Angle between two straight lines Questions in English

(A) The equation of any line passing through $(3, -2)$ is $y + 2 = m(x - 3)$ ..... $(i)$
The slope of the given line $\sqrt{3}x + y = 1$ is $m_1 = -\sqrt{3}$.
The angle $\theta = 60^o$ between the lines is given by $\tan \theta = \left| \frac{m - m_1}{1 + m \cdot m_1} \right|$.
Substituting the values,$\tan 60^o = \left| \frac{m - (-\sqrt{3})}{1 + m(-\sqrt{3})} \right| \implies \sqrt{3} = \left| \frac{m + \sqrt{3}}{1 - m\sqrt{3}} \right|$.
Case $1$: $\frac{m + \sqrt{3}}{1 - m\sqrt{3}} = \sqrt{3} \implies m + \sqrt{3} = \sqrt{3} - 3m \implies 4m = 0 \implies m = 0$.
Case $2$: $\frac{m + \sqrt{3}}{1 - m\sqrt{3}} = -\sqrt{3} \implies m + \sqrt{3} = -\sqrt{3} + 3m \implies 2m = 2\sqrt{3} \implies m = \sqrt{3}$.
Substituting $m = 0$ in $(i)$: $y + 2 = 0(x - 3) \implies y + 2 = 0$.
Substituting $m = \sqrt{3}$ in $(i)$: $y + 2 = \sqrt{3}(x - 3) \implies y + 2 = \sqrt{3}x - 3\sqrt{3} \implies \sqrt{3}x - y - 2 - 3\sqrt{3} = 0$.

(B) The slope of the given line $x - 2y = 3$ is $m_1 = \frac{1}{2}$.
Let the slope of the required lines be $m$.
The angle between the lines is $45^{\circ}$,so $\tan 45^{\circ} = \left| \frac{m - m_1}{1 + m \cdot m_1} \right|$.
$1 = \left| \frac{m - 1/2}{1 + m/2} \right| = \left| \frac{2m - 1}{2 + m} \right|$.
This gives two cases: $2m - 1 = m + 2$ or $2m - 1 = -(m + 2)$.
Case $1$: $m = 3$. The equation is $y - 2 = 3(x - 3)$ $\Rightarrow y - 2 = 3x - 9$ $\Rightarrow 3x - y - 7 = 0$.
Case $2$: $2m - 1 = -m - 2$ $\Rightarrow 3m = -1$ $\Rightarrow m = -1/3$. The equation is $y - 2 = -\frac{1}{3}(x - 3)$ $\Rightarrow 3y - 6 = -x + 3$ $\Rightarrow x + 3y - 9 = 0$.
Thus,the equations are $3x - y - 7 = 0$ and $x + 3y - 9 = 0$.

(B) The given line is $x + y\sqrt{3} + 3\sqrt{3} = 0$,which can be written as $y = -\frac{1}{\sqrt{3}}x - 3$.
The slope of this line is $m_1 = -\frac{1}{\sqrt{3}}$,so the angle it makes with the positive $x$-axis is $\theta_1 = 150^{\circ}$.
Let the slope of the required lines passing through the origin be $m$. The angle between these lines and the given line is $60^{\circ}$.
Using the formula $\tan \theta = |\frac{m - m_1}{1 + m m_1}|$,we have $\tan 60^{\circ} = |\frac{m - (-1/\sqrt{3})}{1 + m(-1/\sqrt{3})}|$.
$\sqrt{3} = |\frac{m\sqrt{3} + 1}{\sqrt{3} - m}|$.
Case $1$: $\sqrt{3}(\sqrt{3} - m) = m\sqrt{3} + 1$ $\Rightarrow 3 - m\sqrt{3} = m\sqrt{3} + 1$ $\Rightarrow 2m\sqrt{3} = 2$ $\Rightarrow m = \frac{1}{\sqrt{3}}$. The line is $y = \frac{1}{\sqrt{3}}x$ or $x - y\sqrt{3} = 0$.
Case $2$: $\sqrt{3}(\sqrt{3} - m) = -(m\sqrt{3} + 1)$ $\Rightarrow 3 - m\sqrt{3} = -m\sqrt{3} - 1$ $\Rightarrow 3 = -1$,which is impossible,implying the line is vertical $(m \to \infty)$. The line is $x = 0$.
Thus,the lines are $x = 0$ and $x - y\sqrt{3} = 0$.

(C) Given lines are $L_1: 2x + 5y = 7$ and $L_2: 2x - 5y = 9$.
Comparing with the slope-intercept form $y = mx + c$,we get:
For $L_1$: $5y = -2x + 7 \implies y = -\frac{2}{5}x + \frac{7}{5}$,so slope $m_1 = -\frac{2}{5}$.
For $L_2$: $-5y = -2x + 9 \implies y = \frac{2}{5}x - \frac{9}{5}$,so slope $m_2 = \frac{2}{5}$.
Since $m_1 \neq m_2$,the lines are not parallel.
Since $m_1 \times m_2 = (-\frac{2}{5}) \times (\frac{2}{5}) = -\frac{4}{25} \neq -1$,the lines are not perpendicular.
Since the slopes are different,the lines must intersect at a unique point.
Therefore,the lines are intersecting.

(B) The slope of the line $y = 3$ is $m_1 = 0$ (since it is a horizontal line).
The slope of the line $y = \sqrt{3}x + 9$ is $m_2 = \sqrt{3}$.
Let $\theta$ be the acute angle between the lines.
Then,$\tan \theta = |\frac{m_2 - m_1}{1 + m_1 m_2}|$.
$\tan \theta = |\frac{\sqrt{3} - 0}{1 + 0 \times \sqrt{3}}| = |\sqrt{3}| = \sqrt{3}$.
Since $\tan \theta = \sqrt{3}$,we have $\theta = 60^o$.

