Slope of line, Equation of line in different forms Questions in English

(A) The equation of the line is given by $ax + (3 - a)y + 7 = 0$.
We know that the slope of a line $Ax + By + C = 0$ is given by $m = -\frac{A}{B}$.
Here,$A = a$ and $B = (3 - a)$.
Given that the slope $m = 7$,we have:
$-\frac{a}{3 - a} = 7$
$\frac{a}{a - 3} = 7$
$a = 7(a - 3)$
$a = 7a - 21$
$6a = 21$
$a = \frac{21}{6} = 3.5$.
The integral part of $a$,denoted as $[a]$,is the greatest integer less than or equal to $a$.
$[3.5] = 3$.
Therefore,the correct option is $A$.

(A) The line passing through points $u(1, -2)$ and $v(-3, 5)$ has the equation: $\frac{x - 1}{-3 - 1} = \frac{y - (-2)}{5 - (-2)} \Rightarrow \frac{x - 1}{-4} = \frac{y + 2}{7} \Rightarrow 7x - 7 = -4y - 8 \Rightarrow 7x + 4y + 1 = 0$.
For point $P(-\frac{1}{7}, 0)$: $7(-\frac{1}{7}) + 4(0) + 1 = -1 + 0 + 1 = 0$. Thus,$P$ lies on the line.
For point $Q(0, -\frac{1}{4})$: $7(0) + 4(-\frac{1}{4}) + 1 = 0 - 1 + 1 = 0$. Thus,$Q$ lies on the line.
For point $R(-2, 3)$: $7(-2) + 4(3) + 1 = -14 + 12 + 1 = -1 \neq 0$. Thus,$R$ does not lie on the line.
Therefore,only points $P$ and $Q$ lie on the line segment.

(D) Let the points of intersection be $A(x_1, y_1)$ and $B(x_2, y_2)$.
Since the point $(1, 2)$ bisects the line segment $AB$,we have:
$\frac{x_1 + x_2}{2} = 1 \Rightarrow x_2 = 2 - x_1$
$\frac{y_1 + y_2}{2} = 2 \Rightarrow y_2 = 4 - y_1$
Since $A$ lies on $3x - 2y - 1 = 0$,we have:
$3x_1 - 2y_1 - 1 = 0$ --- $(i)$
Since $B$ lies on $x + 2y + 1 = 0$,we substitute $x_2$ and $y_2$:
$(2 - x_1) + 2(4 - y_1) + 1 = 0$
$2 - x_1 + 8 - 2y_1 + 1 = 0$
$x_1 + 2y_1 - 11 = 0$ --- (ii)
Solving equations $(i)$ and (ii) by adding them:
$(3x_1 - 2y_1 - 1) + (x_1 + 2y_1 - 11) = 0$
$4x_1 - 12 = 0 \Rightarrow x_1 = 3$
Substituting $x_1 = 3$ in $(i)$:
$3(3) - 2y_1 - 1 = 0 \Rightarrow 8 = 2y_1 \Rightarrow y_1 = 4$
Thus,$A = (3, 4)$.
Then $x_2 = 2 - 3 = -1$ and $y_2 = 4 - 4 = 0$,so $B = (-1, 0)$.
The equation of line $L$ passing through $(3, 4)$ and $(-1, 0)$ is:
$y - 0 = \frac{4 - 0}{3 - (-1)}(x - (-1))$
$y = \frac{4}{4}(x + 1) \Rightarrow y = x + 1 \Rightarrow x - y = -1$
Dividing by $-1$,we get $\frac{x}{-1} + \frac{y}{1} = 1$.
Comparing with $\frac{x}{a} + \frac{y}{b} = 1$,we get $a = -1$ and $b = 1$.
Therefore,$a + 2b + 1 = -1 + 2(1) + 1 = -1 + 2 + 1 = 2$.

(A) The line $L$ passes through $A(-2, 4)$ with an inclination $\theta = 60^{\circ}$.
The coordinates of any point $B(p, q)$ on the line at a distance $r = 6$ from $A(x_1, y_1)$ are given by $p = x_1 + r \cos \theta$ and $q = y_1 + r \sin \theta$.
Since $B$ lies in the $3^{\text{rd}}$ quadrant,we move in the opposite direction along the line,so $r = -6$.
$p = -2 + (-6) \cos 60^{\circ} = -2 - 6(\frac{1}{2}) = -2 - 3 = -5$.
$q = 4 + (-6) \sin 60^{\circ} = 4 - 6(\frac{\sqrt{3}}{2}) = 4 - 3\sqrt{3}$.
We need to calculate $\sqrt{p^2 + q^2 - 8q}$.
Note that $p^2 + q^2 - 8q = p^2 + (q-4)^2 - 16$.
Substituting $p = -5$ and $q = 4 - 3\sqrt{3}$:
$p^2 = (-5)^2 = 25$.
$(q-4)^2 = (4 - 3\sqrt{3} - 4)^2 = (-3\sqrt{3})^2 = 9 \times 3 = 27$.
So,$p^2 + (q-4)^2 - 16 = 25 + 27 - 16 = 52 - 16 = 36$.
Therefore,$\sqrt{p^2 + q^2 - 8q} = \sqrt{36} = 6$.

(A) The equations of the sides are given as:
$AB: x-3y=0$
$AC: 3x-y=0$
$BC: x+y+4=0$
To find the coordinates of vertex $B$,we solve the equations of lines $AB$ and $BC$:
$x-3y=0 \Rightarrow x=3y$
Substitute $x=3y$ into $x+y+4=0$:
$3y+y+4=0$ $\Rightarrow 4y=-4$ $\Rightarrow y=-1$
Then $x=3(-1)=-3$.
So,the vertex $B$ is $(-3, -1)$.
Since the line $3x-y+k=0$ passes through $B(-3, -1)$,we substitute these coordinates into the line equation:
$3(-3)-(-1)+k=0$
$-9+1+k=0$
$-8+k=0 \Rightarrow k=8$.

