Angle between the pair of straight lines, Condition for parallel and perpendicular lines Questions in English

(C) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of lines if $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Comparing $x^2 - 3xy + \lambda y^2 + 3x - 5y + 2 = 0$ with the general form,we have $a=1, b=\lambda, c=2, h=-\frac{3}{2}, g=\frac{3}{2}, f=-\frac{5}{2}$.
Substituting these values into the condition:
$(1)(\lambda)(2) + 2(-\frac{5}{2})(\frac{3}{2})(-\frac{3}{2}) - (1)(-\frac{5}{2})^2 - (\lambda)(\frac{3}{2})^2 - (2)(-\frac{3}{2})^2 = 0$
$2\lambda + \frac{45}{4} - \frac{25}{4} - \frac{9\lambda}{4} - \frac{18}{4} = 0$
$2\lambda - \frac{9\lambda}{4} + \frac{45-25-18}{4} = 0$
$-\frac{\lambda}{4} + \frac{2}{4} = 0 \Rightarrow \lambda = 2$.
Now,the angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} \right|$.
Substituting $a=1, b=2, h=-\frac{3}{2}$:
$\tan \theta = \left| \frac{2\sqrt{(-\frac{3}{2})^2 - (1)(2)}}{1+2} \right| = \left| \frac{2\sqrt{\frac{9}{4} - 2}}{3} \right| = \left| \frac{2\sqrt{\frac{1}{4}}}{3} \right| = \frac{2(\frac{1}{2})}{3} = \frac{1}{3}$.
Therefore,$\csc^2 \theta = 1 + \cot^2 \theta = 1 + (\frac{1}{\tan \theta})^2 = 1 + (3)^2 = 1 + 9 = 10$.

(A) The given equation of the pair of lines is $x^2+\lambda xy-y^2 \tan^2 \theta=0$.
Comparing this with the general form $ax^2+2hxy+by^2=0$,we get $a=1$,$h=\frac{\lambda}{2}$,and $b=-\tan^2 \theta$.
The angle $\alpha$ between the pair of lines is given by $\tan \alpha = \left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$.
Here,$\alpha = 2\theta$,so $\tan 2\theta = \left|\frac{2\sqrt{(\frac{\lambda}{2})^2 - (1)(-\tan^2 \theta)}}{1-\tan^2 \theta}\right|$.
Using the identity $\tan 2\theta = \frac{2\tan \theta}{1-\tan^2 \theta}$,we have $\frac{2\tan \theta}{1-\tan^2 \theta} = \left|\frac{2\sqrt{\frac{\lambda^2}{4}+\tan^2 \theta}}{1-\tan^2 \theta}\right|$.
Squaring both sides,we get $\frac{4\tan^2 \theta}{(1-\tan^2 \theta)^2} = \frac{4(\frac{\lambda^2}{4}+\tan^2 \theta)}{(1-\tan^2 \theta)^2}$.
This simplifies to $\tan^2 \theta = \frac{\lambda^2}{4} + \tan^2 \theta$,which implies $\frac{\lambda^2}{4} = 0$.
Therefore,$\lambda = 0$.

(C) Given equation: $(x \cos \alpha + y \sin \alpha)^2 = (x^2 + y^2) \sin^2 \alpha$
Expanding the left side: $x^2 \cos^2 \alpha + y^2 \sin^2 \alpha + 2xy \sin \alpha \cos \alpha = x^2 \sin^2 \alpha + y^2 \sin^2 \alpha$
Subtracting $y^2 \sin^2 \alpha$ from both sides: $x^2 \cos^2 \alpha + 2xy \sin \alpha \cos \alpha = x^2 \sin^2 \alpha$
Rearranging: $x^2(\cos^2 \alpha - \sin^2 \alpha) + 2xy \sin \alpha \cos \alpha = 0$
This is a homogeneous equation of the form $ax^2 + 2hxy + by^2 = 0$,where $a = \cos^2 \alpha - \sin^2 \alpha$,$h = \sin \alpha \cos \alpha$,and $b = 0$.
For the lines to be perpendicular,the condition is $a + b = 0$.
Substituting the values: $(\cos^2 \alpha - \sin^2 \alpha) + 0 = 0$
$\cos^2 \alpha = \sin^2 \alpha$
$\tan^2 \alpha = 1$
Since $\alpha$ is typically in the first quadrant for such problems,$\alpha = \frac{\pi}{4}$.

(B) The given equation of the pair of lines is $x^2-3xy+\lambda y^2+3x-5y+2=0$.
Comparing this with the general form $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1$,$2h=-3 \Rightarrow h=-\frac{3}{2}$,and $b=\lambda$.
The angle $\theta$ between the lines is given by $\tan \theta = \left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$.
Given $\tan \theta = \frac{1}{3}$,we have $\frac{1}{3} = \left|\frac{2\sqrt{(-\frac{3}{2})^2 - (1)(\lambda)}}{1+\lambda}\right|$.
Squaring both sides,$\frac{1}{9} = \frac{4(\frac{9}{4}-\lambda)}{(1+\lambda)^2}$.
$(1+\lambda)^2 = 36(\frac{9}{4}-\lambda) = 81 - 36\lambda$.
$\lambda^2 + 2\lambda + 1 = 81 - 36\lambda$.
$\lambda^2 + 38\lambda - 80 = 0$.
$(\lambda+40)(\lambda-2) = 0$.
Since $\lambda \geq 0$,we have $\lambda = 2$.