(B) The slopes of the given lines are $m_1 = 2 - \sqrt{3}$ and $m_2 = 2 + \sqrt{3}$.
Let $\theta$ be the angle between the lines.
The formula for the angle between two lines is $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.
Substituting the values: $\tan \theta = |\frac{(2 - \sqrt{3}) - (2 + \sqrt{3})}{1 + (2 - \sqrt{3})(2 + \sqrt{3})}|$.
$\tan \theta = |\frac{-2\sqrt{3}}{1 + (4 - 3)}| = |\frac{-2\sqrt{3}}{2}| = |-\sqrt{3}| = \sqrt{3}$.
Therefore,$\theta = \tan^{-1}(\sqrt{3}) = 60^o$.

(C) The equations of the lines in intercept form $\frac{x}{x_0} + \frac{y}{y_0} = 1$ are:
Line $1$: $\frac{x}{a} + \frac{y}{-b} = 1 \implies \frac{x}{a} - \frac{y}{b} = 1 \implies y = \frac{b}{a}x - b$. Thus,slope $m_1 = \frac{b}{a}$.
Line $2$: $\frac{x}{b} + \frac{y}{-a} = 1 \implies \frac{x}{b} - \frac{y}{a} = 1 \implies y = \frac{a}{b}x - a$. Thus,slope $m_2 = \frac{a}{b}$.
The angle $\theta$ between the lines is given by $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.
Substituting the values: $\tan \theta = |\frac{\frac{b}{a} - \frac{a}{b}}{1 + (\frac{b}{a})(\frac{a}{b})}| = |\frac{\frac{b^2 - a^2}{ab}}{1 + 1}| = |\frac{b^2 - a^2}{2ab}|$.
Therefore,$\theta = \tan^{-1} |\frac{b^2 - a^2}{2ab}|$.

(D) The slope of line $AB$ $(m_1)$ is $\frac{-2 - 2}{12 - (-4)} = \frac{-4}{16} = -\frac{1}{4}$.
The slope of line $BC$ $(m_2)$ is $\frac{6 - (-2)}{8 - 12} = \frac{8}{-4} = -2$.
The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values: $\tan \theta = \left| \frac{-1/4 - (-2)}{1 + (-1/4)(-2)} \right| = \left| \frac{-1/4 + 2}{1 + 1/2} \right| = \left| \frac{7/4}{3/2} \right| = \frac{7}{4} \times \frac{2}{3} = \frac{7}{6}$.
Therefore,$\angle B = \tan^{-1}\left(\frac{7}{6}\right)$.

(A) The given lines are $\frac{x}{a} + \frac{y}{b} = 1$ and $\frac{x}{a} - \frac{y}{b} = 1$.
These can be rewritten as $bx + ay - ab = 0$ and $bx - ay - ab = 0$.
The slopes of these lines are $m_1 = -\frac{b}{a}$ and $m_2 = \frac{b}{a}$.
The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values,$\tan \theta = \left| \frac{-\frac{b}{a} - \frac{b}{a}}{1 + (-\frac{b}{a})(\frac{b}{a})} \right| = \left| \frac{-\frac{2b}{a}}{1 - \frac{b^2}{a^2}} \right| = \left| \frac{-\frac{2b}{a}}{\frac{a^2 - b^2}{a^2}} \right| = \left| \frac{2ab}{a^2 - b^2} \right|$.
Using the identity $2\tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$,we have $2\tan^{-1}\left(\frac{b}{a}\right) = \tan^{-1}\left(\frac{2(b/a)}{1-(b/a)^2}\right) = \tan^{-1}\left(\frac{2ab}{a^2-b^2}\right)$.
Thus,the angle is $2\tan^{-1}\left(\frac{b}{a}\right)$.

(D) The slopes of the given lines are $m_1 = 3$ and $m_2 = \frac{1}{2}$. Let the slope of the third line be $m_3 = m$.
Since the lines are equally inclined to the third line,the angle between the first and third line is equal to the angle between the second and third line.
Using the formula for the angle between two lines,$\tan \theta = |\frac{m_a - m_b}{1 + m_a m_b}|$,we have:
$|\frac{3 - m}{1 + 3m}| = |\frac{m - 1/2}{1 + m/2}|$
$|\frac{3 - m}{1 + 3m}| = |\frac{2m - 1}{2 + m}|$
Case $1$: $\frac{3 - m}{1 + 3m} = \frac{2m - 1}{2 + m}$
$(3 - m)(2 + m) = (2m - 1)(1 + 3m)$
$6 + 3m - 2m - m^2 = 2m + 6m^2 - 1 - 3m$
$6 + m - m^2 = 6m^2 - m - 1$
$7m^2 - 2m - 7 = 0$
Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$m = \frac{2 \pm \sqrt{4 - 4(7)(-7)}}{14} = \frac{2 \pm \sqrt{4 + 196}}{14} = \frac{2 \pm \sqrt{200}}{14} = \frac{2 \pm 10\sqrt{2}}{14} = \frac{1 \pm 5\sqrt{2}}{7}$.

(B) The given lines are in the normal form $x \cos \alpha + y \sin \alpha = p$,where $\alpha$ is the angle that the normal to the line makes with the positive $x$-axis.
For the first line $x \cos \alpha_1 + y \sin \alpha_1 = p_1$,the normal makes an angle $\alpha_1$ with the $x$-axis. Thus,the line itself makes an angle $\theta_1 = \frac{\pi}{2} + \alpha_1$ with the $x$-axis.
For the second line $x \cos \alpha_2 + y \sin \alpha_2 = p_2$,the normal makes an angle $\alpha_2$ with the $x$-axis. Thus,the line itself makes an angle $\theta_2 = \frac{\pi}{2} + \alpha_2$ with the $x$-axis.
The angle $\theta$ between the two lines is given by $|\theta_1 - \theta_2| = |(\frac{\pi}{2} + \alpha_1) - (\frac{\pi}{2} + \alpha_2)| = |\alpha_1 - \alpha_2|$.

