Mix Examples-Straight Line Questions in English

(B) The points $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ are collinear if the area of the triangle formed by them is $0$.
Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$
Substituting the given points $(1, 1)$,$(0, \sec^2 \theta)$,and $(\csc^2 \theta, 0)$:
$\frac{1}{2} |1(\sec^2 \theta - 0) + 0(0 - 1) + \csc^2 \theta(1 - \sec^2 \theta)| = 0$
$\sec^2 \theta + \csc^2 \theta(1 - \sec^2 \theta) = 0$
$\sec^2 \theta + \csc^2 \theta - \csc^2 \theta \sec^2 \theta = 0$
$\frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta} - \frac{1}{\sin^2 \theta \cos^2 \theta} = 0$
$\frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} - \frac{1}{\sin^2 \theta \cos^2 \theta} = 0$
$\frac{1}{\sin^2 \theta \cos^2 \theta} - \frac{1}{\sin^2 \theta \cos^2 \theta} = 0$
This identity $0 = 0$ holds for all $\theta$ where $\sec \theta$ and $\csc \theta$ are defined.
Since $\sec \theta$ is undefined at $\theta = (2n+1)\frac{\pi}{2}$ and $\csc \theta$ is undefined at $\theta = n\pi$,the points are collinear for all $\theta$ except $\theta = \frac{n\pi}{2}$.

(A) Let the line passing through the point $M(1, 5)$ be $\frac{x - 1}{\cos \theta} = \frac{y - 5}{\sin \theta} = r$.
Since the segment is bisected at $M(1, 5)$,the endpoints $A$ and $B$ are at distances $r = d$ and $r = -d$ from $M$ respectively.
Point $A$ is $(1 + d \cos \theta, 5 + d \sin \theta)$ and it lies on $5x - y - 4 = 0$.
Substituting $A$ into the first line: $5(1 + d \cos \theta) - (5 + d \sin \theta) - 4 = 0 \implies 5d \cos \theta - d \sin \theta = 4$.
Point $B$ is $(1 - d \cos \theta, 5 - d \sin \theta)$ and it lies on $3x + 4y - 4 = 0$.
Substituting $B$ into the second line: $3(1 - d \cos \theta) + 4(5 - d \sin \theta) - 4 = 0 \implies 3 - 3d \cos \theta + 20 - 4d \sin \theta - 4 = 0 \implies -3d \cos \theta - 4d \sin \theta = -19 \implies 3d \cos \theta + 4d \sin \theta = 19$.
We have the system:
$5d \cos \theta - d \sin \theta = 4$
$3d \cos \theta + 4d \sin \theta = 19$
Solving for $d \cos \theta$ and $d \sin \theta$:
Multiply the first by $4$: $20d \cos \theta - 4d \sin \theta = 16$.
Adding to the second: $23d \cos \theta = 35 \implies d \cos \theta = \frac{35}{23}$.
Substitute back: $5(\frac{35}{23}) - d \sin \theta = 4 \implies d \sin \theta = \frac{175}{23} - 4 = \frac{175 - 92}{23} = \frac{83}{23}$.
The slope of the line is $m = \frac{\sin \theta}{\cos \theta} = \frac{83}{35}$.
The equation of the line passing through $(1, 5)$ with slope $\frac{83}{35}$ is $y - 5 = \frac{83}{35}(x - 1)$.
$35y - 175 = 83x - 83 \implies 83x - 35y + 92 = 0$.

(A) The median passes through vertex $A$,which is the intersection of the lines $px + qy - 1 = 0$ and $qx + py - 1 = 0$.
Any line passing through the intersection of these two lines is given by the family of lines equation: $(px + qy - 1) + \lambda(qx + py - 1) = 0$,where $\lambda$ is a constant.
Since the median also passes through the midpoint of the base $BC$,which is the point $(p, q)$,we substitute this point into the equation:
$(p(p) + q(q) - 1) + \lambda(q(p) + p(q) - 1) = 0$
$(p^2 + q^2 - 1) + \lambda(2pq - 1) = 0$
$\lambda = -\frac{p^2 + q^2 - 1}{2pq - 1}$.
Substituting $\lambda$ back into the family of lines equation:
$(px + qy - 1) - \frac{p^2 + q^2 - 1}{2pq - 1}(qx + py - 1) = 0$
$(2pq - 1)(px + qy - 1) = (p^2 + q^2 - 1)(qx + py - 1)$.

(C) The given lines are $L_1: 2x + y + 6 = 0$ and $L_2: 4x + 2y - 9 = 0$. Note that $L_2$ can be written as $2(2x + y) - 9 = 0$,meaning $L_1$ and $L_2$ are parallel.
$A$ line perpendicular to these lines will have the form $x - 2y + k = 0$. Since it passes through the origin $(0, 0)$,we have $0 - 2(0) + k = 0$,so $k = 0$. The equation of the line is $x - 2y = 0$.
Find the intersection point $P$ of $x - 2y = 0$ and $2x + y + 6 = 0$:
Substitute $x = 2y$ into $2(2y) + y + 6 = 0$ $\Rightarrow 5y = -6$ $\Rightarrow y = -6/5, x = -12/5$. So $P = (-12/5, -6/5)$.
Find the intersection point $Q$ of $x - 2y = 0$ and $4x + 2y - 9 = 0$:
Substitute $x = 2y$ into $4(2y) + 2y - 9 = 0$ $\Rightarrow 10y = 9$ $\Rightarrow y = 9/10, x = 18/10 = 9/5$. So $Q = (9/5, 9/10)$.
The origin $(0, 0)$ divides the segment $PQ$ in ratio $\lambda : 1$. Using the section formula for the $x$-coordinate:
$0 = \frac{\lambda(9/5) + 1(-12/5)}{\lambda + 1}$ $\Rightarrow 9\lambda - 12 = 0$ $\Rightarrow \lambda = 12/9 = 4/3$.
Thus,the ratio is $4 : 3$.

(C) The line $2x + 3y = 12$ meets the $y$-axis at $B$ where $x=0$,so $B = (0, 4)$.
The slope of the line $2x + 3y = 12$ is $m_1 = -\frac{2}{3}$.
The slope of the line perpendicular to $AB$ is $m_2 = -\frac{1}{m_1} = \frac{3}{2}$.
The equation of the line passing through $(5, 5)$ with slope $\frac{3}{2}$ is $y - 5 = \frac{3}{2}(x - 5)$,which simplifies to $3x - 2y = 5$.
This line meets the $x$-axis at $C$ where $y=0$,so $3x = 5 \implies C = (\frac{5}{3}, 0)$.
To find $E$,we solve the system of equations:
$2x + 3y = 12$ $(i)$
$3x - 2y = 5$ (ii)
Multiplying $(i)$ by $2$ and (ii) by $3$: $4x + 6y = 24$ and $9x - 6y = 15$.
Adding these gives $13x = 39 \implies x = 3$. Substituting $x=3$ into $(i)$,$6 + 3y = 12 \implies 3y = 6 \implies y = 2$. So $E = (3, 2)$.
The area of quadrilateral $OCEB$ can be calculated by splitting it into $\Delta OCE$ and $\Delta OEB$.
Area of $\Delta OCE = \frac{1}{2} |x_O(y_C - y_E) + x_C(y_E - y_O) + x_E(y_O - y_C)| = \frac{1}{2} |0 + \frac{5}{3}(2 - 0) + 3(0 - 0)| = \frac{5}{3}$.
Area of $\Delta OEB = \frac{1}{2} |x_O(y_E - y_B) + x_E(y_B - y_O) + x_B(y_O - y_E)| = \frac{1}{2} |0 + 3(4 - 0) + 0| = 6$.
Total area $= \frac{5}{3} + 6 = \frac{23}{3} \text{ sq. units}$.

(A) Given the lines $3x + 4y = 9$ and $y = mx + 1$.
Substitute $y$ from the second equation into the first:
$3x + 4(mx + 1) = 9$
$3x + 4mx + 4 = 9$
$x(3 + 4m) = 5$
$x = \frac{5}{3 + 4m}$
For $x$ to be an integer,$(3 + 4m)$ must be a divisor of $5$. The divisors of $5$ are $\pm 1$ and $\pm 5$.
Case $1$: $3 + 4m = 1 \implies 4m = -2 \implies m = -0.5$ (not an integer).
Case $2$: $3 + 4m = -1 \implies 4m = -4 \implies m = -1$ (integer).
Case $3$: $3 + 4m = 5 \implies 4m = 2 \implies m = 0.5$ (not an integer).
Case $4$: $3 + 4m = -5 \implies 4m = -8 \implies m = -2$ (integer).
The integral values of $m$ are $-1$ and $-2$. Thus,there are $2$ such values.