(B) Given points are $(3,3)$ and $(7,6)$.
The slope of the line $m = \frac{6-3}{7-3} = \frac{3}{4}$.
The equation of the line is $(y-3) = \frac{3}{4}(x-3)$,which simplifies to $4y - 12 = 3x - 9$,or $3x - 4y = -3$.
To find the $y$-intercept,set $x=0$: $3(0) - 4y = -3 \Rightarrow y = \frac{3}{4}$. The point is $(0, \frac{3}{4})$.
To find the $x$-intercept,set $y=0$: $3x - 4(0) = -3 \Rightarrow x = -1$. The point is $(-1, 0)$.
The length of the segment between $(0, \frac{3}{4})$ and $(-1, 0)$ is given by the distance formula:
$d = \sqrt{(0 - (-1))^2 + (\frac{3}{4} - 0)^2} = \sqrt{1^2 + (\frac{3}{4})^2} = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

(C) The equation of the line passing through $P(1, 2)$ with angle $\theta$ is given by the parametric form: $\frac{x - 1}{\cos \theta} = \frac{y - 2}{\sin \theta} = r$.
Since $PQ = \frac{1}{2}$,the coordinates of $Q$ are $(1 + \frac{1}{2} \cos \theta, 2 + \frac{1}{2} \sin \theta)$.
Since $Q$ lies on the line $x + \sqrt{3}y - 2\sqrt{3} = 0$,we substitute the coordinates of $Q$ into the equation:
$(1 + \frac{1}{2} \cos \theta) + \sqrt{3}(2 + \frac{1}{2} \sin \theta) - 2\sqrt{3} = 0$.
$1 + \frac{1}{2} \cos \theta + 2\sqrt{3} + \frac{\sqrt{3}}{2} \sin \theta - 2\sqrt{3} = 0$.
$\frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta = -1$.
Dividing by $1$,we get $\frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta = -1$.
This can be written as $\cos \theta \cos \frac{\pi}{3} + \sin \theta \sin \frac{\pi}{3} = -1$ is not possible since the range of $\cos(\theta - \frac{\pi}{3})$ is $[-1, 1]$.
Wait,let us re-evaluate: $\frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta = -1$.
$\cos(\theta - \frac{\pi}{3}) = -1$.
$\theta - \frac{\pi}{3} = \pi$.
$\theta = \frac{4\pi}{3}$ or $\theta - \frac{\pi}{3} = -\pi \implies \theta = -\frac{2\pi}{3} = \frac{4\pi}{3}$.
Checking the options,if $\theta = \frac{2\pi}{3}$,$\cos(\frac{2\pi}{3} - \frac{\pi}{3}) = \cos(\frac{\pi}{3}) = \frac{1}{2} \neq -1$.
Re-checking the line equation: $x + \sqrt{3}y - 2\sqrt{3} = 0$.
At $P(1, 2)$,$1 + 2\sqrt{3} - 2\sqrt{3} = 1 \neq 0$.
If $\theta = \frac{2\pi}{3}$,the line is $y - 2 = \tan(\frac{2\pi}{3})(x - 1) \implies y - 2 = -\sqrt{3}(x - 1) \implies \sqrt{3}x + y - (2 + \sqrt{3}) = 0$.
Solving with $x + \sqrt{3}y - 2\sqrt{3} = 0$: $x = 2\sqrt{3} - \sqrt{3}y$.
$\sqrt{3}(2\sqrt{3} - \sqrt{3}y) + y - 2 - \sqrt{3} = 0 \implies 6 - 3y + y - 2 - \sqrt{3} = 0 \implies 2y = 4 - \sqrt{3} \implies y = 2 - \frac{\sqrt{3}}{2}$.
$x = 2\sqrt{3} - \sqrt{3}(2 - \frac{\sqrt{3}}{2}) = 2\sqrt{3} - 2\sqrt{3} + \frac{3}{2} = \frac{3}{2}$.
$PQ^2 = (\frac{3}{2} - 1)^2 + (2 - \frac{\sqrt{3}}{2} - 2)^2 = (\frac{1}{2})^2 + (-\frac{\sqrt{3}}{2})^2 = \frac{1}{4} + \frac{3}{4} = 1$.
For $PQ = \frac{1}{2}$,the correct angle is $\theta = \frac{2\pi}{3}$.

(A) Let the equation of the line $L$ be $\frac{x}{a} + \frac{y}{b} = 1$,where $a$ and $b$ are the $X$ and $Y$ intercepts respectively.
Given that the area of the triangle is $\frac{1}{2} |ab| = 12$,so $|ab| = 24$.
Since the line passes through $(12, 4)$,we have $\frac{12}{a} + \frac{4}{b} = 1$.
Substituting $b = \frac{24}{a}$ or $b = -\frac{24}{a}$ into the equation:
Case $1$: $b = \frac{24}{a} \implies \frac{12}{a} + \frac{4a}{24} = 1 \implies \frac{12}{a} + \frac{a}{6} = 1 \implies a^2 - 6a + 72 = 0$. This has no real roots.
Case $2$: $b = -\frac{24}{a} \implies \frac{12}{a} - \frac{4a}{24} = 1 \implies \frac{12}{a} - \frac{a}{6} = 1 \implies a^2 + 6a - 72 = 0$.
Solving $a^2 + 6a - 72 = 0$,we get $(a+12)(a-6) = 0$,so $a = 6$ or $a = -12$.
If $a = 6$,then $b = -\frac{24}{6} = -4$. The product $P = a \cdot b^2 = 6 \cdot (-4)^2 = 6 \cdot 16 = 96$ (Positive).
If $a = -12$,then $b = -\frac{24}{-12} = 2$. The product $P = a \cdot b^2 = -12 \cdot (2)^2 = -12 \cdot 4 = -48$ (Negative).
Since $P$ is negative,$P = -48$.

