Let the angles made with the positive x-axis by two straight lines drawn from the point P(2, 3) and

Q: Why do we use the parametric form instead of the standard slope-intercept form?

The parametric form is incredibly powerful here because it directly incorporates the distance $r$ and the angle $\theta$ into the coordinates. Using $y - y_1 = m(x - x_1)$ would require you to solve for the slope $m$ and then convert it to $\theta$, which is algebraically more cumbersome. The parametric form lets us solve for $\theta$ in one elegant sweep.

Q: Why are there two angles, $\theta_1$ and $\theta_2$?

Think of the circle centered at $P$ with radius $r$. This circle intersects the line $x + y = 6$ at two distinct points. Each point corresponds to a unique direction (angle) from $P$. Thus, there are two distinct lines, and consequently, two distinct angles $\theta_1$ and $\theta_2$.

Q: Is there a shortcut to finding the sum $\theta_1 + \theta_2$?

While you can find $\theta_1$ and $\theta_2$ individually, the equation $\sin(\theta + \frac{\pi}{4}) = \frac{\sqrt{3}}{2}$ reveals the structure. The two solutions for the argument $\theta + \frac{\pi}{4}$ are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$. Adding these equations directly gives you the sum of the angles without needing to isolate $\theta_1$ and $\theta_2$ first, though calculating them is also perfectly fine!

Question

Let the angles made with the positive $x$-axis by two straight lines drawn from the point $P(2, 3)$ and meeting the line $x + y = 6$ at a distance $\sqrt{\frac{2}{3}}$ from the point P be $	heta_1$ and $	heta_2$. Then the value of $(	heta_1 + 	heta_2)$ is:

Options:

(A) $\frac{\pi}{3}$

(B) $\frac{\pi}{6}$

(C) $\frac{\pi}{2}$

(D) $\frac{\pi}{12}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at point $P(2, 3)$ on a coordinate plane. Before you lies a straight line, defined by the equation $x + y = 6$.

Your task is to draw two lines from $P$ that strike this target line at a very specific distance: $r = \sqrt{\frac{2}{3}}$.

When we talk about a fixed distance from a point, we are implicitly talking about a circle. The points where a circle centered at $P$ with radius $r = \sqrt{\frac{2}{3}}$ intersects the line $x + y = 6$ are the exact locations our lines must hit.

### The Power of Parametric Representation

To find these points, we describe any point $Q$ at a distance $r$ from $P$ at an angle $	heta$ using the parametric form of a line:

$$x = 2 + r \cos 	heta$$
$$y = 3 + r \sin 	heta$$

This serves as our bridge, connecting the physical distance $r$ and the orientation $	heta$ to the Cartesian coordinates $(x, y)$. Since point $Q$ must lie on the line $x + y = 6$, we substitute these parametric expressions into the line equation.

### The Algebraic Unfolding

Substituting our expressions, we get:

$$(2 + r \cos 	heta) + (3 + r \sin 	heta) = 6$$

Combining the constants $2 + 3 = 5$, we have:

$$5 + r(\cos 	heta + \sin 	heta) = 6$$

Subtracting $5$ from both sides, we arrive at:

$$r(\cos 	heta + \sin 	heta) = 1$$

Now, we substitute the given distance $r = \sqrt{\frac{2}{3}}$:

$$\sqrt{\frac{2}{3}}(\cos 	heta + \sin 	heta) = 1$$

Multiplying both sides by $\sqrt{\frac{3}{2}}$, we isolate the trigonometric sum:

$$\cos 	heta + \sin 	heta = \sqrt{\frac{3}{2}}$$

### The Trigonometric Climax

To solve the equation $\cos 	heta + \sin 	heta = \sqrt{\frac{3}{2}}$, we divide by $\sqrt{1^2 + 1^2} = \sqrt{2}$:

$$\frac{1}{\sqrt{2}}\cos 	heta + \frac{1}{\sqrt{2}}\sin 	heta = \frac{\sqrt{3}}{2}$$

Recognizing that $\frac{1}{\sqrt{2}} = \sin(\frac{\pi}{4}) = \cos(\frac{\pi}{4})$, we use the sine addition formula $\sin(A+B) = \sin A \cos B + \cos A \sin B$:

$$\sin(	heta + \frac{\pi}{4}) = \frac{\sqrt{3}}{2}$$

The sine function equals $\frac{\sqrt{3}}{2}$ at two primary angles: $\frac{\pi}{3}$ and $\frac{2\pi}{3}$. Thus, we have two possibilities for our angle $	heta$:

$$	heta_1 + \frac{\pi}{4} = \frac{\pi}{3} \ \Rightarrow 	heta_1 = \frac{\pi}{12}$$

$$	heta_2 + \frac{\pi}{4} = \frac{2\pi}{3} \ \Rightarrow 	heta_2 = \frac{5\pi}{12}$$

### The Final Harmony

The question asks for the sum of these angles, $	heta_1 + 	heta_2$. Adding them together:

$$	heta_1 + 	heta_2 = \frac{\pi}{12} + \frac{5\pi}{12} = \frac{6\pi}{12} = \frac{\pi}{2}$$

The final result is a perfect right angle, **$\frac{\pi}{2}$**. This concludes a journey that proves when you break down complex problems into their fundamental components, the math resolves into something elegant.

Animated Solution for Mathematics - Straight Lines: Let the angles made with the positive $x$ -axis by two straight lines drawn from the point $P (2, 3)$ and meeting the line $x + y = 6$ at a distance $\frac{2}{3}$ from the point P be $θ_{1}$ and $θ_{2}$ . Then the value of $(θ_{1} + θ_{2})$ is:

Select Answer:

Visualized Solution

Visualizing the Setup

The Distance Constraint

Parametric Form of a Line

Coordinates of Intersection $Q$

Applying the Line Equation

Simplifying the Equation

Substituting the Distance $r$

Isolating the Trigonometric Sum

Trigonometric Transformation

Applying the Sine Addition Identity

Finding the Possible Angles

Calculating $θ_{1}$ and $θ_{2}$

The Final Sum

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Power of Parametric Representation

The Algebraic Unfolding

The Trigonometric Climax

The Final Harmony

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