(C) The formula for the angle $\theta$ between the pair of lines $ax^2 + 2hxy + by^2 = 0$ is $\tan \theta = \left| \frac{2 \sqrt{h^2 - ab}}{a + b} \right|$.
Comparing $kx^2 - 4xy + y^2 = 0$ with $ax^2 + 2hxy + by^2 = 0$,we get $a = k$,$2h = -4$ (so $h = -2$),and $b = 1$.
Given $\tan \theta = \frac{1}{2}$,we have $\frac{1}{2} = \left| \frac{2 \sqrt{(-2)^2 - k(1)}}{k + 1} \right|$.
$\frac{1}{2} = \left| \frac{2 \sqrt{4 - k}}{k + 1} \right| \Rightarrow \frac{1}{4} = \frac{4(4 - k)}{(k + 1)^2}$.
$(k + 1)^2 = 16(4 - k) \Rightarrow k^2 + 2k + 1 = 64 - 16k$.
$k^2 + 18k - 63 = 0$.
$(k + 21)(k - 3) = 0$.
Since $\theta$ is an acute angle,the denominator $a+b = k+1$ must be such that the expression is valid. For $k = 3$,$a+b = 4 \neq 0$. Thus,$k = 3$.

(C) The formula for the acute angle $\theta$ between the pair of lines represented by $ax^2+2hxy+by^2=0$ is given by $\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
Given $\theta = \frac{\pi}{4}$,we have $\tan \frac{\pi}{4} = 1$.
Therefore,$1 = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
Squaring both sides,we get $1 = \frac{4(h^2-ab)}{(a+b)^2}$.
This implies $(a+b)^2 = 4h^2 - 4ab$.
Expanding the left side,$a^2 + 2ab + b^2 = 4h^2 - 4ab$.
Rearranging the terms,$4h^2 = a^2 + 6ab + b^2$.

(A) Given the equation of the pair of lines: $mx^2+2hxy+ny^2=0$.
The condition given is $(m+3n)(3m+n)=4h^2$.
Expanding the condition: $3m^2+mn+9mn+3n^2=4h^2 \Rightarrow 3m^2+10mn+3n^2=4h^2$.
The formula for the tangent of the angle $\theta$ between the lines $ax^2+2hxy+by^2=0$ is $\tan \theta = \left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$.
Here,$a=m$ and $b=n$.
Substituting the values: $\tan \theta = \left|\frac{2\sqrt{h^2-mn}}{m+n}\right|$.
From the given condition,$4h^2 = 3m^2+10mn+3n^2$,so $h^2 = \frac{3m^2+10mn+3n^2}{4}$.
Then $h^2-mn = \frac{3m^2+10mn+3n^2-4mn}{4} = \frac{3m^2+6mn+3n^2}{4} = \frac{3(m+n)^2}{4}$.
Therefore,$\sqrt{h^2-mn} = \frac{\sqrt{3}|m+n|}{2}$.
Substituting this into the tangent formula: $\tan \theta = \left|\frac{2 \cdot \frac{\sqrt{3}|m+n|}{2}}{m+n}\right| = \sqrt{3}$.
Thus,$\theta = 60^{\circ} = \frac{\pi}{3}$.

(D) The general equation of a pair of straight lines is $ax^2 + 2hxy + by^2 = 0$.
For the lines to be perpendicular,the condition is $a + b = 0$.
Comparing the given equation $(k^2+2) x^2 + 3xy - 6y^2 = 0$ with the general form,we have $a = (k^2+2)$ and $b = -6$.
Substituting these into the condition $a + b = 0$:
$(k^2+2) + (-6) = 0$
$k^2 - 4 = 0$
$k^2 = 4$
$k = \pm 2$.

(C) The given equation is $(x^2+y^2) \sin \theta + 2xy = 0$,which can be written as $(\sin \theta)x^2 + 2xy + (\sin \theta)y^2 = 0$.
Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = \sin \theta$,$h = 1$,and $b = \sin \theta$.
Let $\alpha$ be the acute angle between the lines. The formula for the angle is $\tan \alpha = \left| \frac{2\sqrt{h^2 - ab}}{a+b} \right|$.
Substituting the values,we get $\tan \alpha = \left| \frac{2\sqrt{1^2 - (\sin \theta)(\sin \theta)}}{\sin \theta + \sin \theta} \right|$.
$\tan \alpha = \left| \frac{2\sqrt{1 - \sin^2 \theta}}{2 \sin \theta} \right| = \left| \frac{2\cos \theta}{2 \sin \theta} \right| = |\cot \theta|$.
Since $\alpha$ is the acute angle,$\tan \alpha = \tan(\frac{\pi}{2} - \theta)$.
Therefore,$\alpha = \frac{\pi}{2} - \theta$.

(B) The acute angle $\theta$ between the pair of lines represented by $ax^{2}+2hxy+by^{2}=0$ is given by $\tan \theta = \left| \frac{2\sqrt{h^{2}-ab}}{a+b} \right|$.
Comparing the given equation $x^{2}-4xy+y^{2}=0$ with $ax^{2}+2hxy+by^{2}=0$,we get $a=1$,$2h=-4$ (so $h=-2$),and $b=1$.
Substituting these values into the formula:
$\tan \theta = \left| \frac{2\sqrt{(-2)^{2}-(1)(1)}}{1+1} \right|$
$\tan \theta = \left| \frac{2\sqrt{4-1}}{2} \right|$
$\tan \theta = \sqrt{3}$.
Since $\theta = \tan^{-1}(k)$,we have $\tan^{-1}(k) = \tan^{-1}(\sqrt{3})$,which implies $k = \sqrt{3}$.

(D) Comparing the given equation $3x^{2}-4\sqrt{3}xy+3y^{2}=0$ with the general form $ax^{2}+2hxy+by^{2}=0$,we get:
$a=3, h=-2\sqrt{3}, b=3$.
We know that the acute angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{h^{2}-ab}}{a+b} \right|$.
Substituting the values,we get $\tan \theta = \left| \frac{2\sqrt{(-2\sqrt{3})^{2}-(3)(3)}}{3+3} \right|$.
$\tan \theta = \left| \frac{2\sqrt{12-9}}{6} \right| = \left| \frac{2\sqrt{3}}{6} \right| = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.
Since $\tan \theta = \frac{1}{\sqrt{3}}$,we have $\theta = 30^{\circ}$.