(A) Let the points be $A(1, 2)$ and $B(-3, 5)$. The point $P(x, y)$ divides the segment $AB$ externally in the ratio $m:n = 4:3$.
Using the section formula for external division: $x = \frac{mx_2 - nx_1}{m - n}$ and $y = \frac{my_2 - ny_1}{m - n}$.
$x = \frac{4(-3) - 3(1)}{4 - 3} = \frac{-12 - 3}{1} = -15$.
$y = \frac{4(5) - 3(2)}{4 - 3} = \frac{20 - 6}{1} = 14$.
So,the point is $(-15, 14)$.
The equation of the line with slope $m = \frac{-2}{3}$ passing through $(-15, 14)$ is given by $y - y_1 = m(x - x_1)$.
$y - 14 = \frac{-2}{3}(x + 15)$.
$3y - 42 = -2x - 30$.
$2x + 3y - 12 = 0$.

(C) The equation of the line passing through $(2, -1)$ and $(5, -3)$ is given by the two-point form: $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.
Substituting the points $(2, -1)$ and $(5, -3)$:
$y - (-1) = \frac{-3 - (-1)}{5 - 2}(x - 2)$
$y + 1 = \frac{-2}{3}(x - 2)$
$3(y + 1) = -2(x - 2)$
$3y + 3 = -2x + 4$
$2x + 3y = 1$ (Equation $L$)
Since $(a, 4)$ lies on $L$:
$2(a) + 3(4) = 1$ $\Rightarrow 2a + 12 = 1$ $\Rightarrow 2a = -11$ $\Rightarrow a = -\frac{11}{2}$.
Since $(-2, b)$ lies on $L$:
$2(-2) + 3(b) = 1$ $\Rightarrow -4 + 3b = 1$ $\Rightarrow 3b = 5$ $\Rightarrow b = \frac{5}{3}$.
We need to check which line contains the point $(a, b) = (-\frac{11}{2}, \frac{5}{3})$.
Testing option $C$: $2x + 6y + 1 = 0$
$2(-\frac{11}{2}) + 6(\frac{5}{3}) + 1 = -11 + 10 + 1 = 0$.
Thus,the point $(a, b)$ lies on the line $2x + 6y + 1 = 0$.

(B) Given the slope $m = \tan \theta = \frac{3}{4}$.
Since $\tan \theta = \frac{3}{4}$,we have $\sin \theta = \frac{3}{5}$ and $\cos \theta = \frac{4}{5}$.
The coordinates of points at a distance $r = 5$ from $A(x_1, y_1) = (3, 2)$ are given by $(x_1 \pm r \cos \theta, y_1 \pm r \sin \theta)$.
For the first point $P$:
$x = 3 + 5 \times \frac{4}{5} = 3 + 4 = 7$
$y = 2 + 5 \times \frac{3}{5} = 2 + 3 = 5$
So,$P = (7, 5)$.
For the second point $Q$:
$x = 3 - 5 \times \frac{4}{5} = 3 - 4 = -1$
$y = 2 - 5 \times \frac{3}{5} = 2 - 3 = -1$
So,$Q = (-1, -1)$.
Thus,the required points are $(7, 5)$ and $(-1, -1)$.

(C) Let the line passing through $P(1, 2)$ make an angle $\theta$ with the positive $X$-axis. The parametric form of the line is given by $x = 1 + r \cos \theta$ and $y = 2 + r \sin \theta$,where $r = \frac{\sqrt{6}}{3}$.
Since the point $(x, y)$ lies on the line $x + y = 4$,we substitute the parametric coordinates:
$(1 + r \cos \theta) + (2 + r \sin \theta) = 4$
$3 + r(\cos \theta + \sin \theta) = 4$
$r(\cos \theta + \sin \theta) = 1$
Substituting $r = \frac{\sqrt{6}}{3}$:
$\frac{\sqrt{6}}{3}(\cos \theta + \sin \theta) = 1$
$\cos \theta + \sin \theta = \frac{3}{\sqrt{6}} = \frac{\sqrt{6}}{2}$
Squaring both sides:
$(\cos \theta + \sin \theta)^2 = \frac{6}{4} = \frac{3}{2}$
$1 + \sin 2\theta = \frac{3}{2}$
$\sin 2\theta = \frac{1}{2}$
Thus,$2\theta = \frac{\pi}{6}$ or $2\theta = \frac{5\pi}{6}$.
Therefore,$\theta = \frac{\pi}{12}$ or $\theta = \frac{5\pi}{12}$.

(C) Let the intercepts be $5C$ and $3C$. The equation is $\frac{x}{5C} + \frac{y}{3C} = 1$. Since it passes through $(-4, 3)$,we have $-\frac{4}{5C} + \frac{3}{3C} = 1$ $\Rightarrow \frac{-4+5}{5C} = 1$ $\Rightarrow 5C = 1$ $\Rightarrow C = \frac{1}{5}$. Thus,$\frac{x}{1} + \frac{y}{3/5} = 1$ $\Rightarrow x + \frac{5y}{3} = 1$ $\Rightarrow 3x + 5y = 3$. So,$A-2$.
$(B)$ Let the line be $\frac{x}{a} + \frac{y}{b} = 1$. The intercepts are $A(a, 0)$ and $B(0, b)$. $P(2, -5)$ is the midpoint,so $\frac{a}{2} = 2 \Rightarrow a = 4$ and $\frac{b}{2} = -5 \Rightarrow b = -10$. The equation is $\frac{x}{4} + \frac{y}{-10} = 1$ $\Rightarrow 5x - 2y = 20$ $\Rightarrow 5x - 2y - 20 = 0$. So,$B-5$.
$(C)$ Line parallel to $2x - 3y + 5 = 0$ is $2x - 3y + k = 0$. It has $x$-intercept $\frac{2}{5}$,so at $y=0, x=\frac{2}{5}$ $\Rightarrow 2(\frac{2}{5}) - 3(0) + k = 0$ $\Rightarrow k = -\frac{4}{5}$. The equation is $2x - 3y - \frac{4}{5} = 0$ $\Rightarrow 10x - 15y - 4 = 0$ $\Rightarrow 10x - 15y = 4$. So,$C-4$.
$(D)$ Line perpendicular to $5x + 2y + 7 = 0$ is $2x - 5y + \mu = 0$. It has $y$-intercept $\frac{4}{5}$,so at $x=0, y=\frac{4}{5}$ $\Rightarrow 2(0) - 5(\frac{4}{5}) + \mu = 0$ $\Rightarrow -4 + \mu = 0$ $\Rightarrow \mu = 4$. The equation is $2x - 5y + 4 = 0$. So,$D-1$.

