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

(B) To determine the position of the point $(x_1, y_1) = (3, -4)$ with respect to the line $ax + by + c = 0$,where $a = 3, b = -4, c = 5$,we evaluate the expression $ax_1 + by_1 + c$.
Substituting the values: $3(3) - 4(-4) + 5 = 9 + 16 + 5 = 30$.
Since the value $30$ is positive and the constant term $c = 5$ is also positive,the point $(3, -4)$ lies on the same side of the line as the origin.

(B) The slope of the line $x - y + 1 = 0$ is $m = 1$. Thus,it makes an angle of $\theta = 45^{\circ}$ with the $x$-axis.
The equation of a line passing through $(2, 3)$ with an angle of $45^{\circ}$ is given by $\frac{x - 2}{\cos 45^{\circ}} = \frac{y - 3}{\sin 45^{\circ}} = r$.
Any point on this line can be represented as $(2 + r \cos 45^{\circ}, 3 + r \sin 45^{\circ}) = (2 + \frac{r}{\sqrt{2}}, 3 + \frac{r}{\sqrt{2}})$.
Since this point lies on the line $2x - 3y + 9 = 0$,we substitute the coordinates:
$2(2 + \frac{r}{\sqrt{2}}) - 3(3 + \frac{r}{\sqrt{2}}) + 9 = 0$.
$4 + \frac{2r}{\sqrt{2}} - 9 - \frac{3r}{\sqrt{2}} + 9 = 0$.
$4 - \frac{r}{\sqrt{2}} = 0 \implies \frac{r}{\sqrt{2}} = 4 \implies r = 4\sqrt{2}$.
Thus,the required distance is $4\sqrt{2}$.

(D) The given lines $x = 3$ and $x = 8$ are parallel to the $Y$-axis.
Any line equidistant from these two lines must also be parallel to the $Y$-axis and will be of the form $x = k$.
Since the line $x = k$ is equidistant from $x = 3$ and $x = 8$,the distance between $x = k$ and $x = 3$ must equal the distance between $x = 8$ and $x = k$.
Thus,$k - 3 = 8 - k$.
$2k = 11$
$k = \frac{11}{2}$
Therefore,the equation of the required line is $x = \frac{11}{2}$,which can be written as $2x - 11 = 0$.

(C) Let the required point be $P(x_1, y_1)$. Since $P$ lies on $x + y = 4$,we have $y_1 = 4 - x_1$.
The perpendicular distance from $P(x_1, y_1)$ to the line $4x + 3y - 10 = 0$ is given as $1$.
Using the formula $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$,we get:
$\frac{|4x_1 + 3(4 - x_1) - 10|}{\sqrt{4^2 + 3^2}} = 1$
$\frac{|4x_1 + 12 - 3x_1 - 10|}{5} = 1$
$|x_1 + 2| = 5$
This gives two cases:
Case $1$: $x_1 + 2 = 5 \implies x_1 = 3$. Then $y_1 = 4 - 3 = 1$. Point is $(3, 1)$.
Case $2$: $x_1 + 2 = -5 \implies x_1 = -7$. Then $y_1 = 4 - (-7) = 11$. Point is $(-7, 11)$.
Thus,the required points are $(3, 1)$ and $(-7, 11)$.

(C) Let the equation of the line be $f(x, y) = x + 2y - 3 = 0$.
Substitute the coordinates of points $A, B,$ and $C$ into $f(x, y)$:
For $A(-1, 3)$: $f(-1, 3) = -1 + 2(3) - 3 = -1 + 6 - 3 = 2 > 0$.
For $B(2, -3)$: $f(2, -3) = 2 + 2(-3) - 3 = 2 - 6 - 3 = -7 < 0$.
For $C(4, 9)$: $f(4, 9) = 4 + 2(9) - 3 = 4 + 18 - 3 = 19 > 0$.
Since $f(A)$ and $f(C)$ have the same sign (positive) and $f(B)$ has a different sign (negative),points $A$ and $C$ lie on one side of the line,while point $B$ lies on the other side.

