Distance between two lines, Perpendicular distance of the line from a point, Position of point w.r.t. line Questions in English

(C) Let $P(\alpha, \beta)$ be the point on the line $2x + 3y + 7 = 0$.
Then $2\alpha + 3\beta + 7 = 0$,so $\beta = \frac{-7-2\alpha}{3}$.
The point $P$ is $\left(\alpha, \frac{-7-2\alpha}{3}\right)$.
The distance between $P(\alpha, \beta)$ and $A(1, -3)$ is $3$.
Using the distance formula: $(\alpha - 1)^2 + \left(\frac{-7-2\alpha}{3} + 3\right)^2 = 3^2$.
$(\alpha - 1)^2 + \left(\frac{-7-2\alpha+9}{3}\right)^2 = 9$.
$(\alpha - 1)^2 + \left(\frac{2-2\alpha}{3}\right)^2 = 9$.
$(\alpha - 1)^2 + \frac{4(1-\alpha)^2}{9} = 9$.
$(\alpha - 1)^2 \left(1 + \frac{4}{9}\right) = 9$.
$(\alpha - 1)^2 \left(\frac{13}{9}\right) = 9$.
$(\alpha - 1)^2 = \frac{81}{13}$.
$\alpha - 1 = \pm \frac{9}{\sqrt{13}}$.
$\alpha = 1 \pm \frac{9}{\sqrt{13}} = \frac{\sqrt{13} \pm 9}{\sqrt{13}}$.
For $\alpha = \frac{\sqrt{13}-9}{\sqrt{13}}$,$\beta = \frac{-7 - 2(\frac{\sqrt{13}-9}{\sqrt{13}})}{3} = \frac{-7\sqrt{13} - 2\sqrt{13} + 18}{3\sqrt{13}} = \frac{-9\sqrt{13} + 18}{3\sqrt{13}} = \frac{-3\sqrt{13} + 6}{\sqrt{13}}$.
Thus,the point is $\left(\frac{\sqrt{13}-9}{\sqrt{13}}, \frac{-3\sqrt{13}+6}{\sqrt{13}}\right)$.

(B) The length of the perpendicular from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.
For the first line $x \sec \theta + y \operatorname{cosec} \theta - k = 0$,we have $p = \frac{|-k|}{\sqrt{\sec^2 \theta + \operatorname{cosec}^2 \theta}} = \frac{|k|}{\sqrt{\frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}}} = \frac{|k| \sin \theta \cos \theta}{\sqrt{\sin^2 \theta + \cos^2 \theta}} = |k| \sin \theta \cos \theta = \frac{|k|}{2} \sin 2\theta$.
Thus,$p^2 = \frac{k^2}{4} \sin^2 2\theta$.
For the second line $x \cos \theta - y \sin \theta - k \cos 2\theta = 0$,we have $q = \frac{|-k \cos 2\theta|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = |k \cos 2\theta|$.
Thus,$q^2 = k^2 \cos^2 2\theta$.
Now,consider $4p^2 + q^2 = 4 \left( \frac{k^2}{4} \sin^2 2\theta \right) + k^2 \cos^2 2\theta = k^2 \sin^2 2\theta + k^2 \cos^2 2\theta = k^2 (\sin^2 2\theta + \cos^2 2\theta) = k^2$.
Therefore,$4p^2 + q^2 = k^2$.

(B) The length of the perpendicular $d$ from a point $(x_1, y_1)$ to the line $ax + by + c = 0$ is given by the formula: $d = \left| \frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}} \right|$.
Given point $(x_1, y_1) = (1, -2)$ and the line $12x + 5y + 63 = 0$.
Substituting the values into the formula:
$d = \left| \frac{12(1) + 5(-2) + 63}{\sqrt{12^2 + 5^2}} \right|$
$d = \left| \frac{12 - 10 + 63}{\sqrt{144 + 25}} \right|$
$d = \left| \frac{65}{\sqrt{169}} \right|$
$d = \left| \frac{65}{13} \right| = 5 \text{ units.}$

