If the point (α, 7√3/3) lies on the curve traced by the mid-points of the line segments of the lines

Q: Why do we need to eliminate $\theta$?

The parameter $\theta$ defines the position of the line at any given moment, but the locus is the path traced by the midpoint, which is a fixed curve in the $xy$-plane. To find the equation of this curve, we need a relationship between $x$ and $y$ that is independent of $\theta$.

Q: Wait, why do we reject the negative root here?

Great catch! The problem states $\theta \in (0, \frac{\pi}{2})$. In this range, both $\sin \theta$ and $\cos \theta$ are positive. Since the coordinates of the midpoint are $h = \frac{7}{2 \cos \theta}$ and $k = \frac{7}{2 \sin \theta}$, both $h$ and $k$ must be positive. Thus, the locus is restricted to the first quadrant, so we must have $x > 0$.

Q: Is there a faster way to find the locus without finding the intercepts?

You could use the normal form of the line directly, but finding the intercepts is the most intuitive and robust method for this type of problem. It builds a clear geometric bridge between the line and its midpoint.

Question

If the point $(\alpha, \frac{7\sqrt{3}}{3})$ lies on the curve traced by the mid-points of the line segments of the lines $x \cos 	heta + y \sin 	heta = 7, 	heta \in (0, \frac{\pi}{2})$ between the co-ordinates axes, then $\alpha$ is equal to

Options:

(A) $-7$

(B) $-7\sqrt{3}$

(C) $7\sqrt{3}$

(D) 7

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in the first quadrant of the Cartesian plane, watching a line segment slide gracefully between the $x$ and $y$ axes. The line is defined by the equation $x \cos 	heta + y \sin 	heta = 7$, where $	heta$ is a parameter that changes, causing the line to shift.

As this line moves, its midpoint traces a path—a locus—that we are tasked to uncover. This is not just an algebraic exercise; it is a geometric dance.

### Finding the Intercepts

To understand the midpoint, we must first understand the endpoints. The line cuts the axes at two points, $P$ and $Q$.

For the $x$-intercept $P$, we set $y = 0$, which gives us $x \cos 	heta = 7$. Thus, the coordinates are:
$$P = \left( \frac{7}{\cos 	heta}, 0 ight)$$

Similarly, for the $y$-intercept $Q$, we set $x = 0$, leading to $y \sin 	heta = 7$. Thus, the coordinates are:
$$Q = \left( 0, \frac{7}{\sin 	heta} ight)$$

### The Midpoint's Journey

Now, let the midpoint of this segment be $M(h, k)$. Using the midpoint formula, we find the coordinates:
$$h = \frac{\frac{7}{\cos 	heta} + 0}{2} = \frac{7}{2 \cos 	heta}$$
$$k = \frac{0 + \frac{7}{\sin 	heta}}{2} = \frac{7}{2 \sin 	heta}$$

These equations describe the position of the midpoint in terms of $	heta$. To find the locus, we must eliminate $	heta$ by rearranging these to isolate the trigonometric functions:
$$\cos 	heta = \frac{7}{2h}, \quad \sin 	heta = \frac{7}{2k}$$

### The Identity and the Locus

Here is where the elegance of trigonometry shines. We know the fundamental identity $\sin^2 	heta + \cos^2 	heta = 1$.

Substituting our expressions, we get:
$$\left( \frac{7}{2h} ight)^2 + \left( \frac{7}{2k} ight)^2 = 1$$

This simplifies to:
$$\frac{49}{4h^2} + \frac{49}{4k^2} = 1$$

Dividing by $49$ and multiplying by $4$, we arrive at the beautiful equation of our locus:
$$\frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{49}$$

### The Final Reveal

The problem states that the point $(\alpha, \frac{7\sqrt{3}}{3})$ lies on this curve. Substituting $x = \alpha$ and $y = \frac{7\sqrt{3}}{3}$, we have:
$$\frac{1}{\alpha^2} + \frac{1}{\left( \frac{7\sqrt{3}}{3} ight)^2} = \frac{4}{49}$$

Squaring the $y$-term gives $\frac{49 	imes 3}{9} = \frac{49}{3}$, so the reciprocal is $\frac{3}{49}$. The equation becomes:
$$\frac{1}{\alpha^2} + \frac{3}{49} = \frac{4}{49}$$

This simplifies to $\frac{1}{\alpha^2} = \frac{1}{49}$, which implies $\alpha^2 = 49$, or $\alpha = \pm 7$. Given our constraint that the midpoint must lie in the first quadrant, we conclude that **$\alpha = 7$**.

Animated Solution for Mathematics - Straight Lines: If the point $(α, \frac{7 3}{3})$ lies on the curve traced by the mid-points of the line segments of the lines $x cos θ + y sin θ = 7, θ \in (0, \frac{π}{2})$ between the co-ordinates axes, then $α$ is equal to

Select Answer:

Visualized Solution

Visualizing the Moving Line

Finding the Intercepts $P$ and $Q$

Defining the Midpoint $M (h, k)$

Isolating Trigonometric Terms

Eliminating the Parameter $θ$

Simplifying the Locus Equation

Substituting the Given Point

Simplifying the $y$ term

Solving for $α$

Final Conclusion

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

Finding the Intercepts

The Midpoint's Journey

The Identity and the Locus

The Final Reveal

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