(D) The given equation of the pair of straight lines is $x^{2}-3xy+\lambda y^{2}+3x-5y+2=0$.
Comparing this with the general form $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$,we get $a=1$,$2h=-3 \Rightarrow h=-\frac{3}{2}$,and $b=\lambda$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{h^{2}-ab}}{a+b} \right|$.
Given $\tan \theta = \frac{1}{3}$,we have $\frac{1}{3} = \left| \frac{2\sqrt{\frac{9}{4}-\lambda}}{1+\lambda} \right|$.
Squaring both sides: $\frac{1}{9} = \frac{4(\frac{9}{4}-\lambda)}{(1+\lambda)^{2}} = \frac{9-4\lambda}{(1+\lambda)^{2}}$.
$(1+\lambda)^{2} = 9(9-4\lambda) = 81-36\lambda$.
$1+2\lambda+\lambda^{2} = 81-36\lambda$.
$\lambda^{2}+38\lambda-80=0$.
$(\lambda+40)(\lambda-2)=0$.
Since $\lambda \geq 0$,we have $\lambda=2$.

(A) The given equation is $x^{2}+2xy \operatorname{cosec} \alpha+y^{2}=0$.
Comparing this with the standard form $ax^{2}+2hxy+by^{2}=0$,we get $a=1$,$h=\operatorname{cosec} \alpha$,and $b=1$.
Let $\theta$ be the angle between the lines.
The formula for the angle between the lines is $\tan \theta = \left| \frac{2\sqrt{h^{2}-ab}}{a+b} \right|$.
Substituting the values,we get $\tan \theta = \left| \frac{2\sqrt{\operatorname{cosec}^{2} \alpha - 1}}{1+1} \right|$.
Since $\operatorname{cosec}^{2} \alpha - 1 = \cot^{2} \alpha$,we have $\tan \theta = \left| \frac{2\sqrt{\cot^{2} \alpha}}{2} \right| = |\cot \alpha|$.
Thus,$\tan \theta = \cot \alpha = \tan \left( \frac{\pi}{2} - \alpha \right)$.
Therefore,$\theta = \frac{\pi}{2} - \alpha$.

(D) The given equation is $y^{2} \sin^{2} \theta - xy \sin^{2} \theta + x^{2}(\cos^{2} \theta - 1) = 0$.
Since $\cos^{2} \theta - 1 = -\sin^{2} \theta$,the equation becomes $y^{2} \sin^{2} \theta - xy \sin^{2} \theta - x^{2} \sin^{2} \theta = 0$.
Dividing by $\sin^{2} \theta$ (assuming $\sin^{2} \theta \neq 0$),we get $y^{2} - xy - x^{2} = 0$.
For a general second-degree equation $ax^{2} + 2hxy + by^{2} = 0$,the angle $\phi$ between the lines is given by $\tan \phi = \left| \frac{2\sqrt{h^{2} - ab}}{a + b} \right|$.
Here,$a = -1$,$2h = -1$ (so $h = -1/2$),and $b = 1$.
Since $a + b = -1 + 1 = 0$,the sum of the coefficients of $x^{2}$ and $y^{2}$ is zero.
This implies that the lines are perpendicular to each other.
Therefore,the angle between the lines is $\frac{\pi}{2}$.

(C) The general equation of a pair of straight lines passing through the origin is given by $ax^2 + 2hxy + by^2 = 0$.
If these lines are perpendicular to each other,the condition is $a + b = 0$.
Given the equation $(1+\sin^2 \theta) x^2 + 2hxy + 2\sin \theta y^2 = 0$,we identify $a = 1+\sin^2 \theta$ and $b = 2\sin \theta$.
Applying the condition $a + b = 0$:
$(1+\sin^2 \theta) + 2\sin \theta = 0$
$(1+\sin \theta)^2 = 0$
$1+\sin \theta = 0$
$\sin \theta = -1$
For $\theta \in [0, 2\pi]$,the value of $\theta$ satisfying $\sin \theta = -1$ is $\theta = \frac{3\pi}{2}$.

(A) The given equation is $x^{2}-xy-6y^{2}-7x+31y-18=0$.
Comparing this with the general equation $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$,we get $a=1$,$b=-6$,and $2h=-1$,which implies $h=-\frac{1}{2}$.
The angle $\theta$ between the pair of lines is given by the formula $\tan \theta = \left| \frac{2\sqrt{h^{2}-ab}}{a+b} \right|$.
Substituting the values,we get $\tan \theta = \left| \frac{2\sqrt{(-\frac{1}{2})^{2} - (1)(-6)}}{1+(-6)} \right|$.
$\tan \theta = \left| \frac{2\sqrt{\frac{1}{4}+6}}{-5} \right| = \left| \frac{2\sqrt{\frac{25}{4}}}{-5} \right|$.
$\tan \theta = \left| \frac{2 \times \frac{5}{2}}{-5} \right| = \left| \frac{5}{-5} \right| = |-1| = 1$.
Therefore,$\theta = \tan^{-1}(1) = \frac{\pi}{4}$.

(A) The angle $\varphi$ between the lines represented by $a x^{2}+2 h x y+b y^{2}=0$ is given by $\tan \varphi = \left| \frac{2 \sqrt{h^{2}-a b}}{a+b} \right|$.
For the equation $x^{2}+2 x y \sec \theta+y^{2}=0$,we have $a=1$,$b=1$,and $h=\sec \theta$.
Substituting these values into the formula:
$\tan \varphi = \left| \frac{2 \sqrt{\sec^{2} \theta - (1)(1)}}{1+1} \right|$
$\tan \varphi = \left| \frac{2 \sqrt{\tan^{2} \theta}}{2} \right|$
$\tan \varphi = \tan \theta$.
Therefore,the angle between the lines is $\theta$.

