Question

If straight lines $ax+by+p\ =\ 0$ and $x\cos \alpha +y\sin \alpha -p\ =\ 0$ include an angle $\dfrac{\pi }{4}$ between them and meet the straight line $x\sin \alpha -y\cos \alpha \ =\ 0$ in the same point, then the value of ${{a}^{2}}+{{b}^{2}}$ is equal to
(a)1
(b)2
(c)3
(d)4

Answer 1

Hint: First find the intersection point of lines without a, b, c. Given that the line $ax+by+c\ =\ 0$ passes through the same point. So, finding the intersection point and then substituting it back into $ax+by+c\ =\ 0$ give us one relation. Let x be angle between 2 lines with slope m, n then:
\[\tan x\ =\ \left| \dfrac{m-n}{1+mn} \right|\]
Use this to find another relation. Solve these both to eliminate \[\alpha \].

Complete step-by-step answer:

Given in question second line equation, which is simple is given by:
$x\sin \alpha -y\cos \alpha \ =\ 0$
$x\ =\ \dfrac{y\cos \alpha }{\sin \alpha }$ …..(1)
By substituting this x value in the straight line whose equation is given by $x\cos \alpha +y\sin \alpha \ =\ p$, we get:
$y\dfrac{{{\cos }^{2}}\alpha }{\sin \alpha }+y\sin \alpha \ =\ p$
By taking least common multiple and cross multiply, we get:
$y{{\cos }^{2}}\alpha +y{{\sin }^{2}}\alpha \ =\ p\sin \alpha $
By taking y as common from left hand side, we get:
$y\left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right)\ =\ p\sin \alpha $
As we know that term inside bracket is always 1, we get:
\[y=\ p\sin \alpha \]
By substituting this value of y in the equation (1), we get value of x to be:
\[x\ =\ p\dfrac{\left( \sin \alpha \right)\left( \cos \alpha \right)}{\sin \alpha }\ =\ p\cos \alpha \]
Substituting this x, y values into equation $ax+by+p\ =\ 0$ , we get:
$a\left( p\cos \alpha \right)+b\left( p\sin \alpha \right)+p\ =\ 0$
By multiplying terms inside the bracket , we get:
$a\cos \alpha +b\sin \alpha +1\ =\ 0$
From above equation, we can say:
$a\cos \alpha +b\sin \alpha \ =\ -1$…..(2)
Slope of line with equation $ax+by+c\ =\ 0$, is given by
\[m\ =\ \dfrac{-a}{b}\]
Slope of line with equation $x\cos \alpha +y\sin \alpha \ =\ p$, is given by as follows:
\[n\ =\ -\cot \alpha \]
The angle between them is \[\dfrac{\pi }{4}\]. So, by applying angle formula
\[\tan x\ =\ \left| \dfrac{m-n}{1+mn} \right|\ \Rightarrow \ \tan \dfrac{\pi }{4}\ =\ \left| \dfrac{m-n}{1+mn} \right|\]
\[\Rightarrow \left| 1+mn \right|\ =\ \left| m-n \right|\ \]
By substituting m, n values into above equation, we get
\[\left| a\sin \alpha -b\cos \alpha \right|\ =\ \left| b\sin \alpha +a\cos \alpha \right|\ =\ 1\]
From above we can say:
\[\left| a\sin \alpha -b\cos \alpha \right|\ =\ 1\]…..(3)
By squaring and adding equation (2) and (3), we get
${{a}^{2}}{{\cos }^{2}}\alpha +{{b}^{2}}{{\sin }^{2}}\alpha +2ab\sin \alpha \cos \alpha +{{b}^{2}}{{\cos }^{2}}\alpha +{{a}^{2}}{{\sin }^{2}}\alpha -2ab\sin \alpha \cos \alpha \ =1+1$
If we take terms of a and b together on left hand side you get the general identity of $\sin ,\ \cos \ \Rightarrow \ {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$, by using this we can simplifying above equation and we can cancel common terms, we get:
${{a}^{2}}+{{b}^{2}}\ =\ 2$.
Therefore option (b) is correct for the value of required expression.

Note: Be careful while applying $\tan x$ formula as there is modulus you can take line as m and remaining line as n. Whenever you see modulus always square the equation. Because while we take square of a number in the modulus we can remove the modulus because by squaring we get the same result for positive and negative so the modulus doesn't play any role here.