(D) The perpendicular bisectors of the sides of a triangle intersect at the circumcenter $O$.
Since the perpendicular bisectors of $AB$ and $AC$ are $x - y + 5 = 0$ and $x + 2y = 0$,their intersection point $O$ is found by solving these equations:
$x - y = -5$ and $x + 2y = 0$.
Subtracting the equations,we get $-3y = -5$,so $y = \frac{5}{3}$.
Then $x = -2y = -\frac{10}{3}$.
Thus,the circumcenter $O$ is $(-\frac{10}{3}, \frac{5}{3})$.
Since $O$ is the circumcenter,$OA = OB = OC$.
Let $B = (x_1, y_1)$. The reflection of $A(1, -2)$ about the line $x - y + 5 = 0$ gives $B$.
The formula for the reflection of $(x_0, y_0)$ about $ax + by + c = 0$ is $\frac{x - x_0}{a} = \frac{y - y_0}{b} = -2 \frac{ax_0 + by_0 + c}{a^2 + b^2}$.
For $A(1, -2)$ and $x - y + 5 = 0$:
$\frac{x_1 - 1}{1} = \frac{y_1 + 2}{-1} = -2 \frac{1 - (-2) + 5}{1^2 + (-1)^2} = -2 \frac{8}{2} = -8$.
So,$x_1 - 1 = -8 \implies x_1 = -7$ and $y_1 + 2 = 8 \implies y_1 = 6$. Thus,$B = (-7, 6)$.
Similarly,for $C(x_2, y_2)$,the reflection of $A(1, -2)$ about $x + 2y = 0$:
$\frac{x_2 - 1}{1} = \frac{y_2 + 2}{2} = -2 \frac{1 + 2(-2)}{1^2 + 2^2} = -2 \frac{-3}{5} = \frac{6}{5}$.
So,$x_2 - 1 = \frac{6}{5} \implies x_2 = \frac{11}{5}$ and $y_2 + 2 = \frac{12}{5} \implies y_2 = \frac{2}{5}$. Thus,$C = (\frac{11}{5}, \frac{2}{5})$.
The equation of line $BC$ passing through $(-7, 6)$ and $(\frac{11}{5}, \frac{2}{5})$ is:
$y - 6 = \frac{\frac{2}{5} - 6}{\frac{11}{5} - (-7)} (x - (-7)) = \frac{-\frac{28}{5}}{\frac{46}{5}} (x + 7) = -\frac{14}{23} (x + 7)$.
$23(y - 6) = -14(x + 7) \implies 23y - 138 = -14x - 98 \implies 14x + 23y - 40 = 0$.

(A) Let the vertex be $A(-1, 2)$ and the base be $BC$ with equation $2x - y - 1 = 0$.
The altitude $AD$ from vertex $A$ to the base $BC$ is the perpendicular distance from $A$ to the line $2x - y - 1 = 0$.
$AD = \left| \frac{2(-1) - (2) - 1}{\sqrt{2^2 + (-1)^2}} \right| = \left| \frac{-2 - 2 - 1}{\sqrt{5}} \right| = \frac{5}{\sqrt{5}} = \sqrt{5}$.
In an equilateral triangle,the altitude $AD$ is related to the side length $s$ by $AD = s \sin(60^o) = s \frac{\sqrt{3}}{2}$.
Therefore,$\sqrt{5} = s \frac{\sqrt{3}}{2}$ $\Rightarrow s = \frac{2\sqrt{5}}{\sqrt{3}} = \sqrt{\frac{4 \times 5}{3}} = \sqrt{\frac{20}{3}}$.

(A) Let the common ratio be $r$. Since ${x_1}, {x_2}, {x_3}$ are in $G$.$P$.,we have ${x_2} = {x_1}r$ and ${x_3} = {x_1}r^2$.
Similarly,since ${y_1}, {y_2}, {y_3}$ are in $G$.$P$. with the same common ratio $r$,we have ${y_2} = {y_1}r$ and ${y_3} = {y_1}r^2$.
The points are $P_1 = ({x_1}, {y_1})$,$P_2 = ({x_1}r, {y_1}r)$,and $P_3 = ({x_1}r^2, {y_1}r^2)$.
Notice that for all points,the ratio of the $y$-coordinate to the $x$-coordinate is constant: $\frac{{y_1}}{{x_1}} = \frac{{y_1}r}{{x_1}r} = \frac{{y_1}r^2}{{x_1}r^2} = m$ (where $m = \frac{{y_1}}{{x_1}}$).
This implies that all three points satisfy the linear equation $y = mx$,which represents a straight line passing through the origin.

(A) The slopes of $AB$ and $BC$ are $m_1 = -4$ and $m_2 = \frac{3}{4}$ respectively. Let $\alpha$ be the angle between $AB$ and $BC$. Then,$\tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{-4 - \frac{3}{4}}{1 + (-4)(\frac{3}{4})} \right| = \left| \frac{-\frac{19}{4}}{1 - 3} \right| = \left| \frac{-\frac{19}{4}}{-2} \right| = \frac{19}{8}$.
Since $AB = AC$,the triangle $ABC$ is isosceles with $\angle ABC = \angle ACB = \alpha$. Thus,the line $AC$ also makes an angle $\alpha$ with the line $BC$. Let $m$ be the slope of line $AC$. The equation of line $AC$ passing through $A(2, -7)$ is $y + 7 = m(x - 2)$.
The angle between $AC$ and $BC$ is $\alpha$,so $\tan \alpha = \left| \frac{m - \frac{3}{4}}{1 + m(\frac{3}{4})} \right| = \frac{19}{8}$.
$\frac{4m - 3}{4 + 3m} = \pm \frac{19}{8}$.
Case $1$: $8(4m - 3) = 19(4 + 3m)$ $\Rightarrow 32m - 24 = 76 + 57m$ $\Rightarrow -25m = 100$ $\Rightarrow m = -4$ (This is the slope of line $AB$).
Case $2$: $8(4m - 3) = -19(4 + 3m)$ $\Rightarrow 32m - 24 = -76 - 57m$ $\Rightarrow 89m = -52$ $\Rightarrow m = -\frac{52}{89}$.
Using $m = -\frac{52}{89}$ in $y + 7 = m(x - 2)$,we get $89(y + 7) = -52(x - 2)$ $\Rightarrow 89y + 623 = -52x + 104$ $\Rightarrow 52x + 89y + 519 = 0$.

(B) The lines $ax + by + p = 0$ and $x \cos \alpha + y \sin \alpha - p = 0$ are inclined at an angle $\pi / 4$.
Therefore,$\tan(\pi / 4) = \left| \frac{(-a/b) - (-\cos \alpha / \sin \alpha)}{1 + (-a/b)(-\cos \alpha / \sin \alpha)} \right| = 1$.
This simplifies to $|-a \sin \alpha + b \cos \alpha| = |b \sin \alpha + a \cos \alpha|$.
Squaring both sides,$a^2 \sin^2 \alpha + b^2 \cos^2 \alpha - 2ab \sin \alpha \cos \alpha = b^2 \sin^2 \alpha + a^2 \cos^2 \alpha + 2ab \sin \alpha \cos \alpha$.
Rearranging gives $(a^2 - b^2)(\sin^2 \alpha - \cos^2 \alpha) = 4ab \sin \alpha \cos \alpha$,which leads to $(a^2 - b^2)(-\cos 2\alpha) = 2ab \sin 2\alpha$.
Since the lines are concurrent with $x \sin \alpha - y \cos \alpha = 0$,the determinant of the coefficients is zero:
$\begin{vmatrix} a & b & p \\ \cos \alpha & \sin \alpha & -p \\ \sin \alpha & -\cos \alpha & 0 \end{vmatrix} = 0$.
Expanding the determinant: $a(-p \cos \alpha) - b(p \sin \alpha) + p(-\cos^2 \alpha - \sin^2 \alpha) = 0$.
$-ap \cos \alpha - bp \sin \alpha - p = 0 \implies a \cos \alpha + b \sin \alpha = -1$.
Squaring this: $a^2 \cos^2 \alpha + b^2 \sin^2 \alpha + 2ab \sin \alpha \cos \alpha = 1$.
Using the concurrency and angle conditions,we find $a^2 + b^2 = 2$.