(A) Let the line be $L(x, y) = 3x - 4y - 8 = 0$.
For point $Q(1, 1)$,$L(1, 1) = 3(1) - 4(1) - 8 = 3 - 4 - 8 = -9$.
Since $L(1, 1) < 0$,for $P(\alpha, \beta)$ to lie on the opposite side,we must have $L(\alpha, \beta) > 0$.
Substituting $P(\alpha, \beta)$ into the line equation: $3\alpha - 4\beta - 8 > 0$.
Since $P$ lies on $3x + y = 0$,we have $\beta = -3\alpha$.
Substituting $\beta = -3\alpha$ into the inequality: $3\alpha - 4(-3\alpha) - 8 > 0$.
$3\alpha + 12\alpha - 8 > 0$ $\Rightarrow 15\alpha > 8$ $\Rightarrow \alpha > \frac{8}{15}$.
Now,using $\alpha = -\frac{\beta}{3}$ in the inequality $3\alpha - 4\beta - 8 > 0$:
$3(-\frac{\beta}{3}) - 4\beta - 8 > 0$ $\Rightarrow -\beta - 4\beta - 8 > 0$ $\Rightarrow -5\beta > 8$ $\Rightarrow \beta < -\frac{8}{5}$.
Thus,$\alpha > \frac{8}{15}$ and $\beta < -\frac{8}{5}$.

(C) Let the point be $(x, y)$ on the line $x+y+3=0$. So,$y = -x-3$. The point is $(x, -x-3)$.
The distance of this point from the line $x+2y+2=0$ is given by $\frac{|x+2(-x-3)+2|}{\sqrt{1^2+2^2}} = \sqrt{5}$.
$\frac{|x-2x-6+2|}{\sqrt{5}} = \sqrt{5} \implies |-x-4| = 5$.
$|x+4| = 5 \implies x+4 = 5$ or $x+4 = -5$.
So,$x_1 = 1$ and $x_2 = -9$.
The value of $|x_1-x_2| = |1 - (-9)| = |10| = 10$.

(A) The given equations of the lines are:
$x \sec \theta - y \operatorname{cosec} \theta = a$ $(i)$
$x \cos \theta + y \sin \theta = a \cos 2 \theta$ $(ii)$
The perpendicular distance $d$ from the origin $(0, 0)$ to a line $Ax + By + C = 0$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.
For line $(i)$:
$p = \frac{|-a|}{\sqrt{\sec^2 \theta + \operatorname{cosec}^2 \theta}} = \frac{a}{\sqrt{\frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}}} = \frac{a}{\sqrt{\frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}}} = a \sin \theta \cos \theta = \frac{a}{2} \sin 2 \theta$.
Thus,$2p = a \sin 2 \theta$.
For line $(ii)$:
$q = \frac{|-a \cos 2 \theta|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = \frac{a \cos 2 \theta}{1} = a \cos 2 \theta$.
Now,calculating $4p^2 + q^2$:
$4p^2 + q^2 = (a \sin 2 \theta)^2 + (a \cos 2 \theta)^2$
$= a^2 (\sin^2 2 \theta + \cos^2 2 \theta)$
$= a^2 (1) = a^2$.

(D) Given points $(1, \pi)$,$(1, 0^{\circ})$ and $(1, \frac{\pi}{2})$ are in polar form.
Converting these to Cartesian coordinates $(x, y) = (r \cos \theta, r \sin \theta)$:
$(1, \pi) \rightarrow (1 \cdot \cos \pi, 1 \cdot \sin \pi) = (-1, 0)$
$(1, 0^{\circ}) \rightarrow (1 \cdot \cos 0^{\circ}, 1 \cdot \sin 0^{\circ}) = (1, 0)$
$(1, \frac{\pi}{2}) \rightarrow (1 \cdot \cos \frac{\pi}{2}, 1 \cdot \sin \frac{\pi}{2}) = (0, 1)$
Now,the equation of the line passing through $(1, 0)$ and $(0, 1)$ is given by the intercept form $\frac{x}{a} + \frac{y}{b} = 1$ or slope-point form:
$y - 0 = \frac{1 - 0}{0 - 1}(x - 1)$ $\Rightarrow y = -x + 1$ $\Rightarrow x + y - 1 = 0$.
The perpendicular distance $d$ from the point $(-1, 0)$ to the line $x + y - 1 = 0$ is calculated using the formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$:
$d = \frac{|1(-1) + 1(0) - 1|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(D) Let $f(x, y) = 5x - 6y + k$.
Since $(2, 3)$ and $(-4, -4/3)$ lie on opposite sides,$f(2, 3) \cdot f(-4, -4/3) < 0$.
$f(2, 3) = 5(2) - 6(3) + k = 10 - 18 + k = k - 8$.
$f(-4, -4/3) = 5(-4) - 6(-4/3) + k = -20 + 8 + k = k - 12$.
So,$(k - 8)(k - 12) < 0$,which implies $8 < k < 12$.
Since $k$ is an integer,$k \in \{9, 10, 11\}$.
Now,$(1, 2)$ and $(4, 5)$ lie on the same side,so $f(1, 2) \cdot f(4, 5) > 0$.
$f(1, 2) = 5(1) - 6(2) + k = k - 7$.
$f(4, 5) = 5(4) - 6(5) + k = k - 10$.
So,$(k - 7)(k - 10) > 0$.
For $k = 9$: $(9 - 7)(9 - 10) = 2(-1) = -2 < 0$ (False).
For $k = 10$: $f(4, 5) = 0$ (Points lie on the line,not same side).
For $k = 11$: $(11 - 7)(11 - 10) = 4(1) = 4 > 0$ (True).
Thus,$k = 11$.
The line is $5x - 6y + 11 = 0$.
The perpendicular distance from origin $(0, 0)$ is $d = \frac{|5(0) - 6(0) + 11|}{\sqrt{5^2 + (-6)^2}} = \frac{11}{\sqrt{25 + 36}} = \frac{11}{\sqrt{61}}$.