(C) The given equation of the line is $\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=\gamma$.
This represents a line passing through $(x_1, y_1)$ with an inclination $\theta$.
The slope of this line is $m_1 = \tan \theta$.
Let the slope of the line perpendicular to this line be $m_2$.
Since the lines are perpendicular,$m_1 \times m_2 = -1$.
Therefore,$m_2 = -\frac{1}{\tan \theta} = -\cot \theta$.
The equation of the required line is given as $\frac{x}{a}+\frac{y}{b}=1$,which can be rewritten as $y = -\frac{b}{a}x + b$.
The slope of this line is $m_2 = -\frac{b}{a}$.
Equating the slopes,we get $-\frac{b}{a} = -\cot \theta$.
Thus,$\frac{b}{a} = \cot \theta$.

(D) Given the line equation: $x + 2y + 1 = 0$.
Converting to slope-intercept form $y = mx + c$:
$2y = -x - 1 \Rightarrow y = -\frac{1}{2}x - \frac{1}{2}$.
Thus,$m = -\frac{1}{2}$ and $c = -\frac{1}{2}$.
Converting to normal form $x \cos \theta + y \sin \theta = p$:
Dividing $x + 2y = -1$ by $\sqrt{1^2 + 2^2} = \sqrt{5}$,we get $\frac{1}{\sqrt{5}}x + \frac{2}{\sqrt{5}}y = -\frac{1}{\sqrt{5}}$.
Since the normal form requires $p > 0$,we multiply by $-1$: $-\frac{1}{\sqrt{5}}x - \frac{2}{\sqrt{5}}y = \frac{1}{\sqrt{5}}$.
Thus,$\cos \theta = -\frac{1}{\sqrt{5}}$ and $\sin \theta = -\frac{2}{\sqrt{5}}$.
Therefore,$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-2/\sqrt{5}}{-1/\sqrt{5}} = 2$.
Now,calculate $\tan^{-1}(\tan \theta + m + c) = \tan^{-1}(2 - \frac{1}{2} - \frac{1}{2}) = \tan^{-1}(1) = \frac{\pi}{4}$.

(C) Let the slope of the line be $m$.
The equation of the line passing through $P(1, 2)$ is $y-2 = m(x-1)$,which simplifies to $mx - y + (2-m) = 0$.
The intersection point of this line with $x+y=4$ is found by solving the system:
$x + (mx + 2 - m) = 4$ $\Rightarrow x(1+m) = m+2$ $\Rightarrow x = \frac{m+2}{m+1}$.
Then $y = 4 - x = 4 - \frac{m+2}{m+1} = \frac{4m+4-m-2}{m+1} = \frac{3m+2}{m+1}$.
The distance between $P(1, 2)$ and the intersection point is given as $\frac{\sqrt{6}}{3}$.
Using the distance formula: $\sqrt{(\frac{m+2}{m+1} - 1)^2 + (\frac{3m+2}{m+1} - 2)^2} = \frac{\sqrt{6}}{3}$.
$\sqrt{(\frac{1}{m+1})^2 + (\frac{m}{m+1})^2} = \frac{\sqrt{6}}{3}$ $\Rightarrow \frac{\sqrt{1+m^2}}{|m+1|} = \frac{\sqrt{6}}{3}$.
Squaring both sides: $\frac{1+m^2}{(m+1)^2} = \frac{6}{9} = \frac{2}{3}$.
$3(1+m^2) = 2(m^2+2m+1)$ $\Rightarrow 3+3m^2 = 2m^2+4m+2$ $\Rightarrow m^2-4m+1 = 0$.
Solving for $m$: $m = \frac{4 \pm \sqrt{16-4}}{2} = 2 \pm \sqrt{3}$.
For $m = 2+\sqrt{3} = \tan(75^\circ) = \tan(\frac{5\pi}{12})$,the angle is $\frac{5\pi}{12}$.
For $m = 2-\sqrt{3} = \tan(15^\circ) = \tan(\frac{\pi}{12})$,the angle is $\frac{\pi}{12}$.

(C) Given equation: $4x + 3y + 2 = 0$ or $4x + 3y = -2$.
To convert to normal form $x \cos \alpha + y \sin \alpha = p$,divide by $\sqrt{4^2 + 3^2} = 5$.
Since $p$ must be positive,divide by $-5$: $\frac{-4}{5}x - \frac{3}{5}y = \frac{2}{5}$.
Thus,$\cos \alpha = \frac{-4}{5}$,$\sin \alpha = \frac{-3}{5}$,and $p = \frac{2}{5}$.
For intercept form $\frac{x}{a} + \frac{y}{b} = 1$,rewrite $4x + 3y = -2$ as $\frac{x}{-2/4} + \frac{y}{-2/3} = 1$.
So,$a = \frac{-1}{2}$ and $b = \frac{-2}{3}$.
Now,calculate $\frac{p \sec \alpha}{ab} = \frac{p}{ab \cos \alpha} = \frac{2/5}{(-1/2 \times -2/3) \times (-4/5)} = \frac{2/5}{(1/3) \times (-4/5)} = \frac{2/5}{-4/15} = \frac{2}{5} \times \frac{-15}{4} = \frac{-3}{2}$.