(B) The given lines are in the normal form $x \cos \alpha + y \sin \alpha = p$.
For the first line $x \cos 30^\circ + y \sin 30^\circ = 3$,the normal makes an angle $\alpha_1 = 30^\circ$ with the positive $x$-axis.
For the second line $x \cos 60^\circ + y \sin 60^\circ = 5$,the normal makes an angle $\alpha_2 = 60^\circ$ with the positive $x$-axis.
The angle $\theta$ between two lines is equal to the angle between their normals.
Therefore,$\theta = |\alpha_2 - \alpha_1| = |60^\circ - 30^\circ| = 30^\circ$.

(C) The given equations of the lines are $y = 2x + 9$ and $y = -\frac{1}{2}x - \frac{7}{2}$.
Comparing these with the slope-intercept form $y = mx + c$,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 two lines are perpendicular to each other.
Therefore,the angle between the lines is $90^o$.

(C) The given lines are $L_1: \frac{x}{a} + \frac{y}{b} = 1$ and $L_2: \frac{x}{b'} + \frac{y}{a'} = 1$.
Rewriting in slope-intercept form $y = mx + c$:
$L_1: y = -\frac{b}{a}x + b$,so slope $m_1 = -\frac{b}{a}$.
$L_2: y = -\frac{a'}{b'}x + a'$,so slope $m_2 = -\frac{a'}{b'}$.
Two lines are perpendicular if $m_1 \times m_2 = -1$.
$m_1 \times m_2 = (-\frac{b}{a}) \times (-\frac{a'}{b'}) = \frac{ba'}{ab'}$.
Given condition: $\frac{1}{ab'} + \frac{1}{ba'} = 0 \implies \frac{ba' + ab'}{abb'a'} = 0 \implies ba' + ab' = 0 \implies ba' = -ab'$.
Substituting this into the product of slopes: $m_1 \times m_2 = \frac{-ab'}{ab'} = -1$.
Since the product of slopes is $-1$,the lines are perpendicular.

(A) Let $L_1 \equiv 2x + 3y - 7 = 0$ and $L_2 \equiv 2x + 3y - 5 = 0$.
The slope of a line $ax + by + c = 0$ is given by $m = -\frac{a}{b}$.
For $L_1$,the slope $m_1 = -\frac{2}{3}$.
For $L_2$,the slope $m_2 = -\frac{2}{3}$.
Since $m_1 = m_2$,the lines have the same slope but different intercepts,therefore they are parallel to each other.

(B) The line $x = 2$ is a vertical line,so its slope $m_1$ is undefined (or $\infty$).
The line $x - 3y = 6$ can be written as $3y = x - 6$,or $y = \frac{1}{3}x - 2$. The slope of this line is $m_2 = \frac{1}{3}$.
The angle $\theta$ between a vertical line and a line with slope $m$ is given by $\tan \theta = \left| \frac{1}{m} \right|$.
Here,$m = \frac{1}{3}$,so $\tan \theta = \left| \frac{1}{1/3} \right| = 3$.
Therefore,$\theta = \tan^{-1}(3)$.

(C) The slopes of the given lines are $m_1 = 2 + \sqrt{3}$ and $m_2 = k$.
Given the angle between the lines is $\theta = 60^{\circ}$,we use the formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values: $\tan 60^{\circ} = \left| \frac{2 + \sqrt{3} - k}{1 + k(2 + \sqrt{3})} \right| = \sqrt{3}$.
Case $1$: $\frac{2 + \sqrt{3} - k}{1 + k(2 + \sqrt{3})} = \sqrt{3} \implies 2 + \sqrt{3} - k = \sqrt{3} + k(2\sqrt{3} + 3) \implies 2 - k = k(2\sqrt{3} + 3) \implies k(3 + 2\sqrt{3} + 1) = 2 \implies k(4 + 2\sqrt{3}) = 2 \implies k = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}$.
Case $2$: $\frac{2 + \sqrt{3} - k}{1 + k(2 + \sqrt{3})} = -\sqrt{3} \implies 2 + \sqrt{3} - k = -\sqrt{3} - k(2\sqrt{3} + 3) \implies 2 + 2\sqrt{3} = k - k(2\sqrt{3} + 3) \implies 2(1 + \sqrt{3}) = k(1 - 2\sqrt{3} - 3) \implies 2(1 + \sqrt{3}) = k(-2 - 2\sqrt{3}) \implies 2(1 + \sqrt{3}) = -2k(1 + \sqrt{3}) \implies k = -1$.

(A) The given line is $(\sqrt{3} - 1)x = (\sqrt{3} + 1)y$,which can be written as $y = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}x$.
We know that $\tan(15^{\circ}) = \tan(45^{\circ} - 30^{\circ}) = \frac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$.
Thus,the slope of the line is $m_1 = \tan(15^{\circ})$.
Let the slope of the required line be $m_2$. The angle between the lines is $75^{\circ}$,so $\tan(75^{\circ}) = |\frac{m_2 - m_1}{1 + m_1 m_2}|$.
Since $\tan(75^{\circ}) = \cot(15^{\circ}) = \frac{1}{\tan(15^{\circ})}$,we have $\frac{1}{\tan(15^{\circ})} = |\frac{m_2 - \tan(15^{\circ})}{1 + m_2 \tan(15^{\circ})}|$.
Solving this,we find $m_2$ corresponds to the slope of the $y$-axis,which is undefined,or $m_2 = \tan(15^{\circ} + 75^{\circ}) = \tan(90^{\circ}) = \infty$.
The line passing through the origin with an undefined slope is the $y$-axis,given by $x = 0$.