(D) Given the system of equations $x+2y=3$ and $3x+6y=a-2$.
For a system of linear equations $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$ to have no solution,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
Comparing the given equations with the standard form,we have $a_1=1, b_1=2, c_1=3$ and $a_2=3, b_2=6, c_2=a-2$.
Substituting these values into the condition:
$\frac{1}{3} = \frac{2}{6} \neq \frac{3}{a-2}$.
Since $\frac{1}{3} = \frac{2}{6}$ is always true,we focus on the inequality $\frac{1}{3} \neq \frac{3}{a-2}$.
Cross-multiplying gives $a-2 \neq 3 \times 3$,which simplifies to $a-2 \neq 9$.
Therefore,$a \neq 11$.

(B) The distance between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.
Let the lines be $L_1: 3x + 4y + 5 = 0$ and $L_2: 3x + 4y - 5 = 0$.
The line $L: 3x + 4y + \lambda = 0$ divides the distance between $L_1$ and $L_2$ in the ratio $3:7$.
The distance between $L$ and $L_1$ is $d_1 = \frac{|\lambda - 5|}{\sqrt{3^2 + 4^2}} = \frac{|\lambda - 5|}{5}$.
The distance between $L$ and $L_2$ is $d_2 = \frac{|\lambda - (-5)|}{\sqrt{3^2 + 4^2}} = \frac{|\lambda + 5|}{5}$.
Given $\frac{d_1}{d_2} = \frac{3}{7}$,we have $\frac{|\lambda - 5|}{|\lambda + 5|} = \frac{3}{7}$.
Assuming $-5 < \lambda < 5$,we get $\frac{5 - \lambda}{\lambda + 5} = \frac{3}{7}$.
$7(5 - \lambda) = 3(\lambda + 5) \Rightarrow 35 - 7\lambda = 3\lambda + 15$.
$10\lambda = 20 \Rightarrow \lambda = 2$.

(C) Since the vertex $A(2,3)$ does not lie on the line $12x+5y-65=0$,the perpendicular distance from $A$ to the base $BC$ is the altitude $AD$ of the equilateral triangle $\triangle ABC$.
The length of the altitude $AD$ is given by the formula for the perpendicular distance from a point $(x_1, y_1)$ to a line $ax+by+c=0$:
$AD = \left|\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}}\right|$
$AD = \left|\frac{12(2)+5(3)-65}{\sqrt{12^2+5^2}}\right| = \left|\frac{24+15-65}{\sqrt{144+25}}\right| = \left|\frac{39-65}{\sqrt{169}}\right| = \left|\frac{-26}{13}\right| = 2$
In an equilateral triangle with side length $s$,the altitude $h$ is given by $h = \frac{\sqrt{3}}{2}s$.
Therefore,$s = \frac{2}{\sqrt{3}} \times h$.
Substituting $h = AD = 2$:
$s = \frac{2}{\sqrt{3}} \times 2 = \frac{4}{\sqrt{3}}$.