(A) The length of the perpendicular from $(0, 0)$ to the line $x \sec \theta + y \csc \theta = a$ is given by:
$P_1 = \frac{|-a|}{\sqrt{\sec^2 \theta + \csc^2 \theta}} = \frac{a}{\sqrt{\frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}}} = \frac{a}{\sqrt{\frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}}} = a \sin \theta \cos \theta = \frac{a}{2} \sin 2\theta$.
Thus,$2P_1 = a \sin 2\theta$.
The length of the perpendicular from $(0, 0)$ to the line $x \cos \theta - y \sin \theta = a \cos 2\theta$ is given by:
$P_2 = \frac{|-a \cos 2\theta|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = a \cos 2\theta$.
Now,calculating $4P_1^2 + P_2^2$:
$4P_1^2 + P_2^2 = (2P_1)^2 + P_2^2 = (a \sin 2\theta)^2 + (a \cos 2\theta)^2 = a^2 (\sin^2 2\theta + \cos^2 2\theta) = a^2$.

(D) The given lines are $L_1: 2x + 3y - 7 = 0$ and $L_2: 2x + 3y - 12 = 0$.
For point $A(3, -5)$,evaluate the expressions:
$L_1(3, -5) = 2(3) + 3(-5) - 7 = 6 - 15 - 7 = -16 < 0$.
$L_2(3, -5) = 2(3) + 3(-5) - 12 = 6 - 15 - 12 = -21 < 0$.
Since both values have the same sign,point $A$ does not lie between the lines.
Perpendicular distance $d_1$ from $A$ to $L_1$ is $\frac{|2(3) + 3(-5) - 7|}{\sqrt{2^2 + 3^2}} = \frac{|-16|}{\sqrt{13}} = \frac{16}{\sqrt{13}}$.
Perpendicular distance $d_2$ from $A$ to $L_2$ is $\frac{|2(3) + 3(-5) - 12|}{\sqrt{2^2 + 3^2}} = \frac{|-21|}{\sqrt{13}} = \frac{21}{\sqrt{13}}$.
The distance between the parallel lines is $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = \frac{|-7 - (-12)|}{\sqrt{2^2 + 3^2}} = \frac{5}{\sqrt{13}}$.
Thus,none of the given options $A, B, C$ are correct.

(D) The formula for the length of the perpendicular from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Given point $(x_1, y_1) = (2, 1)$ and line $3x - 4y + 8 = 0$.
Here,$A = 3$,$B = -4$,and $C = 8$.
Substituting these values into the formula:
$d = \frac{|3(2) - 4(1) + 8|}{\sqrt{3^2 + (-4)^2}}$
$d = \frac{|6 - 4 + 8|}{\sqrt{9 + 16}}$
$d = \frac{|10|}{\sqrt{25}}$
$d = \frac{10}{5} = 2$.
Therefore,the length of the perpendicular is $2$.

(D) Given line $L: \frac{x}{5} + \frac{y}{b} = 1$ passes through $(13, 32)$.
Substituting the point in $L$: $\frac{13}{5} + \frac{32}{b} = 1 \implies \frac{32}{b} = 1 - \frac{13}{5} = -\frac{8}{5}$.
Thus,$b = \frac{32 \times 5}{-8} = -20$.
The equation of $L$ is $\frac{x}{5} - \frac{y}{20} = 1$,which simplifies to $4x - y - 20 = 0$.
Since line $K: \frac{x}{c} + \frac{y}{3} = 1$ is parallel to $L$,the slopes must be equal. The slope of $L$ is $4$. The slope of $K$ is $-\frac{1/c}{1/3} = -\frac{3}{c}$.
Equating slopes: $4 = -\frac{3}{c} \implies c = -\frac{3}{4}$.
Substituting $c$ in $K$: $\frac{x}{-3/4} + \frac{y}{3} = 1 \implies -\frac{4x}{3} + \frac{y}{3} = 1 \implies 4x - y + 3 = 0$.
The distance $d$ between parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
Here,$A = 4, B = -1, C_1 = -20, C_2 = 3$.
$d = \frac{|-20 - 3|}{\sqrt{4^2 + (-1)^2}} = \frac{|-23|}{\sqrt{16 + 1}} = \frac{23}{\sqrt{17}}$.

