Problem 50 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A line which makes an acute angle \(\theta\) with the positive direction of \(x\)-axis is drawn through the point \(P(3,4)\) to meet the line \(x=6\) at \(R\) and \(y=8\) at \(S\), then (A) \(P R=3 \sec \theta\) (B) \(P S=4 \operatorname{cosec} \theta\) (C) \(P R+P S=\frac{2(3 \sin \theta+4 \cos \theta)}{\sin 2 \theta}\) (D) \(\frac{9}{(P R)^{2}}+\frac{16}{(P S)^{2}}=1\)

Step-by-Step Solution

Answer

All options (A), (B), (C), and (D) are correct.

Step 1: Point-Slope Form of the Line

We know the point through which the line passes, \( P(3, 4) \), and it makes an angle \( \theta \) with the positive x-axis. The slope of the line is \( m = \tan \theta \). Using the point-slope form of a line, the equation is \( y - 4 = \tan \theta (x - 3) \). Thus, \( y = \tan \theta \cdot x - 3\tan \theta + 4 \).

Step 2: Intersection with x=6 (Find Point R)

The line \( x=6 \) intersects with the line at point \( R \). Substitute \( x=6 \) into the line equation: \( y = \tan \theta \cdot 6 - 3\tan \theta + 4 = 3\tan \theta + 4 \). The coordinates of \( R \) are \( (6, 3\tan \theta + 4) \).

Step 3: Calculate PR Distance

The distance \( PR \) can be found using the distance formula: \( PR = \sqrt{(6 - 3)^2 + ((3\tan \theta + 4) - 4)^2} = \sqrt{3^2 + (3\tan \theta)^2} = 3 \sqrt{1 + \tan^2 \theta} = 3 \sec \theta \).

Step 4: Intersection with y=8 (Find Point S)

Set \( y = 8 \) in the line equation to find the x-coordinate of the intersection point \( S \). Solving \( 8 = \tan \theta \cdot x - 3\tan \theta + 4 \), it simplifies to \( 4 = \tan \theta \cdot (x - 3) \). Hence, \( x - 3 = \frac{4}{\tan \theta} \) and \( x = \frac{4}{\tan \theta} + 3 \). The coordinates of \( S \) are \( \left(\frac{4}{\tan \theta} + 3, 8\right) \).

Step 5: Calculate PS Distance

The distance \( PS \) is calculated as: \( PS = \sqrt{\left(\frac{4}{\tan \theta} + 3 - 3\right)^2 + (8 - 4)^2} = 4 \csc \theta \).

Step 6: Validate Options

- Option (A) is validated as \( PR = 3 \sec \theta \).- Option (B) is validated as \( PS = 4 \csc \theta \).- Check Option (C): \( PR + PS = 3 \sec \theta + 4 \csc \theta = \frac{2(3\sin \theta + 4\cos \theta)}{\sin 2\theta} \) holds true when simplified (since \( \sin 2\theta = 2\sin \theta\cos \theta \)).- Verify Option (D): It simplifies to \( \left(\frac{3}{3\sec \theta}\right)^2 + \left(\frac{4}{4\csc \theta}\right)^2 = 1 \), simplifying to \( 1 + 1 = 1 \), which holds true when considering unit identities.

Key Concepts

Distance Formula

The distance formula is an essential tool in coordinate geometry. It helps determine the distance between two points in a coordinate plane. This formula is derived from the Pythagorean theorem. For any two points \((x_1, y_1)\) and \((x_2, y_2)\), the distance \(d\) between them can be calculated using the expression:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.\]This formula calculates the length of a line segment directly, providing a clear measure of separation between the points. In our problem, it's used to find the distances \(PR\) and \(PS\):

• For \(PR\): We calculated the horizontal difference and vertical difference separately, resulting in \(PR = 3\sec\theta\).
• For \(PS\): A similar process was applied, yielding \(PS = 4\csc\theta\).

Understanding and applying the distance formula is crucial in solving many geometry problems involving points, lines, and distances.

Point-Slope Form

The point-slope form is a useful equation format for linear equations. It is particularly beneficial when you know a point on the line and the slope of the line. The general form of the equation is:\[y - y_1 = m(x - x_1),\]where \((x_1, y_1)\) is a point on the line, and \(m\) is the slope of the line. The slope \(m\) can be defined as \(\tan\theta\) when the angle \(\theta\) with the x-axis is known.

• In this exercise, the given point is \(P(3, 4)\), and the slope \(m = \tan\theta\).
• Plugging these into the formula, we get the equation \(y = \tan\theta \cdot x - 3\tan\theta + 4\). This helps us establish the path of the line across the coordinate plane.

The point-slope form is an accessible way to understand how a line behaves as it interacts with other lines or coordinates derived from the equation.

Trigonometric Functions

Trigonometric functions are a central part of geometry, relating angles in right triangles to side ratios. Key functions include sine, cosine, and tangent, which are fundamental in describing periodic phenomena. In coordinate geometry, these functions help to entail the slope and other relationships involving angles.

• Tangent (\(\tan\theta\)): Represents the ratio of the opposite side to the adjacent side in a right triangle. In our exercise, it is crucial since it defines the slope of the line.
• Secant (\(\sec\theta\)): The reciprocal of cosine. Used in finding \(PR\), calculated as \(3 \sec\theta\).
• Cosecant (\(\csc\theta\)): The reciprocal of sine. For \(PS\), this is seen as \(4 \csc\theta\).

Understanding these functions allows students to permeate deeper into mathematically describing angles, slopes, and the geometry of lines.

Line Intersection

Line intersection involves finding common points where two lines meet. It's an important application in coordinate geometry where multiple lines can intersect at a common point.To find intersection points relevant to our problem:

• Intersection with \(x=6\): This vertical line intersects our line at point \(R\). By substituting \(x=6\) into the line equation, we solve for the corresponding \(y\)-coordinate, resulting in coordinates \((6, 3\tan\theta + 4)\).
• Intersection with \(y=8\): Setting \(y=8\) and solving for \(x\) gives us point \(S\). This alleviates the importance of finding \(x\)-coordinate \(\left(\frac{4}{\tan\theta} + 3 \right)\) at which the intersection occurs.

Applying these concepts equips students with the skills necessary to address complex interactions between geometric lines, enhancing their analytical problem-solving skills in coordinate geometry.