(A) The slope of line $L$ is $-\cot \alpha$.
The slope of line $x + y + 1 = 0$ is $-1$.
Since the lines are perpendicular,the product of their slopes is $-1$:
$(-\cot \alpha)(-1) = -1$ $\Rightarrow \cot \alpha = -1$ $\Rightarrow \tan \alpha = -1$.
Since $\alpha$ lies in the fourth quadrant,$\alpha = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$.
The perpendicular distance from $(\sqrt{2}, \sqrt{2})$ to $x \cos \alpha + y \sin \alpha - p = 0$ is given by:
$\left| \sqrt{2} \cos \alpha + \sqrt{2} \sin \alpha - p \right| = 5$.
Substituting $\alpha = \frac{7\pi}{4}$:
$\cos \frac{7\pi}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{7\pi}{4} = -\frac{1}{\sqrt{2}}$.
$\left| \sqrt{2} \left( \frac{1}{\sqrt{2}} \right) + \sqrt{2} \left( -\frac{1}{\sqrt{2}} \right) - p \right| = 5$.
$|1 - 1 - p| = 5 \Rightarrow |-p| = 5$.
Since $p$ is positive,$p = 5$.

(B) Since $P(a, 2)$ and $Q(1, b)$ lie on either side of the line $2x - 3y + 1 = 0$,the product of the values obtained by substituting the points into the line equation must be negative:
$(2a - 3(2) + 1)(2(1) - 3b + 1) < 0$
$(2a - 5)(3 - 3b) < 0$
$(2a - 5)(1 - b) < 0$
Since $P(a, 2)$ is the intersection of $4x + 3y + k = 0$ and $3x + 4y + k = 0$:
$4a + 3(2) + k = 0 \Rightarrow 4a + 6 + k = 0$
$3a + 4(2) + k = 0 \Rightarrow 3a + 8 + k = 0$
Subtracting the two equations: $(4a + 6 + k) - (3a + 8 + k) = 0$ $\Rightarrow a - 2 = 0$ $\Rightarrow a = 2$.
Substituting $a = 2$ into the inequality:
$(2(2) - 5)(1 - b) < 0$
$(4 - 5)(1 - b) < 0$
$-(1 - b) < 0$
$b - 1 < 0 \Rightarrow b < 1$.
Thus,$b \in (-\infty, 1)$.

(D) Let the $X$-intercept be $a$ and the $Y$-intercept be $2a$.
The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{2a} = 1$.
Multiplying by $2a$,we get $2x + y = 2a$,or $2x + y - 2a = 0$.
The perpendicular distance from the origin $(0, 0)$ to the line is given as $1$.
The formula for the distance from $(x_1, y_1)$ to $Ax + By + C = 0$ is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Substituting the values: $1 = \frac{|2(0) + 1(0) - 2a|}{\sqrt{2^2 + 1^2}}$.
$1 = \frac{|-2a|}{\sqrt{5}}$.
$|2a| = \sqrt{5}$,which means $2a = \pm \sqrt{5}$.
Substituting $2a$ back into the equation $2x + y = 2a$,we get $2x + y = \pm \sqrt{5}$.

(B) The given lines are $L_1: 2x - 3y + 4 = 0$ and $L_2: 6x - 9y + 7 = 0$.
Note that $L_2$ can be written as $3(2x - 3y) + 7 = 0$,which shows $L_1$ and $L_2$ are parallel.
Let the line $L$ be perpendicular to $L_1$ and $L_2$. The equation of any line perpendicular to $2x - 3y + k = 0$ is $3x + 2y + C = 0$.
Since $P(1, 1)$ lies on $L$,we have $3(1) + 2(1) + C = 0$,which gives $C = -5$. So,$L: 3x + 2y - 5 = 0$.
Point $A$ is the intersection of $L_1$ and $L$:
$2x - 3y = -4$ and $3x + 2y = 5$. Solving this,we get $A = (7/13, 22/13)$.
Point $B$ is the intersection of $L_2$ and $L$:
$6x - 9y = -7$ and $3x + 2y = 5$. Solving this,we get $B = (31/39, 9/39) = (31/39, 3/13)$.
Let $P$ divide $AB$ in ratio $k:1$. Using section formula for $x$-coordinate:
$1 = \frac{k(31/39) + 1(7/13)}{k+1} \implies k+1 = \frac{31k + 21}{39} \implies 39k + 39 = 31k + 21 \implies 8k = -18 \implies k = -18/8 = -9/4$.
The negative sign indicates external division in the ratio $9:4$.