(D) Let the slope of the line be $m$. The equation of the line passing through $(-5, 4)$ is $y-4=m(x+5)$,which simplifies to $mx-y+5m+4=0$.
The distance between the parallel lines $x+2y+1=0$ and $x+2y-1=0$ is $d = \frac{|1 - (-1)|}{\sqrt{1^2 + 2^2}} = \frac{2}{\sqrt{5}}$.
Since the intercept made by the line between these two parallel lines is equal to the distance between them,the line must be perpendicular to the given parallel lines.
The slope of the lines $x+2y+1=0$ and $x+2y-1=0$ is $m_1 = -\frac{1}{2}$.
Therefore,the slope of the required line is $m = -\frac{1}{m_1} = 2$.
Substituting $m=2$ into the equation $y-4=m(x+5)$,we get $y-4=2(x+5)$,which simplifies to $2x-y+14=0$.

(D) The given equation of the line is $x + \sqrt{3} y = -4$.
To find the $X$-intercept $a$,set $y = 0$: $x = -4$,so $a = -4$.
To find the $Y$-intercept $b$,set $x = 0$: $\sqrt{3} y = -4$,so $b = -\frac{4}{\sqrt{3}}$.
The normal form is $x \cos \alpha + y \sin \alpha = p$.
Comparing $x + \sqrt{3} y = -4$ with $x \cos \alpha + y \sin \alpha = p$,we first make the constant term positive: $-x - \sqrt{3} y = 4$.
Dividing by $\sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2$,we get $-\frac{1}{2} x - \frac{\sqrt{3}}{2} y = 2$.
Thus,$\cos \alpha = -\frac{1}{2}$ and $\sin \alpha = -\frac{\sqrt{3}}{2}$,which gives $\alpha = \frac{4 \pi}{3}$ and $p = 2$.
Now,calculate $\sqrt{3} \pi b p - 3 a \alpha$:
$\sqrt{3} \pi \left( -\frac{4}{\sqrt{3}} \right) (2) - 3 (-4) \left( \frac{4 \pi}{3} \right) = -8 \pi + 16 \pi = 8 \pi$.

(B) The line makes an angle of $120^{\circ}$ with the positive direction of the $X$-axis. The slope of the line is $m = \tan(120^{\circ}) = -\sqrt{3}$.
Let the normal to the line make an angle $\alpha$ with the positive $X$-axis. The normal is perpendicular to the line,so the angle of the normal is $\alpha = 120^{\circ} - 90^{\circ} = 30^{\circ}$.
The length of the perpendicular from the origin to the line is given as $p = 4$.
The normal form of the equation of a line is $x \cos \alpha + y \sin \alpha = p$.
Substituting the values,we get $x \cos(30^{\circ}) + y \sin(30^{\circ}) = 4$.
$x(\frac{\sqrt{3}}{2}) + y(\frac{1}{2}) = 4$.
Multiplying by $2$,we get $\sqrt{3}x + y = 8$.

(A) Let the intercepts of the line on the coordinate axes be $a$ and $b$.
Given that $a+b = -1$,so $b = -(1+a)$.
The intercept form of the line is $\frac{x}{a} + \frac{y}{b} = 1$.
Substituting $b$,we get $\frac{x}{a} - \frac{y}{1+a} = 1$.
Since the line passes through $(4,3)$,we have $\frac{4}{a} - \frac{3}{1+a} = 1$.
Multiplying by $a(1+a)$,we get $4(1+a) - 3a = a(1+a)$.
$4 + 4a - 3a = a + a^2$ $\Rightarrow 4 + a = a + a^2$ $\Rightarrow a^2 = 4$.
Thus,$a = 2$ or $a = -2$.
If $a = 2$,then $b = -(1+2) = -3$. The equation is $\frac{x}{2} + \frac{y}{-3} = 1 \Rightarrow 3x - 2y - 6 = 0$.
If $a = -2$,then $b = -(1-2) = 1$. The equation is $\frac{x}{-2} + \frac{y}{1} = 1$ $\Rightarrow -x + 2y = 2$ $\Rightarrow x - 2y + 2 = 0$.
The combined equation is $(3x - 2y - 6)(x - 2y + 2) = 0$.

(C) Let the line intersect the coordinate axes at points $A(a, 0)$ and $B(0, b)$.
The point $P\left(\frac{1}{2}, \frac{1}{3}\right)$ divides the line segment $AB$ in the ratio $2: 3$.
Using the section formula,the coordinates of $P$ are given by:
$P = \left(\frac{2(a) + 3(0)}{2 + 3}, \frac{2(0) + 3(b)}{2 + 3}\right) = \left(\frac{2a}{5}, \frac{3b}{5}\right)$.
Given $P = \left(\frac{1}{2}, \frac{1}{3}\right)$,we equate the coordinates:
$\frac{2a}{5} = \frac{1}{2} \Rightarrow a = \frac{5}{4}$
$\frac{3b}{5} = \frac{1}{3} \Rightarrow b = \frac{5}{9}$
The intercept form of the line equation is $\frac{x}{a} + \frac{y}{b} = 1$.
Substituting the values of $a$ and $b$:
$\frac{x}{5/4} + \frac{y}{5/9} = 1$
$\frac{4x}{5} + \frac{9y}{5} = 1$
$4x + 9y = 5$.