(C) The general equation of a pair of straight lines is given by $Ax^{2}+2Hxy+By^{2}+2Gx+2Fy+C=0$. \\ For the pair of lines to be perpendicular,the sum of the coefficients of $x^{2}$ and $y^{2}$ must be zero. \\ That is,$A+B=0$. \\ Comparing the given equation $12x^{2}+7xy+ay^{2}+13x-y+3=0$ with the general form,we have $A=12$ and $B=a$. \\ Substituting these values into the condition $A+B=0$,we get $12+a=0$. \\ Therefore,$a=-12$.

(D) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of lines if $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Comparing with $x^2 - 3xy + \lambda y^2 + 3x - 5y + 2 = 0$,we have $a=1, h=-3/2, b=\lambda, g=3/2, f=-5/2, c=2$.
Substituting these values: $(1)(\lambda)(2) + 2(-5/2)(3/2)(-3/2) - (1)(-5/2)^2 - (\lambda)(3/2)^2 - (2)(-3/2)^2 = 0$.
$2\lambda + 45/4 - 25/4 - 9\lambda/4 - 9/2 = 0$.
$2\lambda - 9\lambda/4 + 5 - 4.5 = 0 \implies -\lambda/4 + 0.5 = 0 \implies \lambda = 2$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} \right|$.
Here $a=1, b=2, h=-3/2$.
$\tan \theta = \left| \frac{2\sqrt{9/4 - 2}}{1+2} \right| = \frac{2\sqrt{1/4}}{3} = \frac{1}{3}$.
Since $\tan \theta = 1/3$,then $\cot \theta = 3$.
$\operatorname{cosec}^2 \theta = 1 + \cot^2 \theta = 1 + 3^2 = 10$.
Thus,$\frac{\operatorname{cosec}^2 \theta}{\sqrt{10}} = \frac{10}{\sqrt{10}} = \sqrt{10}$.

(B) The given equation is $x^2-3xy+2y^2+3x-5y+2=0$.
Comparing this with the general equation of a pair of straight lines $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get:
$a=1$,$2h=-3 \implies h=-\frac{3}{2}$,$b=2$.
The angle $\theta$ between the pair of lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
Substituting the values:
$\tan \theta = \left| \frac{2\sqrt{(-\frac{3}{2})^2 - (1)(2)}}{1+2} \right| = \left| \frac{2\sqrt{\frac{9}{4}-2}}{3} \right| = \left| \frac{2\sqrt{\frac{1}{4}}}{3} \right| = \frac{2 \times \frac{1}{2}}{3} = \frac{1}{3}$.
Since $\tan \theta = \frac{1}{3}$,we can form a right-angled triangle with opposite side $1$ and adjacent side $3$.
The hypotenuse is $\sqrt{1^2+3^2} = \sqrt{10}$.
Therefore,$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{3}{\sqrt{10}}$.

(C) The general equation of a pair of straight lines passing through the origin is $ax^2 + 2hxy + by^2 = 0$.
Comparing $px^2 - qy^2 = 0$ with the general form,we get $a = p$,$h = 0$,and $b = -q$.
For the lines to be distinct and real,the condition is $h^2 - ab > 0$.
Substituting the values,we get $0^2 - (p)(-q) > 0$.
This simplifies to $pq > 0$.

(A) Comparing the given equation with $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$,we get $a=1, h=-\frac{3}{2}, b=\lambda, g=\frac{3}{2}, f=-\frac{5}{2}, c=2$.
For the equation to represent a pair of lines,the determinant condition must hold: $\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0$.
Substituting the values: $\begin{vmatrix} 1 & -\frac{3}{2} & \frac{3}{2} \\ -\frac{3}{2} & \lambda & -\frac{5}{2} \\ \frac{3}{2} & -\frac{5}{2} & 2 \end{vmatrix} = 0$.
Multiplying by $8$ to simplify: $\begin{vmatrix} 2 & -3 & 3 \\ -3 & 2\lambda & -5 \\ 3 & -5 & 4 \end{vmatrix} = 0$.
Expanding the determinant: $2(8\lambda - 25) + 3(-12 + 15) + 3(15 - 6\lambda) = 0$.
$16\lambda - 50 + 9 + 45 - 18\lambda = 0$ $\Rightarrow -2\lambda + 4 = 0$ $\Rightarrow \lambda = 2$.
Now,the angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{h^{2}-ab}}{a+b} \right|$.
$\tan \theta = \left| \frac{2\sqrt{(-\frac{3}{2})^{2} - (1)(2)}}{1+2} \right| = \left| \frac{2\sqrt{\frac{9}{4}-2}}{3} \right| = \left| \frac{2\sqrt{\frac{1}{4}}}{3} \right| = \frac{2(\frac{1}{2})}{3} = \frac{1}{3}$.
Since $\tan \theta = \frac{1}{3}$,we have $\cot \theta = 3$.
Therefore,$\operatorname{cosec}^{2} \theta = 1 + \cot^{2} \theta = 1 + (3)^{2} = 1 + 9 = 10$.