(A) The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.
Given lines are $3x - 2y + 5 = 0$ and $3x - 2y + C = 0$.
Here,$a = 3$,$b = -2$,$c_1 = 5$,$c_2 = C$,and $d = 5$.
Substituting these values into the formula:
$5 = \frac{|C - 5|}{\sqrt{3^2 + (-2)^2}}$
$5 = \frac{|C - 5|}{\sqrt{9 + 4}}$
$5 = \frac{|C - 5|}{\sqrt{13}}$
$|C - 5| = 5\sqrt{13}$
$C - 5 = \pm 5\sqrt{13}$
$C = 5 \pm 5\sqrt{13} = 5(1 \pm \sqrt{13})$.

(D) The vertices are $O(0,0), B(-3,-1), C(-1,-3)$.
The equation of line $BC$ passing through $(-3,-1)$ and $(-1,-3)$ is $y - (-1) = \frac{-3 - (-1)}{-1 - (-3)}(x - (-3))$,which simplifies to $y + 1 = -1(x + 3)$,or $x + y + 4 = 0$.
The altitude from $O(0,0)$ to $BC$ is the perpendicular distance from $O$ to $x + y + 4 = 0$,which is $h = \frac{|0 + 0 + 4|}{\sqrt{1^2 + 1^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.
The line $DE$ is $2x + 2y + \sqrt{2} = 0$,which can be written as $x + y + \frac{\sqrt{2}}{2} = 0$,or $x + y + \frac{1}{\sqrt{2}} = 0$.
The distance of $DE$ from $O(0,0)$ is $d = \frac{|0 + 0 + \frac{1}{\sqrt{2}}|}{\sqrt{1^2 + 1^2}} = \frac{1/\sqrt{2}}{\sqrt{2}} = \frac{1}{2}$.
The line $DE$ divides the altitude $h$ into two segments of lengths $d$ and $h - d$.
The ratio is $d : (h - d) = \frac{1}{2} : (2\sqrt{2} - \frac{1}{2}) = 1 : (4\sqrt{2} - 1)$.

(C) The distance $d$ between the two parallel lines $\sqrt{3}x + y - 6 = 0$ and $\sqrt{3}x + y + 9 = 0$ is given by:
$d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = \frac{|9 - (-6)|}{\sqrt{(\sqrt{3})^2 + 1^2}} = \frac{15}{\sqrt{3 + 1}} = \frac{15}{2}$.
For an equilateral triangle with height $d$,the side length $s$ satisfies $d = s \sin 60^{\circ} = s \frac{\sqrt{3}}{2}$,so $s = \frac{2d}{\sqrt{3}}$.
The area of the equilateral triangle is $\frac{\sqrt{3}}{4} s^2 = \frac{\sqrt{3}}{4} \left( \frac{2d}{\sqrt{3}} \right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{4d^2}{3} = \frac{d^2}{\sqrt{3}}$.
Substituting $d = \frac{15}{2}$,the area is $\left( \frac{15}{2} \right)^2 \cdot \frac{1}{\sqrt{3}} = \frac{225}{4 \sqrt{3}}$ sq. units.

(A) The given lines are $L_1: 2x+2y+3=0$ and $L_2: 3x+3y+2=0$.
Dividing $L_1$ by $2$,we get $x+y+\frac{3}{2}=0$.
Dividing $L_2$ by $3$,we get $x+y+\frac{2}{3}=0$.
Since the slopes of both lines are $-1$,they are parallel.
The line $L_3: x-y+1=0$ has a slope of $1$.
Since the product of the slopes of $L_1$ (or $L_2$) and $L_3$ is $(-1) \times (1) = -1$,the line $L_3$ is perpendicular to both parallel lines.
The distance $AB$ between the points of intersection is the perpendicular distance between the two parallel lines $x+y+\frac{3}{2}=0$ and $x+y+\frac{2}{3}=0$.
The distance $d$ between two parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$ is given by $d = \frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$.
Here,$a=1, b=1, c_1=\frac{3}{2}, c_2=\frac{2}{3}$.
$AB = \frac{|\frac{3}{2}-\frac{2}{3}|}{\sqrt{1^2+1^2}} = \frac{|\frac{9-4}{6}|}{\sqrt{2}} = \frac{5}{6\sqrt{2}}$.