(B) The slopes of the lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ are $m_1 = -\frac{a_1}{b_1}$ and $m_2 = -\frac{a_2}{b_2}$ respectively.
The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right|$.
Substituting the values of $m_1$ and $m_2$:
$\tan \theta = \left| \frac{-\frac{a_1}{b_1} - (-\frac{a_2}{b_2})}{1 + (-\frac{a_1}{b_1})(-\frac{a_2}{b_2})} \right| = \left| \frac{\frac{a_2b_1 - a_1b_2}{b_1b_2}}{\frac{b_1b_2 + a_1a_2}{b_1b_2}} \right| = \left| \frac{a_2b_1 - a_1b_2}{a_1a_2 + b_1b_2} \right|$.
Thus,$\theta = \tan^{-1} \left| \frac{a_1b_2 - a_2b_1}{a_1a_2 + b_1b_2} \right|$ or $\theta = \cot^{-1} \left| \frac{a_1a_2 + b_1b_2}{a_1b_2 - a_2b_1} \right|$.
Comparing with the given options,option $(B)$ is equivalent to the derived expression.

(A) The slope of the first line $2x - y + 3 = 0$ is $m_1 = -\frac{2}{-1} = 2$.
The slope of the second line $x + 2y + 3 = 0$ is $m_2 = -\frac{1}{2}$.
Since $m_1 \times m_2 = 2 \times (-\frac{1}{2}) = -1$,the product of the slopes is $-1$.
Therefore,the two lines are perpendicular to each other.
Thus,the angle between the lines is $90^\circ$.

(A) The given lines are $L_1: x - y\sqrt{3} = 5$ and $L_2: \sqrt{3}x + y = 7$.
Comparing these with the general form $ax + by + c = 0$,the slopes are:
For $L_1$,$m_1 = -\frac{1}{-\sqrt{3}} = \frac{1}{\sqrt{3}}$.
For $L_2$,$m_2 = -\frac{\sqrt{3}}{1} = -\sqrt{3}$.
Since $m_1 \times m_2 = \left(\frac{1}{\sqrt{3}}\right) \times (-\sqrt{3}) = -1$,the product of the slopes is $-1$.
Therefore,the lines are perpendicular to each other.
The angle between them is $90^\circ$.

(B) The given lines are $2x - y - 15 = 0$ $(i)$ and $3x + y + 4 = 0$ $(ii)$.
Comparing these with the slope-intercept form $y = mx + c$,the slopes are $m_1 = 2$ and $m_2 = -3$.
If $\theta$ is the angle between the lines,then $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values,$\tan \theta = \left| \frac{2 - (-3)}{1 + (2)(-3)} \right| = \left| \frac{5}{1 - 6} \right| = \left| \frac{5}{-5} \right| = 1$.
Since $\tan \theta = 1$,we have $\theta = 45^\circ$.

(B) Let the slope of the first line be $m_1$. The points are $(3, -4)$ and $(-2, 6)$.
$m_1 = \frac{6 - (-4)}{-2 - 3} = \frac{10}{-5} = -2$
Let the slope of the second line be $m_2$. The points are $(-3, 6)$ and $(9, -18)$.
$m_2 = \frac{-18 - 6}{9 - (-3)} = \frac{-24}{12} = -2$
Since $m_1 = m_2 = -2$,the slopes are equal.
Therefore,the lines are parallel.

(C) The slope of the line $2x + 3ay - 1 = 0$ is $m_1 = -\frac{2}{3a}$.
The slope of the line $3x + 4y + 1 = 0$ is $m_2 = -\frac{3}{4}$.
Since the lines are mutually perpendicular,the product of their slopes must be $-1$,i.e.,$m_1 \times m_2 = -1$.
Substituting the values,we get $\left(-\frac{2}{3a}\right) \times \left(-\frac{3}{4}\right) = -1$.
$\frac{6}{12a} = -1$.
$\frac{1}{2a} = -1$.
$2a = -1$.
$a = -\frac{1}{2}$.

(A) Let the points be $A(a \cos \alpha, a \sin \alpha)$ and $B(a \cos \beta, a \sin \beta)$.
The midpoint $P$ of the line segment $AB$ is given by $P\left( \frac{a(\cos \alpha + \cos \beta)}{2}, \frac{a(\sin \alpha + \sin \beta)}{2} \right)$.
The slope of the line $AB$ is $m_1 = \frac{a \sin \beta - a \sin \alpha}{a \cos \beta - a \cos \alpha} = \frac{\sin \beta - \sin \alpha}{\cos \beta - \cos \alpha} = \frac{2 \sin(\frac{\beta - \alpha}{2}) \cos(\frac{\beta + \alpha}{2})}{-2 \sin(\frac{\beta - \alpha}{2}) \sin(\frac{\beta + \alpha}{2})} = -\cot(\frac{\alpha + \beta}{2})$.
The slope of the line $OP$ passing through the origin $(0,0)$ and $P$ is $m_2 = \frac{\frac{a(\sin \alpha + \sin \beta)}{2}}{\frac{a(\cos \alpha + \cos \beta)}{2}} = \frac{\sin \alpha + \sin \beta}{\cos \alpha + \cos \beta} = \frac{2 \sin(\frac{\alpha + \beta}{2}) \cos(\frac{\alpha - \beta}{2})}{2 \cos(\frac{\alpha + \beta}{2}) \cos(\frac{\alpha - \beta}{2})} = \tan(\frac{\alpha + \beta}{2})$.
Now,$m_1 \times m_2 = -\cot(\frac{\alpha + \beta}{2}) \times \tan(\frac{\alpha + \beta}{2}) = -1$.
Since the product of the slopes is $-1$,the lines are perpendicular.

(B) The equations of the lines are $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$.
The slope of the first line is $m_1 = -\frac{a_1}{b_1}$.
The slope of the second line is $m_2 = -\frac{a_2}{b_2}$.
If the lines are perpendicular to each other,the product of their slopes must be $-1$,i.e.,$m_1 \times m_2 = -1$.
Substituting the values,we get $\left(-\frac{a_1}{b_1}\right) \times \left(-\frac{a_2}{b_2}\right) = -1$.
$\Rightarrow \frac{a_1a_2}{b_1b_2} = -1$.
$\Rightarrow a_1a_2 = -b_1b_2$.
$\Rightarrow a_1a_2 + b_1b_2 = 0$.

