Question

& \text{if }f\left( \alpha \right)=x\cos \alpha +y\sin \alpha -p\left( \alpha \right),\text{ } \\\ & \text{then the lines }f\left( \alpha \right)=0\text{ and }f\left( \beta \right)=0\text{ are }\bot \text{ to each other, if} \\\ & \left( a \right)\text{ }\alpha =\beta \text{ }\left( b \right)\alpha +\beta =\dfrac{\pi }{2} \\\ & \left( c \right)|\alpha -\beta |=\dfrac{\pi }{2}\text{ }\left( d \right)\alpha \pm \beta =\dfrac{\pi }{2} \\\ \end{aligned}$$

Answer 1

first find the two equations, by substituting the variables into given function, by substituting first variable you get first line equation similarly when you substitute the second variable you get the second line equation and then apply the condition of perpendicular lines. For finding slope of the line use this formula. If the slope of ax + by + c = 0 is p. Then the value of p is given by:
$p=-\dfrac{a}{b}$
If two straight lines with slope a, b are perpendicular then:
$a\times b=-1$.
Use the given trigonometric identity for simplification purposes:
$\cos \left( A-B \right)=\cos A.\cos B+\sin A.\sin B$

Complete step-by-step answer:
So for the first equation the function is already given.
By equating it to 0, we get:
$x\cos \alpha +y\sin \alpha -p\left( \alpha \right)=0.....\left( 1 \right)$
The equation (1) becomes the first straight line.
Let the slope of straight line 1 is m.
Slope condition:
If the slope of ax + by + c = 0 is p. Then the value of p is given by:
$p=-\dfrac{a}{b}.....\left( 2 \right)$
So by seeing at equation (2) the values of equation (1) are:
p = m
$a=\cos \alpha$
$b=\sin \alpha$
By substituting the values of p, a, b in equation (2), we get:
$m=-\dfrac{\cos \alpha }{\sin \alpha }=-\cot \alpha .....\left( 3 \right)$
So for the second equation the function is already given.
By equating it to 0 and substituting variable, we get:
$x\cos \beta +y\sin \beta -p\left( \beta \right)=0.....\left( 4 \right)$
The equation (4) becomes the second straight line.
Let the slope of straight line 2 is n.
Slope condition:
If the slope of ax + by + c = 0 is p. Then the value of p is given by:
$p=-\dfrac{a}{b}.....\left( 2 \right)$
So by seeing at equation (2) the values of equation (4) are:
p = n
$a=\cos \beta$
$b=\sin \beta$
By substituting the values of p, a, b in equation (2), we get:
$n=-\dfrac{\cos \beta }{\sin \beta }=-\cot \beta .....\left( 5 \right)$
The condition of perpendicular lines:
If two straight lines with slope a, b are perpendicular then
$a\times b=-1$.
By applying the above condition, we get:
m.n = -1
$\left( -\dfrac{\cos \alpha }{\sin \alpha } \right).\left( -\dfrac{\cos \beta }{\sin \beta } \right)=-1$
By cross multiplying, we get:
$\cos \alpha .\cos \beta =-\sin \alpha .\sin \beta$
$\cos \alpha .\cos \beta +\sin \alpha .\sin \beta =0$
We know that:
$\cos \left( A-B \right)=\cos A.\cos B+\sin A.\sin B$
By applying above trigonometric identity, we get:
$\cos \left( \alpha -\beta \right)=0$
$|\alpha -\beta |={{\cos }^{-1}}\left( 0 \right)=\dfrac{\pi }{2}$
$|\alpha -\beta |=\dfrac{\pi }{2}$

So, the correct answer is “Option c”.

Note: Take care of negative signs properly. As there is already a negative sign in the slope already if you confuse it may lead to the wrong answer. While substituting the second variable you must be careful that you must substitute it in x, y both terms. In a hurry, generally students forget to substitute in one of the terms and then lead to the wrong answer. It does not matter if you don’t substitute in the constant because we will calculate slope and then check perpendicular condition which doesn’t depend on the constant term.