The distance between the two points A and A' which lie on y = 2 such that both the line segments AB

Q: Wait, why do we use the absolute value in the angle formula?

The formula $\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$ gives the acute angle between two lines. Since the problem states the angle is $\pi/4$, the absolute value accounts for the fact that the angle could be measured in either direction (clockwise or counter-clockwise), ensuring we capture both possible slopes for line $OA$.

Q: Why are there two points $A$ and $A'$?

The condition $\angle AOB = \pi/4$ defines a locus of points. Since $A$ is restricted to the line $y=2$, the line $OA$ can be rotated by $\pi/4$ relative to $OB$ in two different directions. This rotation intersects the line $y=2$ at two distinct points, $A$ and $A'$, which is why we get two solutions for $x$.

Q: Can I use the dot product instead of slopes?

Yes, you absolutely can! Using $\vec{OA} \cdot \vec{OB} = |\vec{OA}| |\vec{OB}| \cos(\pi/4)$ is a powerful alternative. It leads to the same result, though the slope method is often more direct when dealing with lines passing through the origin.

Question

The distance between the two points $A$ and $A'$ which lie on $y = 2$ such that both the line segments $AB$ and $A'B$ (where $B$ is the point $(2, 3)$) subtend angle $\frac{\pi}{4}$ at the origin, is equal to :

Options:

(A) 10

(B) $\frac{48}{5}$

(C) $\frac{52}{5}$

(D) 3

Sigma Percentile · Accepted Answer

### The Geometry of the Problem

Imagine you are standing at the origin $O(0,0)$, looking out at a coordinate plane. You see a fixed point $B$ at $(2,3)$.

Now, imagine a horizontal line stretching across the plane at $y=2$. Somewhere on this line, there are two points, $A$ and $A'$, such that the angle $\angle AOB$ is exactly $\pi/4$.

We are looking for the distance between these two points, $A$ and $A'$, which requires us to find their respective $x$-coordinates.

### The Slope Connection

To solve this, we must translate the geometric condition into the language of algebra. Let the coordinates of point $A$ be $(x, 2)$.

The slope of the line $OA$, denoted as $m_1$, is:
$$m_1 = \frac{2 - 0}{x - 0} = \frac{2}{x}$$

The slope of the fixed line $OB$, denoted as $m_2$, is:
$$m_2 = \frac{3 - 0}{2 - 0} = \frac{3}{2}$$

### The Angle Formula

The angle $	heta$ between two lines with slopes $m_1$ and $m_2$ is given by the formula:
$$	an 	heta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} ight|$$

Given $	heta = \pi/4$, we know that $	an(\pi/4) = 1$. Substituting our slopes, we obtain:
$$1 = \left| \frac{\frac{3}{2} - \frac{2}{x}}{1 + (\frac{3}{2})(\frac{2}{x})} ight|$$

### Solving the Modulus

Simplifying the expression inside the modulus, the numerator becomes $\frac{3x - 4}{2x}$ and the denominator becomes $\frac{2x + 6}{2x}$.

Canceling the $2x$ terms, we arrive at the core equation:
$$1 = \left| \frac{3x - 4}{2x + 6} ight|$$

To solve for $x$, we must consider both the positive and negative cases of the modulus.

### Case Analysis

**Case 1:**
$$\frac{3x - 4}{2x + 6} = 1$$
$$3x - 4 = 2x + 6 \Rightarrow x = 10$$
Thus, our first point is $A(10, 2)$.

**Case 2:**
$$\frac{3x - 4}{2x + 6} = -1$$
$$3x - 4 = -2x - 6 \Rightarrow 5x = -2 \Rightarrow x = -\frac{2}{5}$$
Thus, our second point is $A'(-\frac{2}{5}, 2)$.

### Final Calculation

Since both points lie on the horizontal line $y=2$, the distance between them is the absolute difference of their $x$-coordinates.

$$AA' = \left| 10 - \left( -\frac{2}{5} ight) ight| = \left| 10 + \frac{2}{5} ight| = \frac{52}{5}$$

The final distance between the two points is **$52/5$** (or **$10.4$**).

Animated Solution for Mathematics - Straight Lines: The distance between the two points $A$ and $A^{'}$ which lie on $y = 2$ such that both the line segments $A B$ and $A^{'} B$ (where $B$ is the point $(2, 3)$ ) subtend angle $\frac{π}{4}$ at the origin, is equal to :

Select Answer:

Visualized Solution

Visualize the Geometry

Define Coordinates and Slopes

Apply Angle Formula

Substitute the Slopes

Simplify the Expression

Solve Case 1: Positive Sign

Solve Case 2: Negative Sign

Calculate Distance $A A^{'}$

The Sigma Insight: Angle Between Two Lines

The Geometry of the Problem

The Slope Connection

The Angle Formula

Solving the Modulus

Case Analysis

Final Calculation

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