(B) The given lines are $y = 2x$ and $y = -\frac{1}{2}x$.
Comparing these with the slope-intercept form $y = mx + c$,the slopes are $m_1 = 2$ and $m_2 = -\frac{1}{2}$.
Since the product of the slopes is $m_1 \times m_2 = 2 \times (-\frac{1}{2}) = -1$,the lines are perpendicular to each other.

(B) The slope of the line $mx + 2y + 1 = 0$ is $m_1 = -\frac{m}{2}$.
The slope of the line $2x + 3y + 5 = 0$ is $m_2 = -\frac{2}{3}$.
Since the lines are perpendicular,the product of their slopes must be $-1$,so $m_1 \times m_2 = -1$.
Substituting the values: $\left( -\frac{m}{2} \right) \times \left( -\frac{2}{3} \right) = -1$.
$\frac{m}{3} = -1$.
Therefore,$m = -3$.

(A) The slope of the line $2x + y + 6 = 0$ is $m_1 = -2$.
The slope of the line $x - 2y + 5 = 0$ is $m_2 = \frac{1}{2}$.
The product of the slopes is $m_1 \times m_2 = (-2) \times (\frac{1}{2}) = -1$,which confirms the lines are perpendicular.
Checking if the second line $x - 2y + 5 = 0$ passes through $(1, 3)$: Substituting $x=1$ and $y=3$ into the equation gives $1 - 2(3) + 5 = 1 - 6 + 5 = 0$. Since the result is $0$,the point $(1, 3)$ lies on the line.
Both statements are true,and the condition of perpendicularity is correctly explained by the product of slopes being $-1$.

(A) The equation of the first line is $7x + 3y + 9 = 0$,which can be written as $3y = -7x - 9$ or $y = -\frac{7}{3}x - 3$. The slope $m_1 = -\frac{7}{3}$.
The equation of the second line is $y = kx + 7$. The slope $m_2 = k$.
Two lines are perpendicular if the product of their slopes is $-1$,i.e.,$m_1 \times m_2 = -1$.
Substituting the values,we get $(-\frac{7}{3}) \times k = -1$.
Therefore,$k = \frac{3}{7}$.

(B) The slope of the line $y = -2$ is $m_1 = 0$.
The slope of the line $y = x + 2$ is $m_2 = 1$.
The angle $\theta$ between two lines is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values: $\tan \theta = \left| \frac{0 - 1}{1 + (0)(1)} \right| = |-1| = 1$.
Thus,$\theta = 45^o$ is the acute angle.
The obtuse angle is $180^o - 45^o = 135^o$.

(C) The given lines are $2x - y + 5 = 0$ and $3x + y + 4 = 0$.
Comparing with $y = mx + c$,the slopes are $m_1 = 2$ and $m_2 = -3$.
The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values: $\tan \theta = \left| \frac{2 - (-3)}{1 + (2)(-3)} \right| = \left| \frac{5}{1 - 6} \right| = \left| \frac{5}{-5} \right| = 1$.
Since $\tan \theta = 1$,we have $\theta = 45^o$.

(C) The first line is $x = 9$,which is a vertical line parallel to the $y$-axis. Its slope is undefined.
The second line is $x - \sqrt{3}y + 7 = 0$,which can be rewritten as $\sqrt{3}y = x + 7$,or $y = \frac{1}{\sqrt{3}}x + \frac{7}{\sqrt{3}}$.
The slope of the second line is $m_2 = \frac{1}{\sqrt{3}}$,which corresponds to an angle of $30^\circ$ with the $x$-axis.
Since the first line is vertical ($90^\circ$ with the $x$-axis),the angle between the two lines is $|90^\circ - 30^\circ| = 60^\circ$.

(B) The equation of the first line is $y = 3$,which is a horizontal line parallel to the $x$-axis. Its slope $m_1 = 0$.
The equation of the second line is $y = \sqrt{3}x + 9$. Comparing this with $y = mx + c$,the slope $m_2 = \sqrt{3}$.
Let $\theta$ be the angle between the lines. Then $\tan \theta = |\frac{m_2 - m_1}{1 + m_1m_2}|$.
$\tan \theta = |\frac{\sqrt{3} - 0}{1 + 0 \times \sqrt{3}}| = |\sqrt{3}| = \sqrt{3}$.
Since $\tan \theta = \sqrt{3}$,we have $\theta = 60^{\circ}$.

(A, B) Let the slope of the given line $2x + 3y = 5$ be $m_1 = -\frac{2}{3}$.
Let the slope of the required line be $m$.
The angle between the two lines is $\theta = \frac{\pi}{4}$.
Using the formula $\tan \theta = |\frac{m - m_1}{1 + m \cdot m_1}|$,we have:
$\tan \frac{\pi}{4} = |\frac{m - (-2/3)}{1 + m(-2/3)}|$
$1 = |\frac{3m + 2}{3 - 2m}|$
This gives two cases:
Case $1$: $\frac{3m + 2}{3 - 2m} = 1 \implies 3m + 2 = 3 - 2m \implies 5m = 1 \implies m = \frac{1}{5}$.
The equation of the line is $y - 3 = \frac{1}{5}(x - 2) \implies 5y - 15 = x - 2 \implies x - 5y + 13 = 0$.
Case $2$: $\frac{3m + 2}{3 - 2m} = -1 \implies 3m + 2 = -3 + 2m \implies m = -5$.
The equation of the line is $y - 3 = -5(x - 2) \implies y - 3 = -5x + 10 \implies 5x + y - 13 = 0$.
Thus,the possible equations are $x - 5y + 13 = 0$ or $5x + y - 13 = 0$.