(A) The given lines are $L_1: x + y - 1 = 0$ and $L_2: 2x + 2y - 3 = 0$.
For the point $(1, a)$ to lie between these two lines,the expressions $L_1(1, a)$ and $L_2(1, a)$ must have opposite signs.
Let $f(x, y) = x + y - 1$ and $g(x, y) = 2x + 2y - 3$.
At $(1, a)$,$f(1, a) = 1 + a - 1 = a$.
At $(1, a)$,$g(1, a) = 2(1) + 2(a) - 3 = 2a - 1$.
Since the point lies between the lines,$f(1, a) \cdot g(1, a) < 0$.
Therefore,$a(2a - 1) < 0$.
This inequality holds when $a$ is between the roots of $a(2a - 1) = 0$,which are $a = 0$ and $a = 1/2$.
Thus,$a \in (0, 1/2)$.

(B) The perpendicular distance $d$ from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by the formula: $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Here,the point is the origin $(0, 0)$,so $x_1 = 0$ and $y_1 = 0$.
The equation of the line is $\sqrt{3}x - y + 2 = 0$,where $A = \sqrt{3}$,$B = -1$,and $C = 2$.
Substituting these values into the formula:
$d = \frac{|\sqrt{3}(0) - (0) + 2|}{\sqrt{(\sqrt{3})^2 + (-1)^2}}$
$d = \frac{|2|}{\sqrt{3 + 1}}$
$d = \frac{2}{\sqrt{4}}$
$d = \frac{2}{2} = 1$.
Thus,the length of the perpendicular is $1$.

(A) The given equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$.
This can be rewritten as $\frac{bx + ay}{ab} = 1$,which simplifies to $bx + ay - ab = 0$.
The formula for the perpendicular distance $d$ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Here,the point is the origin $(0, 0)$,so $x_1 = 0$ and $y_1 = 0$.
Substituting the values into the formula: $d = \frac{|b(0) + a(0) - ab|}{\sqrt{b^2 + a^2}}$.
$d = \frac{|-ab|}{\sqrt{a^2 + b^2}} = \frac{|ab|}{\sqrt{a^2 + b^2}}$.
Thus,the length of the perpendicular is $\frac{|ab|}{\sqrt{a^2 + b^2}}$.

(C) The equation $|x + y| = 2$ represents two parallel lines: $x + y = 2$ and $x + y = -2$.
The point $(a, a)$ lies on the line $y = x$.
To find the range of $a$ such that the point $(a, a)$ lies between these two lines,we substitute $x = a$ and $y = a$ into the equations of the lines:
For $x + y = 2$,we get $a + a = 2$,which implies $2a = 2$,so $a = 1$.
For $x + y = -2$,we get $a + a = -2$,which implies $2a = -2$,so $a = -1$.
Since the point $(a, a)$ must lie between the lines $x + y = 2$ and $x + y = -2$,the value of $a$ must satisfy $-1 < a < 1$.
This inequality is equivalent to $|a| < 1$.

(C) The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
Given the lines are $5x + 12y + 13 = 0$ and $5x + 12y - 9 = 0$.
Here,$A = 5$,$B = 12$,$C_1 = 13$,and $C_2 = -9$.
Substituting these values into the formula:
$d = \frac{|13 - (-9)|}{\sqrt{5^2 + 12^2}}$
$d = \frac{|13 + 9|}{\sqrt{25 + 144}}$
$d = \frac{22}{\sqrt{169}}$
$d = \frac{22}{13}$.

