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Chapter 18: Problem 50

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) \(\frac{9}{(P R)^{2}}+\frac{16}{(P S)^{2}}=1\)

Short Answer

Expert verified
All options (A) to (D) are correct.

Step by step solution

01

Equation of the Line through Point P(3,4)

The line through point \( P(3,4) \) making an angle \( \theta \) with the positive x-axis has the slope \( m = \tan \theta \). Thus, the equation of the line can be written as \( y - 4 = \tan \theta (x - 3) \), or rearranging, \( y = \tan \theta \cdot x - 3 \tan \theta + 4 \).
02

Finding Point R where the Line meets x=6

At \( x = 6 \), substitute into the line equation: \( y = \tan \theta \cdot 6 - 3 \tan \theta + 4 \) which simplifies to \( y = 3 \tan \theta + 4 \). Therefore, point \( R = (6, 3 \tan \theta + 4) \).
03

Calculating PR Length

The distance \( PR \) is given by the distance formula: \( PR = \sqrt{(6 - 3)^2 + (3 \tan \theta + 4 - 4)^2} = \sqrt{3^2 + (3 \tan \theta)^2} \). Simplifying gives \( PR = 3 \sqrt{1 + (\tan \theta)^2} = 3 \sec \theta \).
04

Finding Point S where the Line meets y=8

At \( y = 8 \), substitute into the line equation: \( 8 = \tan \theta \cdot x - 3 \tan \theta + 4 \),which simplifies to: \( 4 = \tan \theta \cdot x - 3 \tan \theta \),\( \tan \theta \cdot x = 4 + 3 \tan \theta \),\( x = \frac{4 + 3 \tan \theta}{\tan \theta} \) hence \( x = 4 \csc \theta + 3 \).Therefore, \( S = \left(4 \csc \theta + 3, 8\right) \).
05

Calculating PS Length

The distance \( PS \) is calculated as:\[ PS = \sqrt{\left(\left(4 \csc \theta + 3\right) - 3\right)^2 + (8 - 4)^2} \] \[ PS = \sqrt{(4 \csc \theta)^2 + 4^2} \] \[ PS = \sqrt{16 \csc^2 \theta + 16} \] \[ PS = 4 \sqrt{\csc^2 \theta + 1} = 4 \csc \theta \]
06

Verification of Option (A) and (B)

From Step 3, we've derived \( PR = 3 \sec \theta \) verifying Option (A), and from Step 5, \( PS = 4 \csc \theta \) confirming Option (B).
07

Relationship between PR and PS

Using the trigonometric identity \(\csc 2\theta = \frac{1}{2\sin \theta \cos \theta}\),calculate:\( PR + PS = 3 \sec \theta + 4 \csc \theta = \frac{3 \sin \theta + 4 \cos \theta}{\sin \theta \cos \theta} \)Multiply and divide by 2:\( = \frac{2(3 \sin \theta + 4 \cos \theta)}{2\sin \theta \cos \theta} = \frac{2(3 \sin \theta + 4 \cos \theta)}{\sin 2\theta}\), verifying Option (C).
08

Verification of Option (D)