(D) The given lines are $y = 1x + 5$ and $y = \sqrt{3}x - 4$.
Comparing these with $y = mx + c$,we get slopes $m_1 = 1$ and $m_2 = \sqrt{3}$.
The angle $\theta$ between two lines is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$.
Substituting the values: $\tan \theta = \left| \frac{1 - \sqrt{3}}{1 + (1)(\sqrt{3})} \right| = \left| \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \right|$.
Rationalizing the expression: $\tan \theta = \left| \frac{(1 - \sqrt{3})^2}{1 - 3} \right| = \left| \frac{1 + 3 - 2\sqrt{3}}{-2} \right| = \left| \frac{4 - 2\sqrt{3}}{-2} \right| = | -2 + \sqrt{3} | = 2 - \sqrt{3}$.
Since $\tan(15^\circ) = 2 - \sqrt{3}$,we have $\theta = 15^\circ$.

(A) The equation of a line passing through $(3, -2)$ is given by $y + 2 = m(x - 3)$.
The slope of the given line $\sqrt{3}x + y = 1$ is $m_1 = -\sqrt{3}$.
The angle $\theta = 60^{\circ}$ between the lines is given by $\tan(60^{\circ}) = \left| \frac{m - m_1}{1 + m \cdot m_1} \right|$.
$\sqrt{3} = \left| \frac{m - (-\sqrt{3})}{1 + m(-\sqrt{3})} \right| = \left| \frac{m + \sqrt{3}}{1 - m\sqrt{3}} \right|$.
Case $1$: $\frac{m + \sqrt{3}}{1 - m\sqrt{3}} = \sqrt{3} \implies m + \sqrt{3} = \sqrt{3} - 3m \implies 4m = 0 \implies m = 0$.
Equation: $y + 2 = 0(x - 3) \implies y + 2 = 0$.
Case $2$: $\frac{m + \sqrt{3}}{1 - m\sqrt{3}} = -\sqrt{3} \implies m + \sqrt{3} = -\sqrt{3} + 3m \implies 2m = 2\sqrt{3} \implies m = \sqrt{3}$.
Equation: $y + 2 = \sqrt{3}(x - 3) \implies y + 2 = \sqrt{3}x - 3\sqrt{3} \implies \sqrt{3}x - y - 2 - 3\sqrt{3} = 0$.

(C) The slope of the first line $3x + 4y - 5 = 0$ is $m_1 = -\frac{3}{4}$.
The slope of the second line $4x + ky - 8 = 0$ is $m_2 = -\frac{4}{k}$.
For two lines to be parallel,their slopes must be equal,so $m_1 = m_2$.
$-\frac{3}{4} = -\frac{4}{k}$.
Cross-multiplying,we get $3k = 16$,which implies $k = \frac{16}{3}$.

(B) Let $m_1$ and $m_2$ be the slopes of $BA$ and $BC$ respectively.
$m_1 = \frac{3 - 1}{2 - (-2)} = \frac{2}{4} = \frac{1}{2}$
$m_2 = \frac{-4 - 3}{-2 - 2} = \frac{-7}{-4} = \frac{7}{4}$
Let $\theta$ be the angle between $BA$ and $BC$. Then,
$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| = \left| \frac{\frac{7}{4} - \frac{1}{2}}{1 + (\frac{7}{4} \times \frac{1}{2})} \right|$
$\tan \theta = \left| \frac{\frac{5}{4}}{\frac{15}{8}} \right| = \left| \frac{5}{4} \times \frac{8}{15} \right| = \frac{2}{3}$
Therefore,$\theta = \tan^{-1}\left(\frac{2}{3}\right)$.

(A) The slope of the given line $x - \sqrt{3}y - 2\sqrt{3} = 0$ is $m = \frac{1}{\sqrt{3}}$.
Let the slope of the required lines be $m'$. The angle between the lines is $\theta = 60^{\circ}$.
Using the formula $\tan \theta = \left| \frac{m' - m}{1 + m'm} \right|$,we have $\tan 60^{\circ} = \left| \frac{m' - 1/\sqrt{3}}{1 + m'/\sqrt{3}} \right|$.
$\sqrt{3} = \left| \frac{\sqrt{3}m' - 1}{\sqrt{3} + m'} \right|$.
Case $1$: $\sqrt{3} = \frac{\sqrt{3}m' - 1}{\sqrt{3} + m'}$ $\Rightarrow 3 + \sqrt{3}m' = \sqrt{3}m' - 1$ $\Rightarrow 3 = -1$ (Impossible,implies vertical line).
$A$ vertical line passing through $(7, 9)$ is $x = 7$.
Case $2$: $-\sqrt{3} = \frac{\sqrt{3}m' - 1}{\sqrt{3} + m'}$ $\Rightarrow -3 - \sqrt{3}m' = \sqrt{3}m' - 1$ $\Rightarrow 2\sqrt{3}m' = -2$ $\Rightarrow m' = -\frac{1}{\sqrt{3}}$.
The equation of the line is $y - 9 = -\frac{1}{\sqrt{3}}(x - 7)$ $\Rightarrow \sqrt{3}y - 9\sqrt{3} = -x + 7$ $\Rightarrow x + \sqrt{3}y = 7 + 9\sqrt{3}$.
Thus,the equations are $x = 7$ and $x + \sqrt{3}y = 7 + 9\sqrt{3}$.

(A) Let the points be $A(0, 0), B(2, 3)$ and $C(2, -2), D(3, 5)$.
The slope of line $AB$ is $m_1 = \frac{3 - 0}{2 - 0} = \frac{3}{2}$.
The slope of line $CD$ is $m_2 = \frac{5 - (-2)}{3 - 2} = \frac{7}{1} = 7$.
Let $\theta$ be the acute angle between the lines. The formula for the angle between two lines is given by $tan\theta = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right|$.
Substituting the values: $tan\theta = \left| \frac{\frac{3}{2} - 7}{1 + (\frac{3}{2})(7)} \right| = \left| \frac{\frac{3 - 14}{2}}{1 + \frac{21}{2}} \right| = \left| \frac{-\frac{11}{2}}{\frac{23}{2}} \right| = \left| -\frac{11}{23} \right| = \frac{11}{23}$.
Therefore,$\theta = tan^{-1}\left(\frac{11}{23}\right)$.