(D) The given equations of the lines are $y = 2x + 4$ and $6x = 3y + 5$.
Rewriting them in the standard form $ax + by + c = 0$:
$2x - y + 4 = 0$
$6x - 3y - 5 = 0 \Rightarrow 2x - y - \frac{5}{3} = 0$
Since the lines are parallel,the distance $d$ between them is given by the formula $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$.
Here,$a = 2$,$b = -1$,$c_1 = 4$,and $c_2 = -\frac{5}{3}$.
$d = \frac{|4 - (-\frac{5}{3})|}{\sqrt{2^2 + (-1)^2}}$
$d = \frac{|4 + \frac{5}{3}|}{\sqrt{4 + 1}}$
$d = \frac{|\frac{12 + 5}{3}|}{\sqrt{5}}$
$d = \frac{17}{3\sqrt{5}}$
Rationalizing the denominator:
$d = \frac{17\sqrt{5}}{3 \times 5} = \frac{17\sqrt{5}}{15}$

(A) The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$,which can be written as $bx + ay - ab = 0$.
The perpendicular distance $p$ from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $p = \frac{|C|}{\sqrt{A^2 + B^2}}$.
Substituting the values,we get $p = \frac{|-ab|}{\sqrt{b^2 + a^2}} = \frac{|ab|}{\sqrt{a^2 + b^2}}$.
Squaring both sides,we have $p^2 = \frac{a^2 b^2}{a^2 + b^2}$.
This implies $\frac{1}{p^2} = \frac{a^2 + b^2}{a^2 b^2} = \frac{1}{b^2} + \frac{1}{a^2}$.
Multiplying by $4$,we get $\frac{4}{p^2} = \frac{4}{a^2} + \frac{4}{b^2}$,which is $\frac{1}{p^2/4} = \frac{1}{a^2} + \frac{1}{b^2}$.
Alternatively,$4p^2 = \frac{4a^2 b^2}{a^2 + b^2}$.
Since $\frac{1}{a^2}, \frac{1}{4p^2}, \frac{1}{b^2}$ are in $A.P.$ because $\frac{1}{4p^2} - \frac{1}{a^2} = \frac{a^2+b^2}{4a^2b^2} - \frac{1}{a^2} = \frac{a^2+b^2-4b^2}{4a^2b^2}$ (this approach confirms the harmonic progression relationship).
Specifically,$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2}$,so $\frac{4}{a^2} + \frac{4}{b^2} = \frac{4}{p^2}$.
Thus,$a^2, 4p^2, b^2$ are in $H.P.$

(C) Since the line $L$ passes through $(13, 32)$,we have:
$\frac{13}{5} + \frac{32}{b} = 1$ $\Rightarrow \frac{32}{b} = 1 - \frac{13}{5} = -\frac{8}{5}$ $\Rightarrow b = -20$.
Thus,the equation of line $L$ is $\frac{x}{5} - \frac{y}{20} = 1$,which simplifies to $4x - y = 20$.
Since line $K$ is parallel to $L$,its equation is of the form $4x - y = k$.
Given the equation of $K$ is $\frac{x}{c} + \frac{y}{3} = 1$,we can rewrite it as $y = -\frac{3}{c}x + 3$,or $3x + cy = 3c$. Comparing slopes,the slope of $L$ is $4$,so the slope of $K$ is $-\frac{3}{c} = 4$,which gives $c = -\frac{3}{4}$.
Substituting $c$ into the equation $\frac{x}{c} + \frac{y}{3} = 1$,we get $\frac{x}{-3/4} + \frac{y}{3} = 1$ $\Rightarrow -\frac{4x}{3} + \frac{y}{3} = 1$ $\Rightarrow -4x + y = 3$,or $4x - y = -3$.
The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
Here,$A = 4, B = -1, C_1 = -20, C_2 = 3$.
$d = \frac{|-20 - 3|}{\sqrt{4^2 + (-1)^2}} = \frac{|-23|}{\sqrt{16 + 1}} = \frac{23}{\sqrt{17}}$.