(A) First,find the point of intersection of the lines $4x + y - 1 = 0$ and $7x - 3y - 35 = 0$.
Multiplying the first equation by $3$,we get $12x + 3y - 3 = 0$.
Adding this to $7x - 3y - 35 = 0$,we get $19x - 38 = 0$,which implies $x = 2$.
Substituting $x = 2$ into $4x + y - 1 = 0$,we get $4(2) + y - 1 = 0$,so $y = -7$.
The point of intersection is $(2, -7)$.
The line joining $(3, 5)$ and $(2, -7)$ has the slope $m = \frac{-7 - 5}{2 - 3} = \frac{-12}{-1} = 12$.
The equation of the line is $y - 5 = 12(x - 3)$,which simplifies to $12x - y - 31 = 0$.
The distance of this line from $(0, 0)$ is $d_1 = \frac{|12(0) - 0 - 31|}{\sqrt{12^2 + (-1)^2}} = \frac{31}{\sqrt{145}}$.
The distance of this line from $(8, 34)$ is $d_2 = \frac{|12(8) - 34 - 31|}{\sqrt{12^2 + (-1)^2}} = \frac{|96 - 65|}{\sqrt{145}} = \frac{31}{\sqrt{145}}$.
Since $d_1 = d_2$,the statement is True.

(B) The coordinates of $A$ are found by setting $x=0$ in $3x + 2y = 24$,giving $A(0, 12)$.
Setting $y=0$ gives $B(8, 0)$.
The midpoint of $AB$ is $M\left(\frac{0+8}{2}, \frac{12+0}{2}\right) = (4, 6)$.
The slope of $AB$ is $m = \frac{0-12}{8-0} = -\frac{3}{2}$.
The slope of the perpendicular bisector is $m' = \frac{2}{3}$.
The equation of the perpendicular bisector is $y - 6 = \frac{2}{3}(x - 4)$,which simplifies to $2x - 3y + 10 = 0$.
The line through $(0, -1)$ parallel to the $x$-axis is $y = -1$.
Substituting $y = -1$ into the bisector equation: $2x - 3(-1) + 10 = 0 \implies 2x + 13 = 0 \implies x = -\frac{13}{2}$.
Thus,$C$ is $\left(-\frac{13}{2}, -1\right)$.
The area of $\triangle ABC$ with vertices $(0, 12), (8, 0), \left(-\frac{13}{2}, -1\right)$ is:
$\text{Area} = \frac{1}{2} |0(0 - (-1)) + 8(-1 - 12) + (-\frac{13}{2})(12 - 0)|$
$= \frac{1}{2} |0 + 8(-13) - \frac{13}{2}(12)|$
$= \frac{1}{2} |-104 - 78| = \frac{1}{2} |-182| = 91 \, sq. \, units$.

(A) Let the lines passing through the origin be $y = mx$. These lines make an angle of $\pm 45^\circ$ with the line $2x + 3y = 6$,which has a slope of $m_1 = -2/3$.
Using the formula $\tan \theta = \left| \frac{m - m_1}{1 + m m_1} \right|$,we have:
$\tan(\pm 45^\circ) = \frac{m - (-2/3)}{1 + m(-2/3)} = \pm 1$
$\frac{3m + 2}{3 - 2m} = \pm 1$
Case $1$: $3m + 2 = 3 - 2m$ $\Rightarrow 5m = 1$ $\Rightarrow m = 1/5$.
The line is $y = \frac{1}{5}x \Rightarrow x - 5y = 0$.
Case $2$: $3m + 2 = -(3 - 2m)$ $\Rightarrow 3m + 2 = -3 + 2m$ $\Rightarrow m = -5$.
The line is $y = -5x \Rightarrow 5x + y = 0$.
The vertices of the triangle are the intersection points of these lines with $2x + 3y = 6$ and the origin $(0,0)$.
Solving $x - 5y = 0$ and $2x + 3y = 6$: $2(5y) + 3y = 6$ $\Rightarrow 13y = 6$ $\Rightarrow y = 6/13, x = 30/13$.
Solving $5x + y = 0$ and $2x + 3y = 6$: $y = -5x$ $\Rightarrow 2x + 3(-5x) = 6$ $\Rightarrow -13x = 6$ $\Rightarrow x = -6/13, y = 30/13$.
The vertices are $(0,0)$,$A(\frac{30}{13}, \frac{6}{13})$,and $B(-\frac{6}{13}, \frac{30}{13})$.
The area $\Delta = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = \frac{1}{2} |0 + \frac{30}{13}(\frac{30}{13} - 0) + (-\frac{6}{13})(0 - \frac{6}{13})| = \frac{1}{2} |\frac{900}{169} + \frac{36}{169}| = \frac{1}{2} \times \frac{936}{169} = \frac{468}{169} = \frac{36}{13}$.

(C) Let the slope of the required lines be $m$. The given line is $x + y = 2$,which has a slope of $m_1 = -1$.
Since the triangle is equilateral,the angle between the sides is $60^\circ$.
Using the formula $\tan \theta = |\frac{m - m_1}{1 + m \cdot m_1}|$,we have:
$\tan 60^\circ = |\frac{m - (-1)}{1 + m(-1)}|$
$\sqrt{3} = |\frac{m + 1}{1 - m}|$
This gives two cases:
$1) \sqrt{3} = \frac{m + 1}{1 - m}$ $\Rightarrow \sqrt{3} - \sqrt{3}m = m + 1$ $\Rightarrow m(1 + \sqrt{3}) = \sqrt{3} - 1$ $\Rightarrow m = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3}$
$2) -\sqrt{3} = \frac{m + 1}{1 - m}$ $\Rightarrow -\sqrt{3} + \sqrt{3}m = m + 1$ $\Rightarrow m(\sqrt{3} - 1) = \sqrt{3} + 1$ $\Rightarrow m = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = 2 + \sqrt{3}$
Using the point-slope form $y - y_1 = m(x - x_1)$ with point $(2, 3)$:
$y - 3 = (2 \pm \sqrt{3})(x - 2)$.

(A) The family of lines passing through the intersection of $L_1: x - 3y + 1 = 0$ and $L_2: 2x + 5y - 9 = 0$ is given by $(x - 3y + 1) + \lambda(2x + 5y - 9) = 0$.
This simplifies to $(1 + 2\lambda)x + (5\lambda - 3)y + (1 - 9\lambda) = 0$.
The perpendicular distance from the origin $(0, 0)$ to this line is given as $\sqrt{5}$.
Using the formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$,we have $\frac{|1 - 9\lambda|}{\sqrt{(1 + 2\lambda)^2 + (5\lambda - 3)^2}} = \sqrt{5}$.
Squaring both sides: $\frac{(1 - 9\lambda)^2}{(1 + 2\lambda)^2 + (5\lambda - 3)^2} = 5$.
$(1 - 81\lambda + 81\lambda^2) = 5(1 + 4\lambda + 4\lambda^2 + 25\lambda^2 - 30\lambda + 9)$.
$(1 - 18\lambda + 81\lambda^2) = 5(29\lambda^2 - 26\lambda + 10) = 145\lambda^2 - 130\lambda + 50$.
$64\lambda^2 - 112\lambda + 49 = 0$.
$(8\lambda - 7)^2 = 0$,which gives $\lambda = \frac{7}{8}$.
Substituting $\lambda = \frac{7}{8}$ into the equation: $(x - 3y + 1) + \frac{7}{8}(2x + 5y - 9) = 0$.
$8x - 24y + 8 + 14x + 35y - 63 = 0$.
$22x + 11y - 55 = 0$.
Dividing by $11$,we get $2x + y - 5 = 0$.

(C) The length of the perpendicular from the origin $(0,0)$ to the line $Ax + By + C = 0$ is given by $p = \frac{|C|}{\sqrt{A^2 + B^2}}$.
For the first line $x \sec \alpha + y \csc \alpha - 2a = 0$,we have $p_1 = \frac{|-2a|}{\sqrt{\sec^2 \alpha + \csc^2 \alpha}} = \frac{2|a|}{\sqrt{\frac{1}{\cos^2 \alpha} + \frac{1}{\sin^2 \alpha}}} = \frac{2|a| \sin \alpha \cos \alpha}{\sqrt{\sin^2 \alpha + \cos^2 \alpha}} = |a \sin 2\alpha|$.
Thus,$p_1^2 = a^2 \sin^2 2\alpha$.
For the second line $x \cos \alpha - y \sin \alpha - a \cos 2\alpha = 0$,we have $p_2 = \frac{|-a \cos 2\alpha|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}} = |a \cos 2\alpha|$.
Thus,$p_2^2 = a^2 \cos^2 2\alpha$.
Now,$\left( \frac{p_1}{p_2} + \frac{p_2}{p_1} \right)^2 = \left( \frac{p_1^2 + p_2^2}{p_1 p_2} \right)^2 = \frac{(a^2 \sin^2 2\alpha + a^2 \cos^2 2\alpha)^2}{a^2 \sin^2 2\alpha \cdot a^2 \cos^2 2\alpha} = \frac{(a^2)^2}{a^4 \sin^2 2\alpha \cos^2 2\alpha} = \frac{1}{\sin^2 2\alpha \cos^2 2\alpha}$.
Multiplying numerator and denominator by $4$,we get $\frac{4}{4 \sin^2 2\alpha \cos^2 2\alpha} = \frac{4}{(2 \sin 2\alpha \cos 2\alpha)^2} = \frac{4}{\sin^2 4\alpha} = 4 \csc^2 4\alpha$.