(B) Let $L_1(x, y) = x+2y-5$ and $L_2(x, y) = 3x-y+1$. The origin $O(0,0)$ gives $L_1(0,0) = -5 < 0$ and $L_2(0,0) = 1 > 0$.
For the point $P(K^2, K+1)$ to lie in the same region as the origin,it must satisfy $L_1(K^2, K+1) < 0$ and $L_2(K^2, K+1) > 0$.
$1$) $L_1(K^2, K+1) = K^2 + 2(K+1) - 5 = K^2 + 2K - 3 < 0$.
$(K+3)(K-1) < 0 \implies K \in (-3, 1)$.
$2$) $L_2(K^2, K+1) = 3K^2 - (K+1) + 1 = 3K^2 - K > 0$.
$K(3K-1) > 0 \implies K \in (-\infty, 0) \cup (1/3, \infty)$.
Taking the intersection of both conditions: $K \in (-3, 0) \cup (1/3, 1)$.
Since $K$ is an integer,the possible values for $K$ are $\{-2, -1\}$.
Thus,there are $2$ such points $P$.

(D) The perpendicular distance $d$ from a point $(x_1, y_1)$ to the line $ax + by + c = 0$ is given by the formula: $d = \left| \frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}} \right|$.
Given the point $(1, 1)$,the line $3x + 4y + c = 0$,and the distance $d = 7$:
$7 = \left| \frac{3(1) + 4(1) + c}{\sqrt{3^2 + 4^2}} \right|$
$7 = \left| \frac{7 + c}{\sqrt{9 + 16}} \right|$
$7 = \left| \frac{7 + c}{5} \right|$
$|7 + c| = 35$
This implies $7 + c = 35$ or $7 + c = -35$.
If $7 + c = 35$,then $c = 28$.
If $7 + c = -35$,then $c = -42$.
Thus,the possible values of $c$ are $28$ and $-42$.

(A) Since the points $A(k, 2k), B(3k, 3k)$,and $C(3, 1)$ are collinear,the slope of $AB$ must be equal to the slope of $BC$.
Slope of $AB = \frac{3k - 2k}{3k - k} = \frac{k}{2k} = \frac{1}{2}$.
Slope of $BC = \frac{1 - 3k}{3 - 3k}$.
Equating the slopes: $\frac{1}{2} = \frac{1 - 3k}{3 - 3k}$.
Cross-multiplying: $3 - 3k = 2(1 - 3k)$ $\Rightarrow 3 - 3k = 2 - 6k$ $\Rightarrow 3k = -1$ $\Rightarrow k = -\frac{1}{3}$.
The equation of the line passing through $B(-1, -1)$ and $C(3, 1)$ is $y - 1 = \frac{1 - (-1)}{3 - (-1)}(x - 3)$.
$y - 1 = \frac{2}{4}(x - 3)$ $\Rightarrow y - 1 = \frac{1}{2}(x - 3)$ $\Rightarrow 2y - 2 = x - 3$ $\Rightarrow x - 2y - 1 = 0$.
The distance from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.
$d = \frac{|-1|}{\sqrt{1^2 + (-2)^2}} = \frac{1}{\sqrt{1 + 4}} = \frac{1}{\sqrt{5}}$.

(A) The distance $d$ between parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
For $3x - 2y + 5 = 0$ and $3x - 2y + 5 + 2\sqrt{13} = 0$,$d = \frac{|5 - (5 + 2\sqrt{13})|}{\sqrt{3^2 + (-2)^2}} = \frac{2\sqrt{13}}{\sqrt{13}} = 2$.
Given $L_1: 3x - 2y + k_1 = 0$ is at distance $\frac{4d}{\sqrt{13}} = \frac{8}{\sqrt{13}}$ from $3x - 2y + 5 = 0$,we have $\frac{|k_1 - 5|}{\sqrt{13}} = \frac{8}{\sqrt{13}} \Rightarrow |k_1 - 5| = 8$. Since $k_1 > 0$,$k_1 - 5 = 8 \Rightarrow k_1 = 13$.
Given $L_2: 3x - 2y + k_2 = 0$ is at distance $\frac{3d}{\sqrt{13}} = \frac{6}{\sqrt{13}}$ from $3x - 2y + 5 = 0$,we have $\frac{|k_2 - 5|}{\sqrt{13}} = \frac{6}{\sqrt{13}} \Rightarrow |k_2 - 5| = 6$. Since $k_2 > 0$,$k_2 - 5 = 6 \Rightarrow k_2 = 11$.
The combined equation is $(3x - 2y + 13)(3x - 2y + 11) = 0$.
Let $u = 3x - 2y$. Then $(u + 13)(u + 11) = u^2 + 24u + 143 = 0$.
Substituting back,we get $(3x - 2y)^2 + 24(3x - 2y) + 143 = 0$.