(A) The given equation of the pair of straight lines is $(\cos^{2} \alpha - 1) x^{2} - 2 \cos^{2} \alpha \cdot xy + \sin^{2} \alpha y^{2} = 0$.
Comparing this with the general equation $ax^{2} + 2hxy + by^{2} = 0$,we get:
$a = \cos^{2} \alpha - 1 = -\sin^{2} \alpha$,
$h = -\cos^{2} \alpha$,
$b = \sin^{2} \alpha$.
Let $\theta$ be the angle between the lines. The formula for the angle is $\tan \theta = \left| \frac{2 \sqrt{h^{2} - ab}}{a + b} \right|$.
Substituting the values:
$a + b = -\sin^{2} \alpha + \sin^{2} \alpha = 0$.
Since the denominator is $0$,the value of $\tan \theta$ is undefined,which implies $\theta = 90^{\circ}$.
Thus,the lines are perpendicular.

(B) The given equation of the pair of lines is $x^{2}+2xy-y^{2}=0$.
Comparing this with the general form $ax^{2}+2hxy+by^{2}=0$,we get $a=1$,$h=1$,and $b=-1$.
We know that the lines represented by $ax^{2}+2hxy+by^{2}=0$ are perpendicular if $a+b=0$.
Here,$a+b = 1 + (-1) = 0$.
Since the sum of the coefficients of $x^{2}$ and $y^{2}$ is zero,the lines are perpendicular to each other.
Therefore,the angle between the lines is $\frac{\pi}{2}$.

(A) The given equation is $[x \cos \theta - y][(\cos \theta + \tan \alpha) x - (1 - \cos \theta \tan \alpha) y] = 0$.
This represents two lines:
$L_1: x \cos \theta - y = 0 \implies y = (\cos \theta) x$,so slope $m_1 = \cos \theta$.
$L_2: (\cos \theta + \tan \alpha) x - (1 - \cos \theta \tan \alpha) y = 0 \implies y = \frac{\cos \theta + \tan \alpha}{1 - \cos \theta \tan \alpha} x$,so slope $m_2 = \frac{\cos \theta + \tan \alpha}{1 - \cos \theta \tan \alpha}$.
The angle $\phi$ between two lines is given by $\tan \phi = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$.
Substituting the values of $m_1$ and $m_2$:
$\tan \phi = \left| \frac{\frac{\cos \theta + \tan \alpha}{1 - \cos \theta \tan \alpha} - \cos \theta}{1 + \left( \frac{\cos \theta + \tan \alpha}{1 - \cos \theta \tan \alpha} \right) \cos \theta} \right|$
$\tan \phi = \left| \frac{\cos \theta + \tan \alpha - \cos \theta (1 - \cos \theta \tan \alpha)}{1 - \cos \theta \tan \alpha + \cos \theta (\cos \theta + \tan \alpha)} \right|$
$\tan \phi = \left| \frac{\cos \theta + \tan \alpha - \cos \theta + \cos^2 \theta \tan \alpha}{1 - \cos \theta \tan \alpha + \cos^2 \theta + \cos \theta \tan \alpha} \right|$
$\tan \phi = \left| \frac{\tan \alpha (1 + \cos^2 \theta)}{1 + \cos^2 \theta} \right| = \tan \alpha$.
Thus,$\phi = \alpha$.

(B) The general equation of a pair of lines is $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$. Comparing this with $9x^2 + axy + 4y^2 + 6x + by - 3 = 0$,we get $A=9, H=a/2, B=4, G=3, F=b/2, C=-3$.
For parallel lines,$H^2 = AB$,so $(a/2)^2 = 9 \times 4 = 36$,which gives $a^2 = 144$,so $a = \pm 12$.
Also,for parallel lines,the ratio of coefficients of $x^2, xy, y^2$ must be the same as the ratio of coefficients of $x, y$ terms,i.e.,$A/G = H/F = G/C$ is not applicable here,but rather $A/H = H/B$ and $G/F = H/B$ or simply using the condition $h^2=ab$ and $af^2=bg^2$.
Using $af^2 = bg^2$,we have $9(b/2)^2 = 4(3)^2$ $\Rightarrow 9(b^2/4) = 36$ $\Rightarrow b^2/4 = 4$ $\Rightarrow b^2 = 16$ $\Rightarrow b = \pm 4$.
Testing $a=12, b=4$ in the equation $9x^2 + 12xy + 4y^2 + 6x + 4y - 3 = 0$,we get $(3x + 2y)^2 + 2(3x + 2y) - 3 = 0$. Let $t = 3x + 2y$,then $t^2 + 2t - 3 = 0 \Rightarrow (t+3)(t-1) = 0$. This represents two parallel lines $3x + 2y + 3 = 0$ and $3x + 2y - 1 = 0$.

(B) The given equation of the pair of lines is:
$(x^2+y^2) \cos^2 \theta = (x \cos \theta + y \sin \theta)^2$
Expanding the right side:
$(x^2+y^2) \cos^2 \theta = x^2 \cos^2 \theta + y^2 \sin^2 \theta + 2xy \sin \theta \cos \theta$
Subtracting $x^2 \cos^2 \theta$ from both sides:
$y^2 \cos^2 \theta = y^2 \sin^2 \theta + 2xy \sin \theta \cos \theta$
Rearranging into the general form $Ax^2 + 2Hxy + By^2 = 0$:
$0x^2 + (2 \sin \theta \cos \theta)xy + (\sin^2 \theta - \cos^2 \theta)y^2 = 0$
For the pair of lines to be perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$A + B = 0$
$0 + (\sin^2 \theta - \cos^2 \theta) = 0$
$\sin^2 \theta = \cos^2 \theta$
$\tan^2 \theta = 1$
$\tan \theta = \pm 1$
Thus,$\theta = \frac{\pi}{4}$ or $\frac{3\pi}{4}$.