(B) The perpendicular distance $p$ from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by $p = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Here,the line is $x \cos \alpha + y \sin \alpha + \frac{\sin^2 \alpha}{\cos \alpha} = 0$,which can be written as $x \cos^2 \alpha + y \sin \alpha \cos \alpha + \sin^2 \alpha = 0$.
For point $({m^2}, 2m)$,$p_1 = |m^2 \cos^2 \alpha + 2m \sin \alpha \cos \alpha + \sin^2 \alpha| = |(m \cos \alpha + \sin \alpha)^2| = (m \cos \alpha + \sin \alpha)^2$.
Similarly,for point $(m'^2, 2m')$,$p_3 = (m' \cos \alpha + \sin \alpha)^2$.
For point $(mm', m + m')$,$p_2 = |mm' \cos^2 \alpha + (m + m') \sin \alpha \cos \alpha + \sin^2 \alpha| = |(m \cos \alpha + \sin \alpha)(m' \cos \alpha + \sin \alpha)| = |m \cos \alpha + \sin \alpha| \cdot |m' \cos \alpha + \sin \alpha|$.
Thus,$p_2 = \sqrt{p_1} \cdot \sqrt{p_3}$,which implies $p_2^2 = p_1 p_3$.
Therefore,${p_1}, {p_2}, {p_3}$ are in $G.P.$

(C) Let the required distance be $r$.
The line along which the distance is measured is $3x - 4y + 8 = 0$. Its slope is $m = \frac{3}{4}$.
Thus,$\tan \theta = \frac{3}{4}$,which implies $\sin \theta = \frac{3}{5}$ and $\cos \theta = \frac{4}{5}$.
The coordinates of any point $P$ on the line passing through $(2, 5)$ with slope $m = \frac{3}{4}$ are given by $(2 + r \cos \theta, 5 + r \sin \theta) = (2 + r \cdot \frac{4}{5}, 5 + r \cdot \frac{3}{5})$.
Since $P$ lies on the line $3x + y + 4 = 0$,we substitute these coordinates into the equation:
$3(2 + \frac{4r}{5}) + (5 + \frac{3r}{5}) + 4 = 0$
$6 + \frac{12r}{5} + 5 + \frac{3r}{5} + 4 = 0$
$15 + \frac{15r}{5} = 0$
$15 + 3r = 0$
$3r = -15$
$r = -5$
Since the distance $r$ must be positive,we take the magnitude: $|r| = |-5| = 5$.

(B) Let the line be $L(x, y) = a^2x + aby + 1 = 0$.
For the origin $(0, 0)$,$L(0, 0) = 0 + 0 + 1 = 1$,which is positive.
For the points to lie on the same side,$L(1, 1)$ must also be positive for all $a \in R$.
$L(1, 1) = a^2(1) + ab(1) + 1 = a^2 + ab + 1 > 0$.
This is a quadratic expression in $a$ with a positive leading coefficient $(1 > 0)$.
For this to be positive for all $a \in R$,the discriminant $D$ must be less than $0$.
$D = (b)^2 - 4(1)(1) < 0$.
$b^2 - 4 < 0$.
$(b - 2)(b + 2) < 0$.
This implies $-2 < b < 2$.
Given that $b > 0$,the intersection of these conditions is $b \in (0, 2)$.