(C) The equation of the line passing through $P(1, 1)$ with angle $\theta$ is $\frac{x - 1}{\cos\theta} = \frac{y - 1}{\sin\theta} = r$.
The coordinates of $A$ (where $y=0$) are $(1 - \cot\theta, 0)$ and $B$ (where $x=0$) are $(0, 1 - \tan\theta)$.
Given $AB^2 = OP^2$,we have $(1 - \cot\theta)^2 + (1 - \tan\theta)^2 = 1^2 + 1^2 = 2$.
Expanding this: $(1 - 2\cot\theta + \cot^2\theta) + (1 - 2\tan\theta + \tan^2\theta) = 2$.
$2 - 2(\cot\theta + \tan\theta) + (\cot^2\theta + \tan^2\theta) = 2$.
Let $u = \tan\theta + \cot\theta$. Since $\cot^2\theta + \tan^2\theta = u^2 - 2$,we get:
$u^2 - 2 - 2u = 0 \Rightarrow u^2 - 2u - 2 = 0$.
Solving for $u$: $u = \frac{2 \pm \sqrt{4 - 4(1)(-2)}}{2} = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$.
Since $|\tan\theta + \cot\theta| \ge 2$,the value $1 - \sqrt{3}$ is rejected as it is approximately $-0.732$.
Thus,$\tan\theta + \cot\theta = 1 + \sqrt{3}$.

(D) For any point $P$ on the line,the value of $|PA - PB|$ is at most $AB$,which occurs when $P, A, B$ are collinear.
To maximize $|PA - PB|$,$P$ must be the intersection of the line $AB$ and the given line $4x + 3y + 9 = 0$.
The equation of line $AB$ passing through $(0, 1)$ and $(2, 0)$ is $\frac{x}{2} + \frac{y}{1} = 1$,which simplifies to $x + 2y = 2$.
We solve the system of equations:
$4x + 3y = -9$ $(1)$
$x + 2y = 2$ $(2)$
From $(2)$,$x = 2 - 2y$. Substituting into $(1)$:
$4(2 - 2y) + 3y = -9$
$8 - 8y + 3y = -9$
$-5y = -17 \implies y = \frac{17}{5}$.
Substituting $y = \frac{17}{5}$ into $x = 2 - 2y$:
$x = 2 - 2\left(\frac{17}{5}\right) = 2 - \frac{34}{5} = -\frac{24}{5}$.
Thus,the point $P$ is $\left( -\frac{24}{5}, \frac{17}{5} \right)$.

(A) Given equation is $16a^2 - 40ab + 25b^2 - c^2 = 0$.
This can be written as $(4a - 5b)^2 - c^2 = 0$.
Using the identity $x^2 - y^2 = (x - y)(x + y)$,we get $(4a - 5b - c)(4a - 5b + c) = 0$.
This implies $4a - 5b - c = 0$ or $4a - 5b + c = 0$.
We compare these with the line equation $ax + by + c = 0$.
Case $1$: $4a - 5b + c = 0$ can be written as $a(4) + b(-5) + c = 0$. Comparing with $ax + by + c = 0$,we get $(x, y) = (4, -5)$.
Case $2$: $4a - 5b - c = 0$ can be written as $a(-4) + b(5) + c = 0$. Comparing with $ax + by + c = 0$,we get $(x, y) = (-4, 5)$.
Thus,the line passes through $(4, -5)$ and $(-4, 5)$.

(C) Given the equations $x + y = |a|$ and $ax - y = 1$.
Adding the two equations,we get $(1 + a)x = |a| + 1$,which implies $x = \frac{|a| + 1}{a + 1}$.
Substituting $x$ into the first equation,$y = |a| - x = |a| - \frac{|a| + 1}{a + 1} = \frac{|a|(a + 1) - (|a| + 1)}{a + 1} = \frac{a|a| + |a| - |a| - 1}{a + 1} = \frac{a|a| - 1}{a + 1}$.
For the intersection point to be in the first quadrant,both $x > 0$ and $y > 0$ must hold.
Since $x = \frac{|a| + 1}{a + 1} > 0$ and $|a| + 1$ is always positive,we must have $a + 1 > 0$,which means $a > -1$.
For $y = \frac{a|a| - 1}{a + 1} > 0$,since $a + 1 > 0$,we must have $a|a| - 1 > 0$,which means $a|a| > 1$.
If $a \in (-1, 0]$,then $a|a| = -a^2$,and $-a^2 > 1$ is impossible.
If $a > 0$,then $a^2 > 1$,which implies $a > 1$.
Thus,the possible values of $a$ are in the interval $(1, \infty)$.

(B) Given that $a, b, c$ are in $A.P.$,we have the relation $2b = a + c$,which can be rewritten as $a - 2b + c = 0$.
The equation of the straight line is $ax + by + c = 0$.
Comparing this with the condition $a(1) + b(-2) + c = 0$,we can see that the line satisfies the equation for the point $(x, y) = (1, -2)$.
Therefore,the line $ax + by + c = 0$ always passes through the point $(1, -2)$.

(A) The slope of the line $AB$ is $m = \frac{1-0}{3-2} = 1$.
Since $m = \tan \theta = 1$,the angle of inclination is $\theta = 45^\circ$.
When the line is rotated anti-clockwise about point $A$ by $15^\circ$,the new angle of inclination becomes $\theta' = 45^\circ + 15^\circ = 60^\circ$.
The slope of the new line is $m' = \tan 60^\circ = \sqrt{3}$.
The line passes through $A(2,0)$,so the equation is given by $y - 0 = \sqrt{3}(x - 2)$.
$y = \sqrt{3}x - 2\sqrt{3}$.
Rearranging the terms,we get $\sqrt{3}x - y - 2\sqrt{3} = 0$.

(A) Given one side is $4x + 7y + 5 = 0.$ The vertices are $A(-3, 1)$ and $B(1, 1).$
Since $A$ and $B$ do not satisfy the given line equation,the side containing $A$ and $B$ is parallel to the given line.
Let the side containing $A$ and $B$ be $4x + 7y + k = 0.$
Since it passes through $(-3, 1),$ $4(-3) + 7(1) + k = 0 \implies -12 + 7 + k = 0 \implies k = 5.$
So,the side is $4x + 7y - 5 = 0.$
Wait,the side containing $A$ and $B$ is $4x + 7y - 5 = 0.$
The other side parallel to this is $4x + 7y + k' = 0.$
Passing through $(1, 1),$ $4(1) + 7(1) + k' = 0 \implies k' = -11.$
So,the opposite side is $4x + 7y - 11 = 0.$
The perpendicular sides have slope $7/4.$
Side through $(-3, 1): y - 1 = \frac{7}{4}(x + 3) \implies 4y - 4 = 7x + 21 \implies 7x - 4y + 25 = 0.$
Side through $(1, 1): y - 1 = \frac{7}{4}(x - 1) \implies 4y - 4 = 7x - 7 \implies 7x - 4y - 3 = 0.$
Thus,the equations are $4x + 7y - 11 = 0, 7x - 4y + 25 = 0$ and $7x - 4y - 3 = 0.$
Comparing with options,option $(A)$ matches the form $7x - 4y + 25 = 0, 4x + 7y - 11 = 0$ and $7x - 4y - 3 = 0.$