(C) The given line is $5x - 12y + 6 = 0$. The slope of this line is $m = \frac{5}{12}$.
Since line $L$ is perpendicular to this line,its slope $m_L$ must satisfy $m_L \times \frac{5}{12} = -1$,so $m_L = -\frac{12}{5}$.
The equation of line $L$ in normal form is $x \cos \theta + y \sin \theta = p$,where $p = 2$ is the distance from the origin.
The slope of this line is $-\cot \theta = -\frac{12}{5}$,which implies $\cot \theta = \frac{12}{5}$.
Thus,$\tan \theta = \frac{5}{12}$.
Since the line makes a positive intercept on the $Y$-axis,we check the intercept form: $y = -\frac{12}{5}x + \frac{2}{\sin \theta}$.
Using $\cot \theta = \frac{12}{5}$,we have $\sin \theta = \frac{5}{13}$ and $\cos \theta = \frac{12}{13}$.
The intercept is $\frac{2}{\sin \theta} = \frac{2}{5/13} = \frac{26}{5} > 0$,which is positive.
Finally,$\tan \theta + \cot \theta = \frac{5}{12} + \frac{12}{5} = \frac{25 + 144}{60} = \frac{169}{60}$.

(D) First,find the point of intersection of the lines $2x + 3y + 1 = 0$ and $x + y - 3 = 0$.
Solving the system:
$x + y = 3 \Rightarrow y = 3 - x$.
Substitute into the first equation: $2x + 3(3 - x) + 1 = 0$ $\Rightarrow 2x + 9 - 3x + 1 = 0$ $\Rightarrow -x + 10 = 0$ $\Rightarrow x = 10$.
Then $y = 3 - 10 = -7$.
The point of intersection is $(10, -7)$.
The slope of the line $L$ is $m = \tan(\tan^{-1} \frac{2}{3}) = \frac{2}{3}$.
The equation of the line passing through $(10, -7)$ with slope $\frac{2}{3}$ is:
$y - (-7) = \frac{2}{3}(x - 10)$ $\Rightarrow 3(y + 7) = 2(x - 10)$ $\Rightarrow 3y + 21 = 2x - 20$ $\Rightarrow 2x - 3y = 41$.
To find the intercepts,write the equation in intercept form $\frac{x}{a} + \frac{y}{b} = 1$:
$\frac{2x}{41} - \frac{3y}{41} = 1 \Rightarrow \frac{x}{41/2} + \frac{y}{-41/3} = 1$.
Thus,$a = \frac{41}{2}$ and $b = -\frac{41}{3}$.
The sum of the intercepts is $a + b = \frac{41}{2} - \frac{41}{3} = \frac{123 - 82}{6} = \frac{41}{6}$.

(C) The equation of the line passing through $P(3,4)$ with an angle of inclination $\theta = \frac{\pi}{6}$ is given by the parametric form: $\frac{x-3}{\cos(\pi/6)} = \frac{y-4}{\sin(\pi/6)} = r$.
Substituting $\cos(\pi/6) = \frac{\sqrt{3}}{2}$ and $\sin(\pi/6) = \frac{1}{2}$,we get: $\frac{x-3}{\sqrt{3}/2} = \frac{y-4}{1/2} = r$.
Thus,any point $Q$ on this line is given by $(3 + \frac{\sqrt{3}r}{2}, 4 + \frac{r}{2})$.
Since $Q$ lies on the line $12x + 5y + 10 = 0$,we substitute the coordinates of $Q$ into this equation:
$12(3 + \frac{\sqrt{3}r}{2}) + 5(4 + \frac{r}{2}) + 10 = 0$.
$36 + 6\sqrt{3}r + 20 + 2.5r + 10 = 0$.
$66 + (6\sqrt{3} + 2.5)r = 0$.
$66 + (\frac{12\sqrt{3} + 5}{2})r = 0$.
$r = -\frac{132}{12\sqrt{3} + 5}$.
The length $PQ$ is $|r| = \frac{132}{12\sqrt{3} + 5}$.

(D) The given line is $x+y-2=0$.
To find the $X$-intercept '$a$',set $y=0$: $x+0-2=0 \Rightarrow x=2$. Thus,$a=2$.
The line can be written in intercept form as $\frac{x}{2} + \frac{y}{2} = 1$.
The normal form of a line is $x \cos \beta + y \sin \beta = p$.
Comparing $x+y=2$ with $x \cos \beta + y \sin \beta = p$,we divide by $\sqrt{1^2+1^2} = \sqrt{2}$:
$\frac{1}{\sqrt{2}}x + \frac{1}{\sqrt{2}}y = \frac{2}{\sqrt{2}} = \sqrt{2}$.
Here,$\cos \beta = \frac{1}{\sqrt{2}}$ and $\sin \beta = \frac{1}{\sqrt{2}}$,so $\beta = 45^{\circ}$.
The perpendicular distance $p = \sqrt{2}$.
Now,calculate $a \tan \beta + p^2$:
$a \tan \beta + p^2 = 2 \tan(45^{\circ}) + (\sqrt{2})^2 = 2(1) + 2 = 4$.

(D) Given $k = \frac{a+b}{ab} = \frac{1}{a} + \frac{1}{b}$.
We check the point $\left(\frac{1}{k}, \frac{1}{k}\right)$ in the equation $\frac{x}{a} + \frac{y}{b} = 1$:
Substitute $x = \frac{1}{k}$ and $y = \frac{1}{k}$:
$\frac{1}{ka} + \frac{1}{kb} = \frac{1}{k} \left(\frac{1}{a} + \frac{1}{b}\right)$.
Since $\frac{1}{a} + \frac{1}{b} = k$,we get:
$\frac{1}{k} \cdot k = 1$.
Thus,the point $\left(\frac{1}{k}, \frac{1}{k}\right)$ satisfies the equation of the line.