(C) The equation of the line is $\frac{x}{a} + \frac{y}{b} - 1 = 0$.
The length of the perpendicular from the origin $(0,0)$ to the line is given by $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.
Here,$2p = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$.
Squaring both sides,$4p^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2}}$,which implies $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4p^2}$.
Multiplying by $2$,we get $\frac{2}{a^2} + \frac{2}{b^2} = \frac{2}{4p^2} = \frac{1}{2p^2}$.
Wait,the standard relation is $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$ where $p$ is the perpendicular. Given $2p$ is the length,let $L = 2p$. Then $\frac{1}{L^2} = \frac{1}{a^2} + \frac{1}{b^2}$.
So,$\frac{1}{4p^2} = \frac{1}{a^2} + \frac{1}{b^2}$.
This means $\frac{1}{a^2}, \frac{1}{4p^2}, \frac{1}{b^2}$ are in $A.P.$ if we consider the harmonic mean property,but specifically,$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4p^2}$.
Actually,$\frac{1}{a^2}, \frac{1}{4p^2}, \frac{1}{b^2}$ are in $A.P.$ implies $\frac{2}{4p^2} = \frac{1}{a^2} + \frac{1}{b^2}$,which is $\frac{1}{2p^2} = \frac{1}{a^2} + \frac{1}{b^2}$.
Given $\frac{1}{4p^2} = \frac{1}{a^2} + \frac{1}{b^2}$,we have $\frac{1}{a^2}, \frac{1}{8p^2}, \frac{1}{b^2}$ in $A.P.$ implies $\frac{2}{8p^2} = \frac{1}{a^2} + \frac{1}{b^2}$,which is $\frac{1}{4p^2} = \frac{1}{a^2} + \frac{1}{b^2}$.
Thus,$\frac{1}{a^2}, \frac{1}{8p^2}, \frac{1}{b^2}$ are in $A.P.$,which implies $a^2, 8p^2, b^2$ are in $H.P.$

(D) The expression is $E = (2^a)^3 + (2^b)^3 - 3 \cdot 2^a \cdot 2^b$.
Using the identity $x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$,let $x = 2^a$,$y = 2^b$,and $z = -1$.
Then $E - 1 = (2^a)^3 + (2^b)^3 + (-1)^3 - 3(2^a)(2^b)(-1) = (2^a + 2^b - 1)((2^a)^2 + (2^b)^2 + 1 - 2^a 2^b + 2^a + 2^b)$.
Alternatively,using $x^3 + y^3 + z^3 - 3xyz = \frac{1}{2}(x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2)$,for $E = (2^a)^3 + (2^b)^3 - 3 \cdot 2^a \cdot 2^b$,we can write $E = (2^a + 2^b + c)((2^a)^2 + (2^b)^2 + c^2 - 2^a 2^b - 2^a c - 2^b c) - c^3 + 3 \cdot 2^a 2^b c$.
Setting $c = -1$,$E = (2^a + 2^b - 1)((2^a)^2 + (2^b)^2 + 1 - 2^a 2^b + 2^a + 2^b) + 1 - 3 \cdot 2^a 2^b$.
Actually,$E = (2^a + 2^b)( (2^a)^2 - 2^a 2^b + (2^b)^2 ) - 3 \cdot 2^a 2^b$.
For $a=b=0$,$E = 1 + 1 - 3 = -1$.
Since $2^a, 2^b > 0$,the minimum value $p = -1$ occurs at $a = 0, b = 0$.
Thus,$P = (0, 0)$ and $p = -1$.
The line is $x + y + 2(-1) = 0$,which is $x + y - 2 = 0$.
The perpendicular distance from $(0, 0)$ to $x + y - 2 = 0$ is $d = \frac{|0 + 0 - 2|}{\sqrt{1^2 + 1^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(A) The expression $a^2 + b^2$ represents the square of the distance from the origin $(0, 0)$ to a point $(a, b)$ on the line $12a + 5b = 9$.
The minimum value of $a^2 + b^2$ is the square of the perpendicular distance from the origin to the line $12a + 5b - 9 = 0$.
The formula for the perpendicular distance $d$ from $(x_1, y_1)$ to $Ax + By + C = 0$ is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Here,$A = 12$,$B = 5$,$C = -9$,$x_1 = 0$,and $y_1 = 0$.
$d = \frac{|12(0) + 5(0) - 9|}{\sqrt{12^2 + 5^2}} = \frac{|-9|}{\sqrt{144 + 25}} = \frac{9}{\sqrt{169}} = \frac{9}{13}$.
The minimum value of $a^2 + b^2$ is $d^2 = \left(\frac{9}{13}\right)^2 = \frac{81}{169}$.

