
What is the equation of the line which cuts off the intercepts $2a\sec \theta $ and $2a\csc \theta $?
A. $x\sin \theta + y\cos \theta - 2a = 0$
B. $x\cos \theta + y\sin \theta - 2a = 0$
C. $x\sec \theta + y\csc \theta - 2a = 0$
D. $x\csc \theta + y\sec \theta - 2a = 0$
Answer
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Hint: We will apply the intercepted form of the equation to find the equation of the line. Then apply trigonometry ratios to simplify the equation.
Formula Used:
The intercept form of a line is $\dfrac{x}{a} + \dfrac{y}{b} = 1$, where $a$ is the x-intercept and $b$ is the y-intercept.
Trigonometry ratio:
$\sec \theta = \dfrac{1}{{\cos \theta }}$
$\csc \theta = \dfrac{1}{{\sin \theta }}$
Complete step by step solution:
Given a line that cuts off the intercepts $2a\sec \theta $ and $2a\csc \theta $.
The x-intercept of the line is $2a\sec \theta $.
So, the value of $a$ is $2a\sec \theta $.
The y-intercept of the line is $2a\csc \theta $.
So, the value of $b$ is $2a\csc \theta $.
Now put the value of $a$ and $b$ in the intercept form of a line.
$\dfrac{x}{{2a\sec \theta }} + \dfrac{y}{{2a\csc \theta }} = 1$
Take common $\dfrac{1}{{2a}}$ from the left side of the equation
$ \Rightarrow \dfrac{1}{{2a}}\left[ {\dfrac{x}{{\sec \theta }} + \dfrac{y}{{\csc \theta }}} \right] = 1$
Multiply both sides by $2a$
$ \Rightarrow 2a \cdot \dfrac{1}{{2a}}\left[ {\dfrac{x}{{\sec \theta }} + \dfrac{y}{{\csc \theta }}} \right] = 1 \cdot 2a$
Cancel out $2a$ from the left side
$ \Rightarrow \dfrac{x}{{\sec \theta }} + \dfrac{y}{{\csc \theta }} = 2a$
Apply the formula of trigonometry ratio formula
$ \Rightarrow \dfrac{x}{{\dfrac{1}{{\cos \theta }}}} + \dfrac{y}{{\dfrac{1}{{\sin \theta }}}} = 2a$
Simplify the above equation
$ \Rightarrow x\cos \theta + y\sin \theta = 2a$
Subtract $2a$ from both sides
$ \Rightarrow x\cos \theta + y\sin \theta - 2a = 2a - 2a$
$ \Rightarrow x\cos \theta + y\sin \theta - 2a = 0$
Option ‘B’ is correct
Note: To this type, you need to know the slope intercepts form of a line that is $\dfrac{x}{a} + \dfrac{y}{b} = 1$ where a is the x-intercept and b is the y-intercept. By substituting the value of a and b, you get the required answer.
Formula Used:
The intercept form of a line is $\dfrac{x}{a} + \dfrac{y}{b} = 1$, where $a$ is the x-intercept and $b$ is the y-intercept.
Trigonometry ratio:
$\sec \theta = \dfrac{1}{{\cos \theta }}$
$\csc \theta = \dfrac{1}{{\sin \theta }}$
Complete step by step solution:
Given a line that cuts off the intercepts $2a\sec \theta $ and $2a\csc \theta $.
The x-intercept of the line is $2a\sec \theta $.
So, the value of $a$ is $2a\sec \theta $.
The y-intercept of the line is $2a\csc \theta $.
So, the value of $b$ is $2a\csc \theta $.
Now put the value of $a$ and $b$ in the intercept form of a line.
$\dfrac{x}{{2a\sec \theta }} + \dfrac{y}{{2a\csc \theta }} = 1$
Take common $\dfrac{1}{{2a}}$ from the left side of the equation
$ \Rightarrow \dfrac{1}{{2a}}\left[ {\dfrac{x}{{\sec \theta }} + \dfrac{y}{{\csc \theta }}} \right] = 1$
Multiply both sides by $2a$
$ \Rightarrow 2a \cdot \dfrac{1}{{2a}}\left[ {\dfrac{x}{{\sec \theta }} + \dfrac{y}{{\csc \theta }}} \right] = 1 \cdot 2a$
Cancel out $2a$ from the left side
$ \Rightarrow \dfrac{x}{{\sec \theta }} + \dfrac{y}{{\csc \theta }} = 2a$
Apply the formula of trigonometry ratio formula
$ \Rightarrow \dfrac{x}{{\dfrac{1}{{\cos \theta }}}} + \dfrac{y}{{\dfrac{1}{{\sin \theta }}}} = 2a$
Simplify the above equation
$ \Rightarrow x\cos \theta + y\sin \theta = 2a$
Subtract $2a$ from both sides
$ \Rightarrow x\cos \theta + y\sin \theta - 2a = 2a - 2a$
$ \Rightarrow x\cos \theta + y\sin \theta - 2a = 0$
Option ‘B’ is correct
Note: To this type, you need to know the slope intercepts form of a line that is $\dfrac{x}{a} + \dfrac{y}{b} = 1$ where a is the x-intercept and b is the y-intercept. By substituting the value of a and b, you get the required answer.
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