $A$ at $(5, 2)$. You draw a line from the origin to $A$, which serves as your baseline.

A second point $B$ is located at $(2, a)$. Because the $x$-coordinate is fixed at $2$, point $B$ is constrained to a vertical line, sliding up and down like a bead on a wire.

We aim to find the positions of $B$ such that the angle $\angle AOB$ is exactly $45^{\circ}$. This geometric constraint defines the relationship between the two lines.

### The Slope Connection

To capture this geometric relationship, we utilize the language of slopes. The slope of our baseline $OA$ is:

$$m_1 = \frac{2 - 0}{5 - 0} = \frac{2}{5}$$

The slope of our dynamic line $OB$ is:

$$m_2 = \frac{a - 0}{2 - 0} = \frac{a}{2}$$

We know the angle between these two lines is $	heta = 45^{\circ}$. The tangent of this angle is given by the formula:

$$	an 	heta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} ight|$$

Since $	an 45^{\circ} = 1$, we set our expression equal to $1$. This transforms the geometry into the following algebraic equation:

$$1 = \left| \frac{\frac{a}{2} - \frac{2}{5}}{1 + (\frac{a}{2})(\frac{2}{5})} ight|$$

### The Modulus Fork in the Road

The modulus bars indicate that there are two paths to take. When we remove the absolute value, we must account for both the positive and negative possibilities.

First, simplify the expression inside the modulus:

$$\frac{\frac{a}{2} - \frac{2}{5}}{1 + \frac{a}{5}} = \frac{\frac{5a - 4}{10}}{\frac{5 + a}{5}} = \frac{5a - 4}{2(5 + a)} = \frac{5a - 4}{10 + 2a}$$

**Case 1:**

$$\frac{5a - 4}{10 + 2a} = 1 \Rightarrow 5a - 4 = 10 + 2a \Rightarrow 3a = 14 \Rightarrow a = \frac{14}{3}$$

**Case 2:**

$$\frac{5a - 4}{10 + 2a} = -1 \Rightarrow 5a - 4 = -10 - 2a \Rightarrow 7a = -6 \Rightarrow a = -\frac{6}{7}$$

These are the two potential locations for point $B$.

### The Final Celebration

The question asks for the absolute value of the product of these two values of $a$. Let us multiply them:

$$\left( \frac{14}{3} ight) 	imes \left( -\frac{6}{7} ight)$$

Notice the elegance of the simplification: $14$ divided by $7$ is $2$, and $-6$ divided by $3$ is $-2$. The product is:

$$2 	imes (-2) = -4$$

Finally, taking the absolute value, we arrive at the result. The final answer is **4**.

Animated Solution for Mathematics - Straight Lines: If the line segment joining the points $(5, 2)$ and $(2, a)$ subtends an angle $\frac{π}{4}$ at the origin, then the absolute value of the product of all possible values of $a$ is :

Select Answer:

Visualized Solution

Visualizing the Fixed Points

Locus of the Dynamic Point $B$

The Angle Subtended at Origin

Calculating the Slopes

Applying the Angle Formula

Simplifying the Equation

Case 1: Positive Modulus

Case 2: Negative Modulus

Product of All Possible Values

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Slope Connection

The Modulus Fork in the Road

The Final Celebration

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