(C) Given the equation of the pair of lines: $4x^2 - 24xy + 11y^2 = 0$.
Factorizing the quadratic equation: $(2x - y)(2x - 11y) = 0$.
Thus,the equations of the two lines are $y = 2x$ and $y = \frac{2}{11}x$.
The angles $\theta_1$ and $\theta_2$ made by these lines with the $X$-axis are $\tan \theta_1 = 2$ and $\tan \theta_2 = \frac{2}{11}$.
We need to find $\tan |\theta_1 - \theta_2|$.
Using the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$:
$\tan(\theta_1 - \theta_2) = \frac{2 - \frac{2}{11}}{1 + 2 \times \frac{2}{11}} = \frac{\frac{22 - 2}{11}}{\frac{11 + 4}{11}} = \frac{20}{15} = \frac{4}{3}$.
Therefore,the absolute value is $\frac{4}{3}$.

(A) The pair of lines $A x^2+2 H x y+B y^2=0$ passes through the origin. For these lines to form an equilateral triangle with the line $a x+b y+c=0$,the angle between the pair of lines must be $60^\circ$ or $\frac{\pi}{3}$ radians.
The angle $\theta$ between the lines $A x^2+2 H x y+B y^2=0$ is given by $\tan \theta = \frac{2 \sqrt{H^2-A B}}{|A+B|}$.
Setting $\theta = \frac{\pi}{3}$,we have $\tan \frac{\pi}{3} = \sqrt{3}$.
So,$\sqrt{3} = \frac{2 \sqrt{H^2-A B}}{|A+B|}$.
Squaring both sides,we get $3 = \frac{4(H^2-A B)}{(A+B)^2}$.
$3(A+B)^2 = 4(H^2-A B)$.
$3(A^2+2 A B+B^2) = 4 H^2-4 A B$.
$3 A^2+6 A B+3 B^2 = 4 H^2-4 A B$.
$3 A^2+10 A B+3 B^2 = 4 H^2$.
Factoring the left side: $3 A^2+9 A B+A B+3 B^2 = 4 H^2$.
$3 A(A+3 B)+B(A+3 B) = 4 H^2$.
$(A+3 B)(3 A+B) = 4 H^2$.
Thus,the correct option is $A$.

(A) The given equation is $2x^2 + 3xy + Ky^2 = 0$.
Dividing by $Ky^2$,we get the quadratic in terms of slope $m = \frac{y}{x}$:
$K(\frac{y}{x})^2 + 3(\frac{y}{x}) + 2 = 0$,which is $Km^2 + 3m + 2 = 0$.
Since one slope is $m_1 = 2$,it must satisfy the equation:
$K(2)^2 + 3(2) + 2 = 0$ $\Rightarrow 4K + 8 = 0$ $\Rightarrow K = -2$.
Substituting $K = -2$ into the equation $Km^2 + 3m + 2 = 0$:
$-2m^2 + 3m + 2 = 0 \Rightarrow 2m^2 - 3m - 2 = 0$.
Factoring the quadratic: $(2m + 1)(m - 2) = 0$.
Thus,the slopes are $m_1 = 2$ and $m_2 = -\frac{1}{2}$.
Since $m_1 \times m_2 = 2 \times (-\frac{1}{2}) = -1$,the product of the slopes is $-1$.
Therefore,the lines are perpendicular to each other,and the angle between them is $\theta = \frac{\pi}{2}$.

(B) The equation $x^2+kxy+y^2=0$ represents a pair of straight lines passing through the origin. Let the slopes of these lines be $m_1$ and $m_2$. The equation of the third line is $x+y=1$,which can be written as $y=-x+1$,so its slope is $m_3=-1$.
Since the lines form an equilateral triangle,the angle between any two lines is $60^{\circ}$.
The angle between the line $y=mx$ and the line $x+y=1$ (slope $-1$) is given by:
$\tan 60^{\circ} = \left| \frac{m - (-1)}{1 + m(-1)} \right|$
$\sqrt{3} = \left| \frac{m+1}{1-m} \right|$
Squaring both sides:
$3 = \frac{(m+1)^2}{(1-m)^2}$
$3(1-2m+m^2) = m^2+2m+1$
$3-6m+3m^2 = m^2+2m+1$
$2m^2-8m+2 = 0$
$m^2-4m+1 = 0$
Since $m = y/x$,we substitute this into the quadratic equation:
$(y/x)^2 - 4(y/x) + 1 = 0$
$y^2 - 4xy + x^2 = 0$
Comparing this with $x^2+kxy+y^2=0$,we get $k=-4$.
Therefore,$k^2 = (-4)^2 = 16$.

(C) The given equation of the pair of straight lines is $ax^2+2hxy+by^2=0$,where $a=1$,$2h=4$ (so $h=2$),and $b=1$.
The formula for the angle $\theta$ between the pair of lines is $\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
Substituting the values,we get $\tan \theta = \left| \frac{2\sqrt{2^2-(1)(1)}}{1+1} \right|$.
$\tan \theta = \frac{2\sqrt{4-1}}{2} = \sqrt{3}$.
Therefore,$\theta = \tan^{-1}(\sqrt{3}) = 60^{\circ}$.

(B) The angle $\theta$ between the lines represented by the homogeneous equation $ax^2+2hxy+by^2=0$ is given by $\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
Given $\theta = \frac{\pi}{4}$,we have $\tan \frac{\pi}{4} = 1$.
Thus,$1 = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
Squaring both sides,we get $1 = \frac{4(h^2-ab)}{(a+b)^2}$.
$(a+b)^2 = 4h^2 - 4ab$.
$a^2 + 2ab + b^2 = 4h^2 - 4ab$.
$4h^2 = a^2 + 6ab + b^2$.