(C) The slope of the line $y=\sqrt{3}x$ is $m=\sqrt{3}$. Since the required line is parallel to this line,its slope is also $\sqrt{3}$.
Thus,$\tan \theta = \sqrt{3}$,which implies $\theta = 60^{\circ}$.
Let the distance $PQ = r$. The coordinates of any point $P$ on the line passing through $Q(2,3)$ at a distance $r$ are given by $(2+r \cos 60^{\circ}, 3+r \sin 60^{\circ}) = (2+\frac{r}{2}, 3+\frac{r\sqrt{3}}{2})$.
Since $P$ lies on the line $2x+4y-27=0$,we substitute these coordinates into the equation:
$2(2+\frac{r}{2}) + 4(3+\frac{r\sqrt{3}}{2}) - 27 = 0$
$4 + r + 12 + 2\sqrt{3}r - 27 = 0$
$r(1+2\sqrt{3}) - 11 = 0$
$r = \frac{11}{2\sqrt{3}+1}$
Rationalizing the denominator:
$r = \frac{11(2\sqrt{3}-1)}{(2\sqrt{3}+1)(2\sqrt{3}-1)} = \frac{11(2\sqrt{3}-1)}{12-1} = \frac{11(2\sqrt{3}-1)}{11} = 2\sqrt{3}-1$.
Therefore,the length of the line segment $PQ$ is $2\sqrt{3}-1$.

(D) Let the angle made by line $L$ with the positive $X$-axis be $\theta$. The coordinates of point $B$ can be written as $(2 + 4 \cos \theta, 3 + 4 \sin \theta)$.
Since $B$ lies on the line $4x - 3y - 19 = 0$,we substitute these coordinates into the equation:
$4(2 + 4 \cos \theta) - 3(3 + 4 \sin \theta) - 19 = 0$
$8 + 16 \cos \theta - 9 - 12 \sin \theta - 19 = 0$
$16 \cos \theta - 12 \sin \theta = 20$
Dividing by $4$,we get $4 \cos \theta - 3 \sin \theta = 5$.
This can be written as $4 \cos \theta - 3 \sin \theta = 5(\cos^2 \theta + \sin^2 \theta)$.
Alternatively,using $R \cos(\theta + \alpha) = 5$ where $R = \sqrt{4^2 + (-3)^2} = 5$,we have $5 \cos(\theta + \alpha) = 5$,where $\cos \alpha = 4/5$ and $\sin \alpha = 3/5$.
Thus,$\cos(\theta + \alpha) = 1$,which implies $\theta + \alpha = 0$ or $2\pi$.
Since $\tan \alpha = 3/4$,$\theta = -\alpha = -\operatorname{Tan}^{-1}(3/4)$.
In the anti-clockwise direction,the angle is $2\pi - \operatorname{Tan}^{-1}(3/4)$ or simply the angle corresponding to the slope $m = \tan \theta$. Solving $4 \cos \theta - 3 \sin \theta = 5$ gives $\tan \theta = -3/4$. The angle is $\pi - \operatorname{Tan}^{-1}(3/4)$.

(C) Let the slope of the required line be $m$. The slope of the line joining $(1, 1)$ and $(2, 3)$ is $m_1 = \frac{3-1}{2-1} = 2$.
The angle between the two lines is $45^{\circ}$,so $\tan(45^{\circ}) = |\frac{m - m_1}{1 + m \cdot m_1}|$.
$1 = |\frac{m - 2}{1 + 2m}|$.
This gives two cases: $1 + 2m = m - 2$ or $1 + 2m = -(m - 2)$.
Case $1$: $m = -3$. Since the slope is negative,this is a valid solution.
Case $2$: $1 + 2m = -m + 2 \implies 3m = 1 \implies m = \frac{1}{3}$. This is positive,so we reject it.
The equation of the line with slope $m = -3$ passing through $(4, -3)$ is $y - (-3) = -3(x - 4)$,which simplifies to $y + 3 = -3x + 12$,or $3x + y = 9$.
Dividing by $9$,we get $\frac{x}{3} + \frac{y}{9} = 1$.
The $x$-intercept is $a = 3$ and the $y$-intercept is $b = 9$.
The sum of the intercepts is $a + b = 3 + 9 = 12$.

(C) The given polar equation is $\cos \theta + 7 \sin \theta = \frac{1}{r}$.
Multiplying both sides by $r$,we get $r \cos \theta + 7 r \sin \theta = 1$.
Using the standard conversion formulas $x = r \cos \theta$ and $y = r \sin \theta$,the equation becomes $x + 7y = 1$.
This is a linear equation in $x$ and $y$,which represents a straight line.

(A) The given polar equation of the line is $\sin \theta - \cos \theta = \frac{1}{r}$.
Multiplying by $r$,we get $r \sin \theta - r \cos \theta = 1$.
Using the conversion $x = r \cos \theta$ and $y = r \sin \theta$,the Cartesian equation is $y - x = 1$,or $x - y + 1 = 0$.
The slope of this line is $m_1 = 1$.
The line perpendicular to this line will have a slope $m_2 = -1$.
The point given is $\left(2, \frac{\pi}{6}\right)$. Converting to Cartesian coordinates:
$x = 2 \cos \left(\frac{\pi}{6}\right) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$
$y = 2 \sin \left(\frac{\pi}{6}\right) = 2 \cdot \frac{1}{2} = 1$
So the point is $(\sqrt{3}, 1)$.
The equation of the line with slope $-1$ passing through $(\sqrt{3}, 1)$ is:
$y - 1 = -1(x - \sqrt{3})$
$y - 1 = -x + \sqrt{3}$
$x + y = \sqrt{3} + 1$
Converting back to polar form using $x = r \cos \theta$ and $y = r \sin \theta$:
$r \cos \theta + r \sin \theta = \sqrt{3} + 1$
$r(\sin \theta + \cos \theta) = \sqrt{3} + 1$
$\sin \theta + \cos \theta = \frac{\sqrt{3} + 1}{r}$

(A) Given that $a, b$ and $c$ are in $AP$.
Therefore,$2b = a + c$ or $c = 2b - a$.
The equation of the straight line is $ax + 2by + c = 0$.
Substituting the value of $c$,we get $ax + 2by + (2b - a) = 0$.
Rearranging the terms,we have $a(x - 1) + 2b(y + 1) = 0$.
For this line to pass through a fixed point for all values of $a$ and $b$,the coefficients must be zero independently.
Thus,$x - 1 = 0 \Rightarrow x = 1$ and $y + 1 = 0 \Rightarrow y = -1$.
Hence,the fixed point is $(1, -1)$.