(A) Let the vertex opposite to the hypotenuse be $B(2, 2)$. Since the triangle is an isosceles right-angled triangle,the angles at the base are $45^\circ$.
Let the slope of one of the sides passing through $B(2, 2)$ be $m$.
The equation of this side is $y - 2 = m(x - 2)$,which simplifies to $mx - y + (2 - 2m) = 0$.
The angle between this line and the hypotenuse $3x + 4y - 4 = 0$ is $45^\circ$.
Using the formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$,where $m_1 = m$ and $m_2 = -\frac{3}{4}$:
$\tan 45^\circ = \left| \frac{m - (-3/4)}{1 + m(-3/4)} \right| = 1$
$\left| \frac{4m + 3}{4 - 3m} \right| = 1$
This gives two cases:
Case $1$: $\frac{4m + 3}{4 - 3m} = 1$ $\Rightarrow 4m + 3 = 4 - 3m$ $\Rightarrow 7m = 1$ $\Rightarrow m = \frac{1}{7}$.
Substituting $m = \frac{1}{7}$ into $y - 2 = m(x - 2)$:
$y - 2 = \frac{1}{7}(x - 2)$ $\Rightarrow 7y - 14 = x - 2$ $\Rightarrow x - 7y + 12 = 0$.
Case $2$: $\frac{4m + 3}{4 - 3m} = -1$ $\Rightarrow 4m + 3 = -4 + 3m$ $\Rightarrow m = -7$.
Substituting $m = -7$ into $y - 2 = m(x - 2)$:
$y - 2 = -7(x - 2)$ $\Rightarrow y - 2 = -7x + 14$ $\Rightarrow 7x + y - 16 = 0$.
Comparing with the given options,$x - 7y + 12 = 0$ is the correct equation.

(B) The equation of the lines is $y = \sqrt{3}|x| + 2$.
This represents two lines: $y = \sqrt{3}x + 2$ for $x \ge 0$ and $y = -\sqrt{3}x + 2$ for $x < 0$.
The point of intersection $A$ is $(0, 2)$.
The angle between the lines and the $y$-axis (which is the angle bisector) can be found from the slope. For $y = \sqrt{3}x + 2$,the slope is $\sqrt{3}$,so the angle with the $x$-axis is $60^\circ$. The angle with the $y$-axis is $90^\circ - 60^\circ = 30^\circ$.
Let $P$ be a point on the line at a distance $AP = 5$.
The foot of the perpendicular $M$ from $P$ onto the $y$-axis lies on the $y$-axis.
The distance $AM = AP \cos(30^\circ) = 5 \times \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2}$.
Since $A$ is at $(0, 2)$ and the lines extend upwards,the $y$-coordinate of $M$ is $2 + \frac{5\sqrt{3}}{2} = \frac{4 + 5\sqrt{3}}{2}$.
Thus,the coordinates of $M$ are $\left(0, \frac{4 + 5\sqrt{3}}{2}\right)$.

(B) Let the vertices of the square be $O(0,0)$,$A(a \cos \alpha, a \sin \alpha)$,$B$,and $C$.
Since the side $OA$ makes an angle $\alpha$ with the $x$-axis,its coordinates are $A(a \cos \alpha, a \sin \alpha)$.
The diagonal $OB$ makes an angle of $\alpha + \frac{\pi}{4}$ with the $x$-axis.
The slope of $OB$ is $\tan(\alpha + \frac{\pi}{4})$.
The diagonal $AC$ is perpendicular to $OB$.
The slope of $AC$ is $-\cot(\alpha + \frac{\pi}{4}) = -\frac{1}{\tan(\alpha + \frac{\pi}{4})} = -\frac{1 - \tan \alpha \tan(\frac{\pi}{4})}{\tan \alpha + \tan(\frac{\pi}{4})} = -\frac{1 - \tan \alpha}{1 + \tan \alpha} = \frac{\tan \alpha - 1}{\tan \alpha + 1} = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha}$.
The equation of the line $AC$ passing through $A(a \cos \alpha, a \sin \alpha)$ is:
$y - a \sin \alpha = \frac{\sin \alpha - \cos \alpha}{\sin \alpha + \cos \alpha} (x - a \cos \alpha)$
$y(\sin \alpha + \cos \alpha) - a \sin \alpha(\sin \alpha + \cos \alpha) = x(\sin \alpha - \cos \alpha) - a \cos \alpha(\sin \alpha - \cos \alpha)$
$y(\sin \alpha + \cos \alpha) - x(\sin \alpha - \cos \alpha) = a \sin^2 \alpha + a \sin \alpha \cos \alpha - a \sin \alpha \cos \alpha + a \cos^2 \alpha$
$y(\sin \alpha + \cos \alpha) - x(\sin \alpha - \cos \alpha) = a(\sin^2 \alpha + \cos^2 \alpha) = a$.
Thus,the equation is $y(\cos \alpha + \sin \alpha) - x(\sin \alpha - \cos \alpha) = a$.

(B) The vertices of the triangle are $(0, 0)$,$(0, 21)$,and $(21, 0)$.
The interior points $(x, y)$ must satisfy $x > 0$,$y > 0$,and $x + y < 21$.
For a fixed $x$,$y$ can take integer values from $1$ to $21 - x - 1 = 20 - x$.
Since $y > 0$,we must have $20 - x \geq 1$,so $x$ can range from $1$ to $19$.
For each $x \in \{1, 2, \dots, 19\}$,the number of possible values for $y$ is $20 - x$.
The total number of integral points is $\sum_{x=1}^{19} (20 - x) = 19 + 18 + \dots + 1$.
Using the sum formula $\frac{n(n+1)}{2}$ with $n = 19$,we get $\frac{19 \times 20}{2} = 190$.

(D) The given lines are $y - mx = 0$,$y - mx - 1 = 0$,$y - nx = 0$,and $y - nx - 1 = 0$.
These lines form a parallelogram.
The area of a parallelogram formed by lines $a_1x + b_1y + c_1 = 0$,$a_1x + b_1y + c_2 = 0$,$a_2x + b_2y + d_1 = 0$,and $a_2x + b_2y + d_2 = 0$ is given by the formula:
$\text{Area} = \left| \frac{(c_1 - c_2)(d_1 - d_2)}{a_1b_2 - a_2b_1} \right|$
Here,the equations are:
$mx - y + 0 = 0$
$mx - y + 1 = 0$
$nx - y + 0 = 0$
$nx - y + 1 = 0$
Comparing with the formula,we have $c_1 = 0, c_2 = 1, d_1 = 0, d_2 = 1, a_1 = m, b_1 = -1, a_2 = n, b_2 = -1$.
$\text{Area} = \left| \frac{(0 - 1)(0 - 1)}{m(-1) - n(-1)} \right| = \left| \frac{(-1)(-1)}{-m + n} \right| = \left| \frac{1}{n - m} \right| = \frac{1}{|m - n|}$.

(B) The given family of lines passes through the intersection of $2x + y + 4 = 0$ and $x - 2y - 3 = 0$.
Solving these equations: $2(2x + y + 4) + (x - 2y - 3) = 0$ $\Rightarrow 5x + 5 = 0$ $\Rightarrow x = -1$.
Substituting $x = -1$ in $2x + y + 4 = 0$,we get $-2 + y + 4 = 0 \Rightarrow y = -2$.
So,all lines in the family pass through the fixed point $P(-1, -2)$.
The distance between $P(-1, -2)$ and $M(2, -3)$ is $PM = \sqrt{(2 - (-1))^2 + (-3 - (-2))^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$.
Any line passing through $P$ at a distance $d$ from $M$ must satisfy $d \le PM$. Since the required distance is exactly $\sqrt{10} = PM$,the only line at this distance is the one perpendicular to the segment $PM$ at point $P$.
Therefore,there is exactly $1$ such line.

(D) The two parallel lines are $L_1: x + 2y + 3 = 0$ and $L_2: x + 2y - 7 = 0$.
The distance $d$ between these two parallel lines is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = \frac{|3 - (-7)|}{\sqrt{1^2 + 2^2}} = \frac{10}{\sqrt{5}}$.
Since these lines form two sides of a square,the distance between the other two parallel sides must also be $d = \frac{10}{\sqrt{5}}$.
The third side is $L_3: 2x - y - 4 = 0$. The fourth side $L_4$ must be parallel to $L_3$,so let $L_4: 2x - y + \lambda = 0$.
The distance between $L_3$ and $L_4$ is $\frac{|\lambda - (-4)|}{\sqrt{2^2 + (-1)^2}} = \frac{|\lambda + 4|}{\sqrt{5}}$.
Equating the distances: $\frac{|\lambda + 4|}{\sqrt{5}} = \frac{10}{\sqrt{5}}$,which implies $|\lambda + 4| = 10$.
This gives $\lambda + 4 = 10$ or $\lambda + 4 = -10$.
So,$\lambda = 6$ or $\lambda = -14$.
The equations of the fourth sides are $2x - y + 6 = 0$ and $2x - y - 14 = 0$.