(A) The given equation is of the form $Ax^2 + 2Hxy + By^2 = 0$,where $A = \cos \theta(\cos \theta + 1)$,$2H = -(2 \cos \theta + \sin^2 \theta)$,and $B = 1 - \cos \theta$.
The angle $\alpha$ between the lines is given by $\tan \alpha = \left| \frac{2\sqrt{H^2 - AB}}{A + B} \right|$.
First,calculate $A + B = \cos^2 \theta + \cos \theta + 1 - \cos \theta = \cos^2 \theta + 1$.
Next,calculate $H^2 - AB = \frac{(2 \cos \theta + \sin^2 \theta)^2}{4} - \cos \theta(\cos \theta + 1)(1 - \cos \theta)$.
$= \frac{4 \cos^2 \theta + 4 \cos \theta \sin^2 \theta + \sin^4 \theta}{4} - \cos \theta(1 - \cos^2 \theta) = \cos^2 \theta + \cos \theta \sin^2 \theta + \frac{\sin^4 \theta}{4} - \cos \theta + \cos^3 \theta$.
$= \cos^2 \theta + \cos \theta(1 - \cos^2 \theta) + \frac{\sin^4 \theta}{4} - \cos \theta + \cos^3 \theta = \cos^2 \theta + \frac{\sin^4 \theta}{4}$.
Thus,$\tan \alpha = \frac{2 \sqrt{\cos^2 \theta + \frac{\sin^4 \theta}{4}}}{\cos^2 \theta + 1} = \frac{\sqrt{4 \cos^2 \theta + \sin^4 \theta}}{\cos^2 \theta + 1} = \frac{\sqrt{4 \cos^2 \theta + (1 - \cos^2 \theta)^2}}{\cos^2 \theta + 1}$.
$= \frac{\sqrt{4 \cos^2 \theta + 1 + \cos^4 \theta - 2 \cos^2 \theta}}{\cos^2 \theta + 1} = \frac{\sqrt{\cos^4 \theta + 2 \cos^2 \theta + 1}}{\cos^2 \theta + 1} = \frac{\sqrt{(\cos^2 \theta + 1)^2}}{\cos^2 \theta + 1} = 1$.
Therefore,$\alpha = \tan^{-1}(1) = \frac{\pi}{4}$.

(B) The acute angle $\theta$ between the pair of straight lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} \right|$.
Comparing the given equation $6x^2 + 11xy - 10y^2 = 0$ with $ax^2 + 2hxy + by^2 = 0$,we get $a = 6$,$b = -10$,and $2h = 11$,so $h = \frac{11}{2}$.
Substituting these values into the formula:
$\tan \theta = \left| \frac{2\sqrt{(\frac{11}{2})^2 - (6)(-10)}}{6 + (-10)} \right| = \left| \frac{2\sqrt{\frac{121}{4} + 60}}{-4} \right| = \left| \frac{2\sqrt{\frac{121 + 240}{4}}}{-4} \right| = \left| \frac{2\sqrt{\frac{361}{4}}}{-4} \right| = \left| \frac{2 \cdot \frac{\sqrt{361}}{2}}{-4} \right| = \left| \frac{\sqrt{361}}{-4} \right| = \frac{\sqrt{361}}{4}$.
Therefore,$\theta = \tan^{-1}\left(\frac{\sqrt{361}}{4}\right)$.

(B) The given equation of the pair of lines is $ax^2+6xy+by^2-10x+10y-6=0$.
Comparing this with the general equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$,we get $A=a, H=3, B=b, G=-5, F=5, C=-6$.
For the lines to be perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero,i.e.,$a+b=0$,which implies $b=-a$.
Also,the condition for the general second-degree equation to represent a pair of lines is $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Substituting the values: $a(b)(-6) + 2(5)(3)(-5) - a(5)^2 - b(-5)^2 - (-6)(3)^2 = 0$.
$-6ab - 150 - 25a - 25b + 54 = 0$.
$-6ab - 25(a+b) - 96 = 0$.
Since $a+b=0$,we have $-6a(-a) - 25(0) - 96 = 0$.
$6a^2 = 96 \Rightarrow a^2 = 16$.
Therefore,$|a| = \sqrt{16} = 4$.

(A) The angle $\theta$ between the pair of lines $ax^2 + 2hxy + by^2 = 0$ is given by $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} \right|$.
For the equation $x^2 + 2xy \sec \theta + y^2 = 0$,we have $a = 1$,$h = \sec \theta$,and $b = 1$.
Let $\phi$ be the angle between these lines.
Then $\tan \phi = \left| \frac{2\sqrt{(\sec \theta)^2 - (1)(1)}}{1+1} \right|$.
$\tan \phi = \left| \frac{2\sqrt{\sec^2 \theta - 1}}{2} \right|$.
Since $\sec^2 \theta - 1 = \tan^2 \theta$,we have $\tan \phi = \sqrt{\tan^2 \theta} = \tan \theta$.
Therefore,$\phi = \theta$.

(C) The given equation is $(\sin ^2 \alpha) y^2 - 2xy(\cos ^2 \alpha) + (\cos ^2 \alpha - 1) x^2 = 0$.
Comparing this with the standard form $ax^2 + 2hxy + by^2 = 0$,we get:
$a = \cos ^2 \alpha - 1 = -\sin ^2 \alpha$
$h = -\cos ^2 \alpha$
$b = \sin ^2 \alpha$
The angle $\theta$ between the pair of lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$.
Calculating the denominator: $a + b = -\sin ^2 \alpha + \sin ^2 \alpha = 0$.
Since the denominator is $0$,the lines are perpendicular to each other.
Therefore,$\tan \theta = \infty$,which implies $\theta = 90^{\circ}$.