(B) The given parallel lines are $4x + 2y - 9 = 0$ and $2x + y + 6 = 0$.
We can rewrite the first equation as $2x + y = \frac{9}{2}$.
The second equation is $2x + y = -6$.
Let the line through the origin be $y = mx$. Substituting this into the equations:
For $P$: $2x + mx = \frac{9}{2} \Rightarrow x_P = \frac{9}{2(2+m)}$,$y_P = \frac{9m}{2(2+m)}$.
For $Q$: $2x + mx = -6 \Rightarrow x_Q = \frac{-6}{2+m}$,$y_Q = \frac{-6m}{2+m}$.
The ratio in which $O(0,0)$ divides $PQ$ is given by $\frac{OP}{OQ} = \frac{|x_P|}{|x_Q|} = \frac{9/2(2+m)}{6/(2+m)} = \frac{9}{12} = \frac{3}{4}$.
Thus,the point $O$ divides $PQ$ in the ratio $3: 4$.

(B) Let $(h, k)$ be any point on the line $7x - 9y + 10 = 0$. Then,$7h - 9k + 10 = 0$,which implies $h = \frac{9k - 10}{7}$.
Perpendicular distance $d_{1}$ from $(h, k)$ to $3x + 4y - 5 = 0$ is given by $d_{1} = \frac{|3h + 4k - 5|}{\sqrt{3^{2} + 4^{2}}} = \frac{|3h + 4k - 5|}{5}$.
Perpendicular distance $d_{2}$ from $(h, k)$ to $12x + 5y - 7 = 0$ is given by $d_{2} = \frac{|12h + 5k - 7|}{\sqrt{12^{2} + 5^{2}}} = \frac{|12h + 5k - 7|}{13}$.
Substituting $h = \frac{9k - 10}{7}$ into the expressions:
$3h + 4k - 5 = 3(\frac{9k - 10}{7}) + 4k - 5 = \frac{27k - 30 + 28k - 35}{7} = \frac{55k - 65}{7} = \frac{5(11k - 13)}{7}$.
So,$d_{1} = \frac{|5(11k - 13)|}{5 \times 7} = \frac{|11k - 13|}{7}$.
$12h + 5k - 7 = 12(\frac{9k - 10}{7}) + 5k - 7 = \frac{108k - 120 + 35k - 49}{7} = \frac{143k - 169}{7} = \frac{13(11k - 13)}{7}$.
So,$d_{2} = \frac{|13(11k - 13)|}{13 \times 7} = \frac{|11k - 13|}{7}$.
Since $d_{1} = \frac{|11k - 13|}{7}$ and $d_{2} = \frac{|11k - 13|}{7}$,we have $d_{1} = d_{2}$.

(B, D) Let $(h, k)$ be a point on the line $x+y+1=0.$
Since the point lies on the line,we have $h+k+1=0,$ which implies $h = -k-1.$
The perpendicular distance from $(h, k)$ to the line $3x+4y+2=0$ is given by $\frac{|3h+4k+2|}{\sqrt{3^2+4^2}} = \frac{1}{5}.$
Substituting $h = -k-1$ into the distance formula:
$\frac{|3(-k-1)+4k+2|}{5} = \frac{1}{5}$
$|-3k-3+4k+2| = 1$
$|k-1| = 1$
This gives two cases:
$k-1 = 1 \implies k=2,$ then $h = -2-1 = -3.$ Point is $(-3, 2).$
$k-1 = -1 \implies k=0,$ then $h = 0-1 = -1.$ Point is $(-1, 0).$
Thus,the required points are $(-3, 2)$ and $(-1, 0).$ Both options $B$ and $D$ are correct.