(D) The given equation is $y + \sqrt{3} |x| = 2$. This represents two lines: $L_1: y + \sqrt{3}x = 2$ (for $x \ge 0$) and $L_2: y - \sqrt{3}x = 2$ (for $x < 0$).
Both lines intersect at the point $P(0, 2)$.
The angle bisectors of these lines are the $y$-axis $(x=0)$ and the line $y=2$.
The point $A$ lies on the lines at a distance of $d = \frac{4}{\sqrt{3}}$ from $P(0, 2)$.
For $L_1$,the slope is $m = -\sqrt{3}$,so the angle with the $x$-axis is $120^\circ$. The coordinates of $A$ are $(0 + d \cos 120^\circ, 2 + d \sin 120^\circ) = \left( \frac{4}{\sqrt{3}} \cdot \left( -\frac{1}{2} \right), 2 + \frac{4}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} \right) = \left( -\frac{2}{\sqrt{3}}, 4 \right)$.
However,checking the geometry,the foot of the perpendicular from $A$ to the angle bisector $x=0$ is $(0, 4)$.

(D) The line $x + y = p$ intersects the axes at $A(p, 0)$ and $B(0, p)$. The area of $\triangle OAB = \frac{1}{2} \times p \times p = \frac{p^2}{2}$.
Let $Q$ divide $AB$ in the ratio $\lambda : 1$. The coordinates of $Q$ are $\left(\frac{p}{1+\lambda}, \frac{p\lambda}{1+\lambda}\right)$.
Since $PQ \perp AB$,the slope of $PQ$ is $1$ (as the slope of $AB$ is $-1$).
The equation of line $PQ$ is $y - \frac{p\lambda}{1+\lambda} = 1(x - \frac{p}{1+\lambda})$.
Setting $x = 0$ to find $P$ on the $y$-axis,we get $y = \frac{p\lambda - p}{1+\lambda} = \frac{p(\lambda - 1)}{\lambda + 1}$.
Area of $\triangle APQ = \frac{1}{2} \times AQ \times PQ$. Using the coordinates,the area is $\frac{p^2 \lambda}{2(\lambda + 1)^2}$.
Given $\frac{\text{Area}(\triangle APQ)}{\text{Area}(\triangle OAB)} = \frac{3}{8}$ $\Rightarrow \frac{p^2 \lambda / 2(\lambda + 1)^2}{p^2 / 2} = \frac{3}{8}$ $\Rightarrow \frac{\lambda}{(\lambda + 1)^2} = \frac{3}{8}$.
$8\lambda = 3(\lambda^2 + 2\lambda + 1) \Rightarrow 3\lambda^2 - 2\lambda + 3 = 0$. This quadratic has no real roots. Re-evaluating the geometry,if $Q$ divides $AB$ in ratio $k:1$,then $\frac{AQ}{BQ} = k$. The ratio of areas is $\frac{k}{(k+1)^2} = \frac{3}{8}$ is incorrect. The correct ratio is $\frac{AQ}{BQ} = 3$ or $1/3$.

(B) The line $AB$ is $2x + 3y = 12$. The slope of $AB$ is $m_1 = -\frac{2}{3}$.
Since the line $ED$ is perpendicular to $AB$,its slope $m_2 = \frac{3}{2}$.
The equation of line $ED$ passing through $(5, 5)$ is $y - 5 = \frac{3}{2}(x - 5)$,which simplifies to $3x - 2y = 5$.
To find point $E$,solve the system of equations $2x + 3y = 12$ and $3x - 2y = 5$. Multiplying the first by $2$ and the second by $3$,we get $4x + 6y = 24$ and $9x - 6y = 15$. Adding them gives $13x = 39$,so $x = 3$. Substituting $x = 3$ into $2x + 3y = 12$ gives $6 + 3y = 12$,so $y = 2$. Thus,$E = (3, 2)$.
Point $C$ is the intersection of $3x - 2y = 5$ with the $x$-axis $(y=0)$,so $3x = 5$,$x = \frac{5}{3}$. Thus,$C = (\frac{5}{3}, 0)$.
Point $B$ is the intersection of $2x + 3y = 12$ with the $y$-axis $(x=0)$,so $3y = 12$,$y = 4$. Thus,$B = (0, 4)$.
The area of quadrilateral $OCEB$ can be calculated as the area of $\Delta OBC$ minus the area of $\Delta OEC$ is not correct; rather,it is the area of $\Delta OBC$ plus the area of $\Delta BEC$ is not correct. The quadrilateral $OCEB$ is composed of $\Delta OBC$ and $\Delta BEC$ is incorrect. The area is best calculated as Area$(\Delta OBC)$ - Area$(\Delta OEC)$.
Area$(\Delta OBC)$ = $\frac{1}{2} \times OB \times OC$ is wrong. The vertices are $O(0,0), C(\frac{5}{3}, 0), E(3, 2), B(0, 4)$.
Using the shoelace formula: Area = $\frac{1}{2} |(0 \times 0 + \frac{5}{3} \times 2 + 3 \times 4 + 0 \times 0) - (0 \times \frac{5}{3} + 0 \times 3 + 2 \times 0 + 4 \times 0)| = \frac{1}{2} |(\frac{10}{3} + 12) - 0| = \frac{1}{2} \times \frac{46}{3} = \frac{23}{3}$ sq. units.

(D) The given lines are $L_1: 2x + y - 5 = 0$ and $L_2: x - 2y - 3 = 0$.
Since the product of their slopes is $( -2 ) \times ( 1/2 ) = -1$,the lines are perpendicular.
Let $A$ be the intersection point. Since $AB = AC$,$\triangle ABC$ is an isosceles right-angled triangle with the right angle at $A$.
The line $BC$ must make an angle of $45^\circ$ or $135^\circ$ with the given lines.
The slopes of the angle bisectors of the lines $L_1$ and $L_2$ are given by $\frac{2x+y-5}{\sqrt{5}} = \pm \frac{x-2y-3}{\sqrt{5}}$,which simplifies to $x + 3y - 2 = 0$ and $3x - y - 8 = 0$.
The slopes of these bisectors are $-1/3$ and $3$.
Since $BC$ is perpendicular to the angle bisector,the slope of $BC$ is either $3$ or $-1/3$.
For slope $m = 3$ passing through $(2, 3)$: $y - 3 = 3(x - 2) \implies 3x - y - 3 = 0$.
For slope $m = -1/3$ passing through $(2, 3)$: $y - 3 = -1/3(x - 2) \implies x + 3y - 11 = 0$.
Thus,both lines are possible.

(A) The slope of line $L_2$ is $4$. Since $L_1 \perp L_2$,the slope of $L_1$ is $m = -\frac{1}{4}$.
Let the equation of line $L_1$ be $y = -\frac{1}{4}x + k$,which can be written as $x + 4y = 4k$.
Let $c = 4k$. The intercepts are $(c, 0)$ and $(0, \frac{c}{4})$.
The area of the right-angled triangle $T$ formed by the origin $(0, 0)$ and these intercepts is $\frac{1}{2} \times |c| \times |\frac{c}{4}| = 8$.
$|c^2| = 64$,so $c = \pm 8$.
Taking $c = 8$,the vertices are $(0, 0), (8, 0),$ and $(0, 2)$.
The lengths of the sides are $8, 2,$ and $\sqrt{8^2 + 2^2} = \sqrt{64 + 4} = \sqrt{68}$.
The perimeter is $8 + 2 + \sqrt{68} = 10 + \sqrt{68}$.

(A) Let the line passing through $P(3, 1)$ have slope $m$. The equation of the line is $(y - 1) = m(x - 3)$,or $mx - y + (1 - 3m) = 0$.
The distance $d$ of this line from the origin $O(0, 0)$ is given by $d = \frac{|1 - 3m|}{\sqrt{m^2 + 1}}$.
For the distance to be maximum,the line $AB$ must be perpendicular to $OP$. The slope of $OP$ is $m_{OP} = \frac{1}{3}$. Thus,the slope of $AB$ is $m = -3$.
The equation of line $AB$ is $(y - 1) = -3(x - 3)$,which simplifies to $3x + y = 10$.
To find the intercepts,set $y = 0$ to get $x = \frac{10}{3}$ (point $A$) and set $x = 0$ to get $y = 10$ (point $B$).
The area of triangle $OAB$ is $\frac{1}{2} \times OA \times OB = \frac{1}{2} \times \frac{10}{3} \times 10 = \frac{50}{3} \text{ sq. units}$.