(D) Given the polar equation: $r \cos \left(\theta-\frac{\pi}{3}\right)=2$
Using the trigonometric identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$:
$r \left( \cos \theta \cos \frac{\pi}{3} + \sin \theta \sin \frac{\pi}{3} \right) = 2$
Substituting $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$:
$r \left( \cos \theta \cdot \frac{1}{2} + \sin \theta \cdot \frac{\sqrt{3}}{2} \right) = 2$
Multiplying by $2$:
$r \cos \theta + \sqrt{3} r \sin \theta = 4$
Using the conversion formulas $x = r \cos \theta$ and $y = r \sin \theta$:
$x + \sqrt{3} y = 4$
This is a linear equation in $x$ and $y$,which represents a straight line.

(C) Let the equation of the line in intercept form be $\frac{x}{a} + \frac{y}{b} = 1$.
Given that the sum of intercepts is $a + b = -1$,so $b = -1 - a$.
The line passes through $(4, 3)$,so $\frac{4}{a} + \frac{3}{b} = 1$.
Substituting $b = -(1 + a)$,we get $\frac{4}{a} - \frac{3}{1 + a} = 1$.
$4(1 + a) - 3a = a(1 + a) \Rightarrow 4 + 4a - 3a = a + a^2$.
$a^2 = 4 \Rightarrow a = 2$ or $a = -2$.
If $a = 2$,then $b = -1 - 2 = -3$. The equation is $\frac{x}{2} + \frac{y}{-3} = 1$.
If $a = -2$,then $b = -1 - (-2) = 1$. The equation is $\frac{x}{-2} + \frac{y}{1} = 1$.

(C) Let the given points be $A(a, b)$ and $B(-a, -b)$.
The slope $m$ of the line passing through $A$ and $B$ is given by:
$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-b - b}{-a - a} = \frac{-2b}{-2a} = \frac{b}{a}$.
The equation of the line passing through $(a, b)$ with slope $\frac{b}{a}$ is:
$y - b = \frac{b}{a}(x - a)$
$ay - ab = bx - ab$
$bx = ay$
Now,we check the given options to see which point satisfies the equation $bx = ay$:
For option $(C)$,substituting $x = a^2$ and $y = ab$:
$b(a^2) = a(ab)$
$a^2b = a^2b$
Since the equation is satisfied,the line passes through $(a^2, ab)$.

(C) Let the equation of the line in intercept form be $\frac{x}{a} + \frac{y}{b} = 1$.
The coordinates of the points where the line intersects the axes are $A(0, b)$ and $B(a, 0)$.
Since the point $(\alpha, \beta)$ is the midpoint of the segment $AB$,we have:
$\alpha = \frac{0 + a}{2} \Rightarrow a = 2\alpha$
$\beta = \frac{b + 0}{2} \Rightarrow b = 2\beta$
Substituting these values into the intercept form equation:
$\frac{x}{2\alpha} + \frac{y}{2\beta} = 1$
Multiplying both sides by $2$:
$\frac{x}{\alpha} + \frac{y}{\beta} = 2$

(B) The initial line passes through $A(2,0)$ and makes an angle of $30^{\circ}$ with the positive $x$-axis.
When the line is rotated clockwise by $15^{\circ}$,the new angle $\theta$ with the positive $x$-axis becomes $30^{\circ} - 15^{\circ} = 15^{\circ}$.
The slope of the new line is $m = \tan 15^{\circ} = \tan(45^{\circ} - 30^{\circ}) = \frac{\tan 45^{\circ} - \tan 30^{\circ}}{1 + \tan 45^{\circ} \tan 30^{\circ}} = \frac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$.
Rationalizing the denominator: $m = \frac{(\sqrt{3}-1)^2}{3-1} = \frac{3+1-2\sqrt{3}}{2} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3}$.
The equation of the line passing through $(2,0)$ with slope $m = 2-\sqrt{3}$ is $y - 0 = (2-\sqrt{3})(x - 2)$.
$y = (2-\sqrt{3})x - 4 + 2\sqrt{3}$.
Rearranging the terms,we get $(2-\sqrt{3})x - y - 4 + 2\sqrt{3} = 0$.

(B) The given equation is $y-y_1=m(x-x_1)$,which can be rewritten as $y=mx+(y_1-mx_1)$.
Since $m$ and $x_1$ are fixed,the slope of the lines is constant $(m)$.
Lines with the same slope are parallel to each other.
Therefore,for different values of $y_1$,we obtain a family of parallel lines.
Thus,option $B$ is correct.

(C) Let the points be $A(3q, 0)$,$B(0, 3p)$,and $C(1, 1)$.
Since the points are collinear,the slope of $AC$ must be equal to the slope of $BC$.
Slope of $AC = \frac{1 - 0}{1 - 3q} = \frac{1}{1 - 3q}$.
Slope of $BC = \frac{3p - 1}{0 - 1} = \frac{3p - 1}{-1} = 1 - 3p$.
Equating the slopes: $\frac{1}{1 - 3q} = 1 - 3p$.
$1 = (1 - 3p)(1 - 3q)$.
$1 = 1 - 3q - 3p + 9pq$.
$3p + 3q = 9pq$.
Dividing both sides by $3pq$,we get $\frac{3p}{3pq} + \frac{3q}{3pq} = \frac{9pq}{3pq}$.
$\frac{1}{q} + \frac{1}{p} = 3$.