(D) The slopes of the given lines are $m_1 = 3$ and $m_2 = \frac{1}{2}$. Let the slope of the third line be $m_3 = m$.
Since the line $y = mx + 4$ makes equal angles with the other two lines,we have:
$|\frac{m - 3}{1 + 3m}| = |\frac{m - 1/2}{1 + m/2}|$
$|\frac{m - 3}{1 + 3m}| = |\frac{2m - 1}{2 + m}|$
Case $1$: $\frac{m - 3}{1 + 3m} = \frac{2m - 1}{2 + m}$
$(m - 3)(m + 2) = (2m - 1)(3m + 1)$
$m^2 - m - 6 = 6m^2 - m - 1$
$5m^2 = -5 \Rightarrow m^2 = -1$ (No real solution).
Case $2$: $\frac{m - 3}{1 + 3m} = -(\frac{2m - 1}{2 + m})$
$(m - 3)(m + 2) = -(2m - 1)(3m + 1)$
$m^2 - m - 6 = -(6m^2 - m - 1)$
$m^2 - m - 6 = -6m^2 + m + 1$
$7m^2 - 2m - 7 = 0$
Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$m = \frac{2 \pm \sqrt{4 - 4(7)(-7)}}{14} = \frac{2 \pm \sqrt{4 + 196}}{14} = \frac{2 \pm \sqrt{200}}{14} = \frac{2 \pm 10\sqrt{2}}{14} = \frac{1 \pm 5\sqrt{2}}{7}$.

(B) The given line is $\sqrt{3}x + y = 1$,which can be written as $y = -\sqrt{3}x + 1$. The slope of this line is $m_2 = -\sqrt{3}$.
Let the slope of line $L$ be $m_1$. The angle between the lines is $60^{\circ}$.
Using the formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$,we have:
$\tan 60^{\circ} = \left| \frac{m_1 - (-\sqrt{3})}{1 + m_1(-\sqrt{3})} \right|$
$\sqrt{3} = \left| \frac{m_1 + \sqrt{3}}{1 - \sqrt{3}m_1} \right|$
This gives two cases:
Case $1$: $\frac{m_1 + \sqrt{3}}{1 - \sqrt{3}m_1} = \sqrt{3}$ $\Rightarrow m_1 + \sqrt{3} = \sqrt{3} - 3m_1$ $\Rightarrow 4m_1 = 0$ $\Rightarrow m_1 = 0$.
The equation of line $L$ with slope $0$ passing through $(3, -2)$ is $y - (-2) = 0(x - 3) \Rightarrow y + 2 = 0$.
Case $2$: $\frac{m_1 + \sqrt{3}}{1 - \sqrt{3}m_1} = -\sqrt{3}$ $\Rightarrow m_1 + \sqrt{3} = -\sqrt{3} + 3m_1$ $\Rightarrow 2m_1 = 2\sqrt{3}$ $\Rightarrow m_1 = \sqrt{3}$.
The equation of line $L$ with slope $\sqrt{3}$ passing through $(3, -2)$ is $y - (-2) = \sqrt{3}(x - 3)$ $\Rightarrow y + 2 = \sqrt{3}x - 3\sqrt{3}$ $\Rightarrow y - \sqrt{3}x + 2 + 3\sqrt{3} = 0$.

(B) Let the slope of the required line be $m_1$. The given line is $y = mx + c$,so its slope is $m$. The angle between the two lines is given as $\theta = tan^{-1} m$.
Using the formula for the angle between two lines: $tan \theta = |\frac{m_1 - m}{1 + m_1 m}|$.
Substituting $\theta = tan^{-1} m$,we get $m = |\frac{m_1 - m}{1 + m_1 m}|$.
Case $1$: $\frac{m_1 - m}{1 + m_1 m} = m \implies m_1 - m = m + m_1 m^2 \implies m_1(1 - m^2) = 2m \implies m_1 = \frac{2m}{1 - m^2}$.
The equation of the line passing through the origin $(0, 0)$ with slope $m_1$ is $y = m_1 x$,so $y = \frac{2m}{1 - m^2} x$,which simplifies to $(1 - m^2)y = 2mx$ or $2mx + (m^2 - 1)y = 0$.
Case $2$: $\frac{m_1 - m}{1 + m_1 m} = -m \implies m_1 - m = -m - m_1 m^2 \implies m_1(1 + m^2) = 0 \implies m_1 = 0$.
The equation of the line is $y = 0x$,which is $y = 0$.
Thus,the equations are $y = 0$ and $2mx + (m^2 - 1)y = 0$.

(B) The given line is $x - \sqrt{3}y + 5 = 0$.
Rearranging it into slope-intercept form $y = mx + c$,we get $\sqrt{3}y = x + 5$,which implies $y = \frac{1}{\sqrt{3}}x + \frac{5}{\sqrt{3}}$.
The slope of this line is $m_1 = \frac{1}{\sqrt{3}}$.
The $y$-axis is a vertical line,so its slope $m_2$ is undefined (or we can consider the angle it makes with the $x$-axis as $90^o$).
The angle $\theta$ between a line with slope $m$ and the $y$-axis is given by $\tan \theta = |\frac{1}{m}|$.
Here,$\tan \theta = |\frac{1}{1/\sqrt{3}}| = \sqrt{3}$.
Therefore,$\theta = \tan^{-1}(\sqrt{3}) = 60^o$.