(D) Given the equation $x^3 - 3x^2 + x + \lambda = 0$ with roots $a, b, c$.
By Vieta's formulas,the sum of the roots is $a + b + c = 3$,which implies $a + b = 3 - c$.
The line equation is $x(a + b) + y = 1$,which can be written in intercept form as $\frac{x}{1/(a+b)} + \frac{y}{1} = 1$.
The area $A$ of the triangle formed with the coordinate axes is $A = \frac{1}{2} \times |\text{base}| \times |\text{height}| = \frac{1}{2} \times \frac{1}{|a+b|} \times 1 = \frac{1}{2|3-c|}$.
Since $c \in [1, 2]$,we have $3 - c \in [1, 2]$,so $a + b > 0$.
Thus,$A = \frac{1}{2(3-c)}$.
To maximize $A$,we need to maximize the denominator $2(3-c)$,which occurs when $c$ is at its maximum value in the interval $[1, 2]$.
For $c = 2$,$A = \frac{1}{2(3-2)} = \frac{1}{2}$.
Therefore,the maximum value of $A$ is $1/2$ sq. units.

(B) $|PA - PB|$ is maximum when $P, A, B$ are collinear. The equation of line $AB$ is $x + y = 2$.
Solving $x + y = 2$ and $2x + 3y + 1 = 0$ gives $P(7, -5)$.
$|QA - QB|$ is minimum when $Q$ is the intersection of the line $2x + 3y + 1 = 0$ and the perpendicular bisector of $AB$.
The perpendicular bisector of $AB$ is $y - x = 0$ (since the slope of $AB$ is $-1$).
Solving $y = x$ and $2x + 3y + 1 = 0$ gives $5x = -1$,so $x_2 = -1/5$ and $y_2 = -1/5$.
Thus,$x_1 - y_1 + x_2 - y_2 = (7 - (-5)) + (-1/5 - (-1/5)) = 12 + 0 = 12$.

(A) For option $(A)$: Let the line be $ax + by + c = 0$. The sum of perpendicular distances is $\frac{4a+2b+c}{\sqrt{a^2+b^2}} + \frac{2a+4b+c}{\sqrt{a^2+b^2}} + \frac{c}{\sqrt{a^2+b^2}} = 0$. This implies $6a + 6b + 3c = 0$,or $2a + 2b + c = 0$. This is the condition for the line to pass through $(2,2)$. Thus,$(A)$ is true.
For option $(B)$: The condition $Ar(\Delta PAB) + Ar(\Delta PBC) + Ar(\Delta PAC) = Ar(\Delta ABC)$ is satisfied for any point $P$ inside $\Delta ABC$,not just the orthocenter. Thus,$(B)$ is false.
For option $(C)$: The angle bisectors of a pair of lines are perpendicular to each other,they do not bisect the angle between themselves. Thus,$(C)$ is false.
For option $(D)$: The harmonic conjugate of the midpoint of $A$ and $B$ is at infinity. Thus,$(D)$ is false.

(D) Let the equation of the line passing through $(2, 5)$ be $\frac{x}{a} + \frac{y}{b} = 1$.
Since the line passes through $(2, 5)$,we have $\frac{2}{a} + \frac{5}{b} = 1$.
The area of the triangle formed with the coordinate axes is $\frac{1}{2} |ab| = 24$,so $|ab| = 48$.
Case $1$: $ab = 48$. Then $b = \frac{48}{a}$. Substituting into the line equation: $\frac{2}{a} + \frac{5a}{48} = 1$ $\Rightarrow 96 + 5a^2 = 48a$ $\Rightarrow 5a^2 - 48a + 96 = 0$. The discriminant $D = (-48)^2 - 4(5)(96) = 2304 - 1920 = 384 > 0$. This gives $2$ lines.
Case $2$: $ab = -48$. Then $b = -\frac{48}{a}$. Substituting: $\frac{2}{a} - \frac{5a}{48} = 1$ $\Rightarrow 96 - 5a^2 = 48a$ $\Rightarrow 5a^2 + 48a - 96 = 0$. The discriminant $D = (48)^2 - 4(5)(-96) = 2304 + 1920 = 4224 > 0$. This gives $2$ lines.
Total number of lines $= 2 + 2 = 4$.

(B) The distance between the two parallel lines $3x + 4y - 11 = 0$ and $3x + 4y - 1 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.
$d = \frac{|11 - 1|}{\sqrt{3^2 + 4^2}} = \frac{10}{5} = 2 \text{ units}$.
Since the intercept cut by the line between these two parallel lines is exactly equal to the distance between them,the required line must be perpendicular to the given parallel lines.
The slope of the given lines is $m = -\frac{3}{4}$.
Therefore,the slope of the required line is $m' = \frac{4}{3}$.
The equation of the line passing through $(3, 2)$ with slope $\frac{4}{3}$ is $y - 2 = \frac{4}{3}(x - 3)$.
$3(y - 2) = 4(x - 3) \implies 3y - 6 = 4x - 12 \implies 4x - 3y - 6 = 0$ or $3y - 4x + 6 = 0$.

(B) Let $C$ be $(\alpha, \beta)$.
Since $AC = BC$,we have $AC^2 = BC^2$.
$(\alpha + 2)^2 + (\beta - 3)^2 = (\alpha - 2)^2 + (\beta - 0)^2$
$\alpha^2 + 4\alpha + 4 + \beta^2 - 6\beta + 9 = \alpha^2 - 4\alpha + 4 + \beta^2$
$8\alpha - 6\beta + 9 = 0$ ......$(1)$
Slope of $AB = \frac{0 - 3}{2 - (-2)} = -\frac{3}{4}$.
The equation of a line parallel to $AB$ is $y = -\frac{3}{4}x + c$.
Given $y$-intercept $c = \frac{43}{12}$,the equation is $y = -\frac{3}{4}x + \frac{43}{12}$,which simplifies to $9x + 12y = 43$.
Since this line passes through $C(\alpha, \beta)$,we have $9\alpha + 12\beta = 43$ ......$(2)$.
Solving equations $(1)$ and $(2)$:
From $(1)$,$6\beta = 8\alpha + 9 \Rightarrow 12\beta = 16\alpha + 18$.
Substituting into $(2)$: $9\alpha + 16\alpha + 18 = 43$ $\Rightarrow 25\alpha = 25$ $\Rightarrow \alpha = 1$.
Then $12\beta = 43 - 9(1) = 34 \Rightarrow \beta = \frac{34}{12} = \frac{17}{6}$.
Thus,$C = \left(1, \frac{17}{6}\right)$.

(C) The given lines are $L_1: 4x + 3y - 10 = 0$ and $L_2: 8x + 6y + 5 = 0$.
Note that $L_2$ can be written as $2(4x + 3y) + 5 = 0$,which means $4x + 3y = -2.5$.
Since the lines $4x + 3y = 10$ and $4x + 3y = -2.5$ are parallel,any line through the origin $O(0,0)$ will intersect them at points $A$ and $B$ such that the ratio of the distances $OA$ and $OB$ is equal to the ratio of the perpendicular distances from the origin to these lines.
The perpendicular distance $d_1$ from $(0,0)$ to $4x + 3y - 10 = 0$ is $d_1 = \frac{|-10|}{\sqrt{4^2 + 3^2}} = \frac{10}{5} = 2$.
The perpendicular distance $d_2$ from $(0,0)$ to $8x + 6y + 5 = 0$ is $d_2 = \frac{|5|}{\sqrt{8^2 + 6^2}} = \frac{5}{10} = 0.5 = \frac{1}{2}$.
Since the lines are on opposite sides of the origin (as the constant terms have opposite signs relative to the origin),the origin $O$ divides the segment $AB$ internally in the ratio $OA:OB = d_1:d_2 = 2 : \frac{1}{2} = 4:1$.

(A) Let the intercepts of line $L$ on the $x$-axis and $y$-axis be $a$ and $b$ respectively. The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$.
Since the segment is bisected at $P(1, 2)$,we have $\frac{a}{2} = 1 \implies a = 2$ and $\frac{b}{2} = 2 \implies b = 4$.
Thus,the equation of line $L$ is $\frac{x}{2} + \frac{y}{4} = 1$,which simplifies to $2x + y = 4 \quad (1)$.
The slope of line $L$ is $m = -2$. The slope of the line $L_1$ perpendicular to $L$ is $m_1 = -\frac{1}{m} = \frac{1}{2}$.
Since $L_1$ passes through $(-2, 1)$,its equation is $y - 1 = \frac{1}{2}(x + 2) \implies 2y - 2 = x + 2 \implies x - 2y = -4 \quad (2)$.
Solving equations $(1)$ and $(2)$:
From $(1)$,$y = 4 - 2x$. Substituting into $(2)$:
$x - 2(4 - 2x) = -4$
$x - 8 + 4x = -4$
$5x = 4 \implies x = \frac{4}{5}$.
Then $y = 4 - 2(\frac{4}{5}) = 4 - \frac{8}{5} = \frac{12}{5}$.
The point of intersection is $\left( \frac{4}{5}, \frac{12}{5} \right)$.