(A) The given equation is $ab(x^2 - y^2) + (a^2 - b^2)xy = 0$.
Expanding this,we get $abx^2 + (a^2 - b^2)xy - aby^2 = 0$.
This is a homogeneous equation of the second degree in $x$ and $y$ of the form $Ax^2 + 2Hxy + By^2 = 0$,where $A = ab$,$2H = a^2 - b^2$,and $B = -ab$.
The condition for the lines to be perpendicular is $A + B = 0$.
Here,$A + B = ab + (-ab) = 0$.
Since the sum of the coefficients of $x^2$ and $y^2$ is zero,the lines represented by the equation are perpendicular to each other.
Therefore,the angle between the lines is $\frac{\pi}{2}$.

(C) The given equation is $x^2-7xy+12y^2=0$. Comparing this with the general form $ax^2+2hxy+by^2=0$,we have $a=1$,$2h=-7$,and $b=12$.
The angle $\theta$ between the pair of lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
Substituting the values,we get $\tan \theta = \left| \frac{2\sqrt{(-7/2)^2 - 1 \cdot 12}}{1+12} \right| = \left| \frac{2\sqrt{49/4 - 12}}{13} \right| = \left| \frac{2\sqrt{1/4}}{13} \right| = \frac{2 \cdot (1/2)}{13} = \frac{1}{13}$.
Since $\tan \theta = \frac{1}{13}$,we can construct a right-angled triangle with opposite side $1$ and adjacent side $13$. The hypotenuse is $\sqrt{1^2+13^2} = \sqrt{1+169} = \sqrt{170}$.
Therefore,$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{170}}$.
Hence,option $(C)$ is correct.

(C) The given equation is $2x^2 + 5xy + 3y^2 + 6x + 7y + 4 = 0$.
Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = 2$,$2h = 5 \Rightarrow h = \frac{5}{2}$,and $b = 3$.
The angle $\theta$ between the pair of straight lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$.
Substituting the values: $\tan \theta = \left| \frac{2\sqrt{(\frac{5}{2})^2 - (2)(3)}}{2 + 3} \right| = \left| \frac{2\sqrt{\frac{25}{4} - 6}}{5} \right| = \left| \frac{2\sqrt{\frac{1}{4}}}{5} \right| = \left| \frac{2 \times \frac{1}{2}}{5} \right| = \frac{1}{5}$.
Since the angle is $\tan^{-1}(k)$,we have $\tan \theta = k$.
Thus,$k = \pm \frac{1}{5}$.

(B) Given the equation of the pair of straight lines is $6x^2 - 5xy + y^2 = 0$.
Dividing by $x^2$ (assuming $x \neq 0$),we get:
$\left(\frac{y}{x}\right)^2 - 5\left(\frac{y}{x}\right) + 6 = 0$.
Let $m = \frac{y}{x}$,then $m^2 - 5m + 6 = 0$.
Factoring the quadratic equation:
$(m - 3)(m - 2) = 0$.
Thus,the slopes of the lines are $m_1 = \tan \alpha = 3$ and $m_2 = \tan \beta = 2$.
Using the formula for the tangent of the difference of two angles:
$\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$.
Substituting the values:
$\tan(\alpha - \beta) = \frac{3 - 2}{1 + (3)(2)} = \frac{1}{1 + 6} = \frac{1}{7}$.

(A) The condition for a pair of straight lines represented by the general equation $ax^2 + 2hxy + by^2 = 0$ to be perpendicular is that the sum of the coefficients of $x^2$ and $y^2$ must be zero,i.e.,$a + b = 0$.
For option $A$: $2 x^2 = y(x + 2 y)$
$\Rightarrow 2 x^2 - xy - 2 y^2 = 0$
Here,$a = 2$ and $b = -2$.
Sum of coefficients: $a + b = 2 + (-2) = 0$.
Since the condition $a + b = 0$ is satisfied,the lines represented by this equation are perpendicular.

(B) The pair of straight lines passing through the origin is given by the equation $ax^2 + 2hxy + by^2 = 0$.
The angle $\theta$ between these lines is given by the formula $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$.
When the lines are perpendicular,the angle $\theta = 90^{\circ}$.
Since $\tan 90^{\circ}$ is undefined (approaches $\infty$),the denominator must be zero.
Therefore,$a + b = 0$.

(A) The acute angle $(\theta)$ between the pair of lines represented by the homogeneous equation $ax^2+2hxy+by^2=0$ is given by the formula:
$\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$
If the lines are at right angles (perpendicular),then $\theta = 90^\circ$.
Since $\tan 90^\circ$ is undefined,the denominator must be zero.
Therefore,$a+b=0$.

(C) The given equation of the pair of straight lines is $x^2+4xy+y^2=0$.
Comparing this with the general equation $ax^2+2hxy+by^2=0$,we get $a=1$,$2h=4 \Rightarrow h=2$,and $b=1$.
The formula for the angle $\theta$ between the pair of lines is $\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
Substituting the values,we get $\tan \theta = \left| \frac{2\sqrt{2^2-(1)(1)}}{1+1} \right|$.
$\tan \theta = \frac{2\sqrt{4-1}}{2} = \sqrt{3}$.
Since $\tan \theta = \sqrt{3}$,we have $\theta = 60^{\circ}$.

(B) The given equation of the pair of straight lines is $3dx^2 - 5xy + (d^2 - 2)y^2 = 0$.
For a general equation of a pair of lines $Ax^2 + 2Hxy + By^2 = 0$,the lines are perpendicular if $A + B = 0$.
Here,$A = 3d$ and $B = d^2 - 2$.
Setting $A + B = 0$,we get $3d + d^2 - 2 = 0$,which is $d^2 + 3d - 2 = 0$.
Using the quadratic formula $d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we find $d = \frac{-3 \pm \sqrt{3^2 - 4(1)(-2)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 8}}{2} = \frac{-3 \pm \sqrt{17}}{2}$.
Since there are two distinct real values for $d$,the condition is satisfied for $2$ values of $d$.
Thus,option $B$ is correct.