(B) Let the point on the line $x+y=4$ be $P(h, 4-h)$.
The perpendicular distance from $P(h, 4-h)$ to the line $4x+3y-10=0$ is given by $d = \frac{|4h + 3(4-h) - 10|}{\sqrt{4^2 + 3^2}}$.
Given $d=1$,we have $\frac{|4h + 12 - 3h - 10|}{5} = 1$.
$|h + 2| = 5$.
This gives two cases:
Case $1$: $h+2 = 5 \implies h = 3$. The point is $(3, 4-3) = (3, 1)$.
Case $2$: $h+2 = -5 \implies h = -7$. The point is $(-7, 4-(-7)) = (-7, 11)$.
Thus,the coordinates are $(3, 1)$ and $(-7, 11)$.

(A) The given lines are $L_1: x+y-4=0$ and $L_2: 2x+2y-5=0$,which can be rewritten as $x+y-2.5=0$.
Since the slopes of both lines are equal $(-1)$,the lines are parallel.
The perpendicular distance $d$ between two parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$ is given by $d = \frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$.
Here,$d = \frac{|-4 - (-2.5)|}{\sqrt{1^2+1^2}} = \frac{|-1.5|}{\sqrt{2}} = \frac{1.5}{\sqrt{2}} = \frac{3}{2\sqrt{2}} \approx 1.06$.
Since the distance between the two lines is $\approx 1.06$,which is greater than $1$,there is no point on the line $x+y=4$ that is at a distance of $1$ unit from the line $2x+2y=5$.
Thus,the number of such points is $0$.

(D) Let the point $P = (2, -3)$ and the point $Q = (-1, 2)$.
Any line passing through $P$ will have a distance from $Q$ that is at most the distance $PQ$.
The distance $PQ = \sqrt{(2 - (-1))^2 + (-3 - 2)^2} = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}$.
Since $\sqrt{34} \approx 5.83$,and we are looking for a line at a distance of $8$ from $Q$,we note that $8 > \sqrt{34}$.
Because the perpendicular distance from a point to a line cannot exceed the distance between the point and a fixed point on the line,no such line exists.
Thus,the number of such lines is $0$.

(C) Let the line passing through $P(2,3)$ with angle $\theta$ be represented in parametric form as $(x, y) = (2 + r\cos\theta, 3 + r\sin\theta)$,where $r = \sqrt{\frac{2}{3}}$.
Since this point lies on the line $x+y=6$,we substitute the coordinates:
$(2 + r\cos\theta) + (3 + r\sin\theta) = 6$
$r(\cos\theta + \sin\theta) + 5 = 6$
$\sqrt{\frac{2}{3}}(\cos\theta + \sin\theta) = 1$
$\cos\theta + \sin\theta = \sqrt{\frac{3}{2}}$
Squaring both sides:
$(\cos\theta + \sin\theta)^2 = \frac{3}{2}$
$1 + \sin(2\theta) = \frac{3}{2}$
$\sin(2\theta) = \frac{1}{2}$
Thus,$2\theta = \frac{\pi}{6}$ or $2\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.
Therefore,$\theta_1 = \frac{\pi}{12}$ and $\theta_2 = \frac{5\pi}{12}$.
Summing the angles: $\theta_1 + \theta_2 = \frac{\pi}{12} + \frac{5\pi}{12} = \frac{6\pi}{12} = \frac{\pi}{2}$.

(D) The rectangle is bounded by $x=0, x=3, y=0, y=4$. The center of the rectangle is $(\frac{3}{2}, 2)$.
Any line that divides the rectangle into two equal areas must pass through its center $(\frac{3}{2}, 2)$.
The line $L$ is perpendicular to $3x+y+6=0$. The slope of $3x+y+6=0$ is $-3$,so the slope of line $L$ is $\frac{1}{3}$.
The equation of line $L$ passing through $(\frac{3}{2}, 2)$ with slope $\frac{1}{3}$ is:
$y - 2 = \frac{1}{3}(x - \frac{3}{2})$
$3y - 6 = x - \frac{3}{2}$
$6y - 12 = 2x - 3$
$2x - 6y + 9 = 0$.
The distance of the point $(\frac{1}{2}, -5)$ from the line $2x - 6y + 9 = 0$ is:
$d = \frac{|2(\frac{1}{2}) - 6(-5) + 9|}{\sqrt{2^2 + (-6)^2}}$
$d = \frac{|1 + 30 + 9|}{\sqrt{4 + 36}}$
$d = \frac{40}{\sqrt{40}} = \sqrt{40} = 2\sqrt{10}$.