(D) If two lines are perpendicular to a common line,they must be parallel to each other.
The slope of the first line $m_1$ is given by $m_1 = p(p^2 + 1)$.
The slope of the second line $m_2$ is given by $m_2 = -\frac{(p^2 + 1)^2}{p^2 + 1} = -(p^2 + 1)$.
For the lines to be parallel,$m_1 = m_2$.
$p(p^2 + 1) = -(p^2 + 1)$.
Since $p^2 + 1 \neq 0$ for any real $p$,we can divide both sides by $(p^2 + 1)$:
$p = -1$.
Thus,the lines are perpendicular to a common line for exactly one value of $p$.

(A) Let the slope of the required line be $m$.
The given lines are $3x - 4y - 7 = 0$ and $12x - 5y + 6 = 0$.
The slopes of these lines are $m_1 = \frac{3}{4}$ and $m_2 = \frac{12}{5}$.
Since the required line makes equal angles with the given lines,the angle between the required line and the given lines is equal.
Using the formula $\tan \theta = |\frac{m - m_1}{1 + m m_1}| = |\frac{m - m_2}{1 + m m_2}|$,we have:
$|\frac{m - 3/4}{1 + 3m/4}| = |\frac{m - 12/5}{1 + 12m/5}|$
$|\frac{4m - 3}{4 + 3m}| = |\frac{5m - 12}{5 + 12m}|$
$(4m - 3)(5 + 12m) = \pm (5m - 12)(4 + 3m)$
Case $1$: $(4m - 3)(5 + 12m) = (5m - 12)(4 + 3m)$
$20m + 48m^2 - 15 - 36m = 20m + 15m^2 - 48 - 36m$
$33m^2 = -33 \Rightarrow m^2 = -1$ (No real solution).
Case $2$: $(4m - 3)(5 + 12m) = -(5m - 12)(4 + 3m)$
$48m^2 - 16m - 15 = -(15m^2 - 16m - 48)$
$48m^2 - 16m - 15 = -15m^2 + 16m + 48$
$63m^2 - 32m - 63 = 0$
Solving for $m$: $m = \frac{32 \pm \sqrt{1024 - 4(63)(-63)}}{126} = \frac{32 \pm \sqrt{1024 + 15876}}{126} = \frac{32 \pm 130}{126}$
$m_1 = \frac{162}{126} = \frac{9}{7}$ and $m_2 = \frac{-98}{126} = -\frac{7}{9}$.
Using $y - 5 = m(x - 4)$:
For $m = 9/7$: $7y - 35 = 9x - 36 \Rightarrow 9x - 7y = 1$.
For $m = -7/9$: $9y - 45 = -7x + 28 \Rightarrow 7x + 9y = 73$.

(A) The given lines are $L_1: y = x - 5$ and $L_2: y = \sqrt{3}x + 7$.
Comparing with $y = mx + c$,the slopes are $m_1 = 1$ and $m_2 = \sqrt{3}$.
The angle $\theta$ between two lines is given by $\tan \theta = |\frac{m_2 - m_1}{1 + m_1 m_2}|$.
Substituting the values,$\tan \theta = |\frac{\sqrt{3} - 1}{1 + (1)(\sqrt{3})}| = |\frac{\sqrt{3} - 1}{\sqrt{3} + 1}|$.
Rationalizing the denominator,$\tan \theta = |\frac{(\sqrt{3} - 1)^2}{3 - 1}| = |\frac{3 + 1 - 2\sqrt{3}}{2}| = |\frac{4 - 2\sqrt{3}}{2}| = |2 - \sqrt{3}|$.
Since $\tan 15^o = 2 - \sqrt{3}$,we have $\theta = 15^o$.

(A) The slopes of the lines are as follows:
$L_1: x - y = 1 \implies y = x - 1,$ slope $m_1 = 1.$
$L_2: x + y = 1 \implies y = -x + 1,$ slope $m_2 = -1.$
$L_3: 2x + 2y = 5 \implies y = -x + 2.5,$ slope $m_3 = -1.$
$L_4: 2x - 2y = 7 \implies y = x - 3.5,$ slope $m_4 = 1.$
Since $m_1 \times m_2 = 1 \times (-1) = -1,$ $L_1 \perp L_2.$
Since $m_2 = m_3 = -1,$ $L_2 \parallel L_3.$
Since $m_1 = m_4 = 1,$ $L_1 \parallel L_4.$
Checking the options:
Option $A$: $L_1 \perp L_2$ (True),$L_2 \parallel L_3$ (True),$L_1$ intersects $L_4$ (False,they are parallel).
Option $B$: $L_1 \perp L_2$ (True),$L_1 \parallel L_3$ (False).
Option $C$: $L_1 \perp L_2$ (True),$L_1 \parallel L_3$ (False).
Option $D$: $L_1 \parallel L_4$ (True),$L_2 \parallel L_3$ (True),$L_2$ intersects $L_3$ (False,they are parallel).
Wait,re-evaluating the options provided in the prompt. Given the standard form,$L_2 \parallel L_3$ is correct. Let's re-check the intersection logic. $L_2$ and $L_3$ are parallel,so they do not intersect. The question likely contains a typo in the options. Based on the slopes,$L_2 \parallel L_3$ and $L_1 \parallel L_4$ and $L_1 \perp L_2$ are the geometric properties.

(A) If two lines are perpendicular to a common line,they must be parallel to each other.
Let the slopes of the lines be $m_1$ and $m_2$.
For the first line $p(p^2 + 1)x - y + q = 0$,the slope $m_1 = \frac{p(p^2 + 1)}{1} = p(p^2 + 1)$.
For the second line $(p^2 + 1)^2x + (p^2 + 1)y + 2q = 0$,the slope $m_2 = -\frac{(p^2 + 1)^2}{p^2 + 1} = -(p^2 + 1)$.
Since the lines are parallel,$m_1 = m_2$.
$p(p^2 + 1) = -(p^2 + 1)$.
Since $p^2 + 1 \neq 0$ for any real $p$,we can divide by $(p^2 + 1)$.
$p = -1$.
Thus,there is exactly one value of $p$.