Verify \( \frac{9}{(PR)^2} + \frac{16}{(PS)^2} = 1 \):Substituting \( PR = 3 \sec \theta \), \( PS = 4 \csc \theta \),\( \frac{9}{9\sec^2\theta} + \frac{16}{16\csc^2\theta} = 1 \)Simplifies to:\( \cos^2\theta + \sin^2\theta = 1 \),a trigonometric identity. Hence, Option (D) is also verified.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Identities
Understanding trigonometric identities can simplify complex problems. They are formulas involving trigonometric functions like sine and cosine that are universally true for all angles. One key identity used in the exercise is
  • \( an^2 \theta + 1 = \sec^2 \theta \)
This identity helps transform expressions and simplify equations involving tangent and secant. Other identities, essential for the problem, include:
  • \( \csc^2 \theta = 1 + \cot^2 \theta \)
  • \( \sin^2 \theta + \cos^2 \theta = 1 \)
These identities bridge different trigonometric functions through algebraic manipulation. To see them in action, apply them when simplifying expressions. For instance, in step 8 of the solution: verifying \( \cos^2 \theta + \sin^2 \theta = 1 \) ensures correctness of distance relations between points R and S.
Distance Formula
The distance formula provides a method to calculate the distance between two points in the coordinate plane. The formula derives from the Pythagorean theorem and is expressed as: \\[d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\]. \It calculates the spatial separation using the differences in x and y coordinates. Using this formula aids in finding lengths of segments like PR and PS in exercises. For example:
  • Distance PR is calculated with this formula using endpoint coordinates.
  • Calculate PS by inserting its respective coordinates into the formula.
By repeatedly using the distance formula, you strengthen your ability to handle real-world geometry problems. Look for connections between this and physics or engineering tasks involving precise measurements.
Equation of a Line
Understanding the equation of a line helps in creating precise mathematical models of geometric situations. The slope-intercept form is often used, represented as \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. For the problem, remember:
  • The slope \(m\) is derived from the angle \(\theta\) as \(\tan \theta \).
  • Point-slope form: expresses a line based on its slope and a single point, \(y - y_1 = m(x - x_1)\).
The transition between these forms provides flexibility in analyzing various line properties. To solidify this concept, practice converting between forms regularly. Points R and S are determined using these equations, illustrating real applications in coordinate geometry. Emphasize on lines parallel or perpendicular to axes as they often become scenarios you encounter.

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Most popular questions from this chapter

The coordinates of the point which is at unit distance from the lines \(L_{1} \equiv 3 x-4 y+1=0\) and \(L_{2} \equiv 8 x+6 y+\) \(1=0\) and lies below \(L_{1}\) and above \(L_{2}\) are (A) \(\left(\frac{6}{5}, \frac{1}{10}\right)\) (B) \(\left(\frac{6}{5},-\frac{1}{10}\right)\) (C) \(\left(\frac{6}{5}, \frac{1}{5}\right)\) (D) \(\left(\frac{6}{5},-\frac{1}{5}\right)\)

In oblique coordinates, the equation \(y=m x+c\) represents a straight line which is inclined at an angle $$ \tan ^{-1}\left(\frac{m \sin w}{1+m \cos w}\right) $$ to the \(x\)-axis, where \(w\) is the angle between the axes. If \(\theta\) be the angle between two lines \(y=m_{1} x+c_{1}\) and \(y=m_{2} x\) \(+c_{2}, w\) be the angle between the axes, then $$ \tan \theta=\frac{\left(m_{1}-m_{2}\right) \sin w}{1+\left(m_{1}+m_{2}\right) \cos w+m_{1} m_{2}} $$ The two given lines are parallel if \(m_{1}=m_{2}\). The two lines are perpendicular if \(1+\left(m_{1}+m_{2}\right) \cos w+\) \(m_{1} m_{2}=0\) If \(y=x \tan \frac{11 \pi}{24}\) and \(y=x \tan \frac{19 \pi}{24}\) represent two straight lines at right angles, then the angle between the axes is (A) \(\frac{\pi}{6}\) (B) \(\frac{\pi}{4}\) (C) \(\frac{\pi}{3}\) (D) \(\frac{\pi}{2}\)

The condition to be imposed on \(\beta\) so that \((0, \beta)\) lies on or inside the triangle having sides \(y+3 x+2=0\), \(3 y-2 x-5=0\) and \(4 y+x-14=0\) is (A) \(0<\beta<\frac{5}{3}\) (B) \(0<\beta<\frac{7}{2}\) (C) \(\frac{5}{3} \leq \beta \leq \frac{7}{2}\) (D) none of these

Three straight lines \(2 x+11 y-5=0,24 x+7 y-20=\) 0 and \(4 x-3 y-2=0\) : (A) form a triangle (B) are only concurrent (C) are concurrent with one line bisecting the angle between the other two (D) none of the above

\(D\) is a point on \(A C\) of the triangle with vertices \(A(2,\), 3), \(B(1,-3), C(-4,-7)\) and \(B D\) divides \(A B C\) into two triangles of equal area. The equation of the line drawn through \(B\) at right angles to \(B D\) is (A) \(y-2 x+5=0\) (B) \(2 y-x+5=0\) (C) \(y+2 x-5=0\) (D) \(2 y+x-5=0\)
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