(D) Two lines are perpendicular,so the product of their slopes is $-1$.
Slope of $L_1: x + (a - 1)y = 1$ is $m_1 = -\frac{1}{a - 1}$.
Slope of $L_2: 2x + a^2y = 1$ is $m_2 = -\frac{2}{a^2}$.
Since $m_1 m_2 = -1$,we have $\left(-\frac{1}{a - 1}\right) \left(-\frac{2}{a^2}\right) = -1$.
$\Rightarrow \frac{2}{a^2(a - 1)} = -1$ $\Rightarrow a^3 - a^2 + 2 = 0$.
Factoring the cubic equation: $(a + 1)(a^2 - 2a + 2) = 0$.
Since $a^2 - 2a + 2 = (a - 1)^2 + 1 > 0$,the only real solution is $a = -1$.
Substituting $a = -1$ into the line equations:
$L_1: x + (-1 - 1)y = 1 \Rightarrow x - 2y = 1$.
$L_2: 2x + (-1)^2y = 1 \Rightarrow 2x + y = 1$.
Solving the system:
$x - 2y = 1$ $(1)$
$2x + y = 1$ $(2)$
Multiplying $(2)$ by $2$: $4x + 2y = 2$.
Adding $(1)$ and $(2)$: $5x = 3 \Rightarrow x = \frac{3}{5}$.
Substituting $x = \frac{3}{5}$ into $(2)$: $2(\frac{3}{5}) + y = 1 \Rightarrow y = 1 - \frac{6}{5} = -\frac{1}{5}$.
The point of intersection is $P(\frac{3}{5}, -\frac{1}{5})$.
The distance from the origin $(0, 0)$ is $OP = \sqrt{(\frac{3}{5})^2 + (-\frac{1}{5})^2} = \sqrt{\frac{9}{25} + \frac{1}{25}} = \sqrt{\frac{10}{25}} = \sqrt{\frac{2}{5}}$.

(D) Given lines are:
$5x - y + 4 = 0$ ..... $(1)$
$3x + 4y - 4 = 0$ ..... $(2)$
Let the required line intersect the lines $(1)$ and $(2)$ at points $P(\alpha_{1}, \beta_{1})$ and $Q(\alpha_{2}, \beta_{2})$ respectively.
Since $P$ lies on $(1)$,we have $\beta_{1} = 5\alpha_{1} + 4$.
Since $Q$ lies on $(2)$,we have $\beta_{2} = \frac{4 - 3\alpha_{2}}{4}$.
The midpoint of $PQ$ is $(1, 5)$,so $\frac{\alpha_{1} + \alpha_{2}}{2} = 1$ and $\frac{\beta_{1} + \beta_{2}}{2} = 5$.
From the first,$\alpha_{2} = 2 - \alpha_{1}$.
Substituting into the second: $\frac{(5\alpha_{1} + 4) + \frac{4 - 3(2 - \alpha_{1})}{4}}{2} = 5$.
$20\alpha_{1} + 16 + 4 - 6 + 3\alpha_{1} = 40 \implies 23\alpha_{1} = 26 \implies \alpha_{1} = \frac{26}{23}$.
Then $\beta_{1} = 5(\frac{26}{23}) + 4 = \frac{130 + 92}{23} = \frac{222}{23}$.
The line passes through $(1, 5)$ and $(\frac{26}{23}, \frac{222}{23})$.
The slope $m = \frac{\frac{222}{23} - 5}{\frac{26}{23} - 1} = \frac{222 - 115}{26 - 23} = \frac{107}{3}$.
The equation is $y - 5 = \frac{107}{3}(x - 1) \implies 3y - 15 = 107x - 107$.
Thus,the equation is $107x - 3y - 92 = 0$.

(B) Let $P \equiv (x_1, mx_1)$ and $Q \equiv (x_2, mx_2)$.
The area of $\Delta ABC$ is given by:
$A_1 = \frac{1}{2} |3(0 - 1) + 2(1 - 4) + (-1)(4 - 0)| = \frac{1}{2} |-3 - 6 - 4| = \frac{13}{2}$.
The area of $\Delta PQC$ with vertices $(x_1, mx_1)$,$(x_2, mx_2)$,and $(2, 0)$ is:
$A_2 = \frac{1}{2} |x_1(mx_2 - 0) + x_2(0 - mx_1) + 2(mx_1 - mx_2)| = \frac{1}{2} |mx_1x_2 - mx_1x_2 + 2m(x_1 - x_2)| = m|x_1 - x_2|$.
Given $A_1 = 3A_2$,we have $\frac{13}{2} = 3m|x_1 - x_2|$,so $|x_1 - x_2| = \frac{13}{6m}$.
The equation of line $AC$ is $y - 0 = \frac{1 - 0}{-1 - 2}(x - 2)$ $\Rightarrow y = -\frac{1}{3}(x - 2)$ $\Rightarrow x + 3y = 2$.
Intersection $P$ with $y = mx$: $x + 3(mx) = 2 \Rightarrow x_1 = \frac{2}{1 + 3m}$.
The equation of line $BC$ is $y - 0 = \frac{4 - 0}{3 - 2}(x - 2)$ $\Rightarrow y = 4(x - 2)$ $\Rightarrow y = 4x - 8$.
Intersection $Q$ with $y = mx$: $mx = 4x - 8$ $\Rightarrow x(4 - m) = 8$ $\Rightarrow x_2 = \frac{8}{4 - m}$.
$|x_1 - x_2| = |\frac{2}{1 + 3m} - \frac{8}{4 - m}| = |\frac{8 - 2m - 8 - 24m}{(1 + 3m)(4 - m)}| = \frac{26m}{(3m + 1)(4 - m)}$.
Equating to $\frac{13}{6m}$:
$\frac{26m}{(3m + 1)(4 - m)} = \frac{13}{6m} \Rightarrow 12m^2 = (3m + 1)(4 - m) = -3m^2 + 11m + 4$.
$15m^2 - 11m - 4 = 0 \Rightarrow (15m + 4)(m - 1) = 0$.
Since $m > 0$,we get $m = 1$.

(A) The first line is $x \operatorname{cosec} \alpha - y \sec \alpha = k \cot 2 \alpha$,which can be written as $\frac{x}{\sin \alpha} - \frac{y}{\cos \alpha} = \frac{k \cos 2 \alpha}{\sin 2 \alpha}$.
Multiplying by $\sin \alpha \cos \alpha$,we get $x \cos \alpha - y \sin \alpha = \frac{k \cos 2 \alpha}{\sin 2 \alpha} \cdot \sin \alpha \cos \alpha = \frac{k \cos 2 \alpha}{2 \sin \alpha \cos \alpha} \cdot \sin \alpha \cos \alpha = \frac{k}{2} \cos 2 \alpha$.
The perpendicular distance $p$ from the origin $(0, 0)$ to the line $x \cos \alpha - y \sin \alpha - \frac{k}{2} \cos 2 \alpha = 0$ is $p = \left| \frac{-\frac{k}{2} \cos 2 \alpha}{\sqrt{\cos^{2} \alpha + \sin^{2} \alpha}} \right| = \left| \frac{k}{2} \cos 2 \alpha \right|$.
Thus,$2p = |k \cos 2 \alpha|$,which implies $4p^{2} = k^{2} \cos^{2} 2 \alpha$ $(i)$.
The second line is $x \sin \alpha + y \cos \alpha = k \sin 2 \alpha$. The perpendicular distance $q$ from the origin is $q = \left| \frac{-k \sin 2 \alpha}{\sqrt{\sin^{2} \alpha + \cos^{2} \alpha}} \right| = |k \sin 2 \alpha|$.
Thus,$q^{2} = k^{2} \sin^{2} 2 \alpha$ $(ii)$.
Adding $(i)$ and $(ii)$,we get $4p^{2} + q^{2} = k^{2} (\cos^{2} 2 \alpha + \sin^{2} 2 \alpha) = k^{2}$.