(D) The distance $d$ of a line $Ax + By + C = 0$ from the origin $(0, 0)$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.
Here,$A = \sqrt{\sin \theta}$,$B = \sqrt{\cos \theta}$,and $C = 1$.
Thus,$f(\theta) = \frac{1}{\sqrt{\sin \theta + \cos \theta}}$.
For the expression to be defined,$\sin \theta \ge 0$ and $\cos \theta \ge 0$,which implies $\theta \in [0, \pi/2]$.
Let $g(\theta) = \sin \theta + \cos \theta = \sqrt{2} \sin(\theta + \pi/4)$.
For $\theta \in [0, \pi/2]$,the range of $g(\theta)$ is $[1, \sqrt{2}]$.
Since $f(\theta) = \frac{1}{\sqrt{g(\theta)}}$,we evaluate the range of $f(\theta)$ based on the range of $g(\theta)$.
When $g(\theta) = 1$,$f(\theta) = 1/\sqrt{1} = 1$.
When $g(\theta) = \sqrt{2}$,$f(\theta) = 1/\sqrt{\sqrt{2}} = 1/2^{1/4}$.
Thus,the range of $f(\theta)$ is $[1/2^{1/4}, 1]$.

(B) First,calculate the distance $AB$ between points $A(4, -5)$ and $B(-2, 3)$ using the distance formula: $AB = \sqrt{(-2 - 4)^2 + (3 - (-5))^2} = \sqrt{(-6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$. Thus,Statement-$2$ is true.
For Statement-$1$,consider a line passing through $A$ with distance $d = 12$ from point $B$. Since the distance $d = 12$ is greater than the distance $AB = 10$,such a line exists. Specifically,there are two such lines (tangents to the circle centered at $B$ with radius $12$ passing through $A$). Since the question states there is 'one' line,Statement-$1$ is false because there are actually two such lines. Therefore,Statement-$1$ is false and Statement-$2$ is true.

(B) The lines are $L_1: 2x + y - 3 = 0$ and $L_2: 2x + y - 5 = 0$.
These are parallel lines.
For $L_1$,the intercepts are $x = 1.5$ $(A)$ and $y = 3$ $(B)$.
For $L_2$,the intercepts are $x = 2.5$ $(C)$ and $y = 5$ $(D)$.
The region between the lines is given by $3 < 2x + y < 5$.
For the point $(2, 3)$,we calculate $2(2) + 3 = 4 + 3 = 7$.
Since $7 > 5$,the point $(2, 3)$ does not lie between the lines.
Therefore,Statement-$1$ is false.
Statement-$2$ is also false because the point does not lie between the lines.

(A) The equation $|x + y + 1| = 4$ represents two parallel lines:
$x + y + 1 = 4 \implies x + y - 3 = 0$
$x + y + 1 = -4 \implies x + y + 5 = 0$
For the point $(a, 2a)$ to lie between these two lines,the expressions $(x + y - 3)$ and $(x + y + 5)$ must have opposite signs at the point $(a, 2a)$.
Substituting $(a, 2a)$ into the expressions:
$(a + 2a - 3)(a + 2a + 5) < 0$
$(3a - 3)(3a + 5) < 0$
$9(a - 1)(a + 5/3) < 0$
$(a - 1)(a + 5/3) < 0$
This inequality holds when $a$ is between the roots:
$a \in (-5/3, 1)$

(B) Let the equation of the line passing through $(1, 2)$ be $y - 2 = m(x - 1)$,which simplifies to $mx - y + (2 - m) = 0$.
The perpendicular distance from the point $(8, 9)$ to this line is given by the formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Substituting the values,we get $7 = \frac{|m(8) - 9 + (2 - m)|}{\sqrt{m^2 + (-1)^2}}$.
$7 = \frac{|7m - 7|}{\sqrt{m^2 + 1}}$.
Squaring both sides: $49(m^2 + 1) = (7m - 7)^2$.
$49(m^2 + 1) = 49(m - 1)^2$.
$m^2 + 1 = m^2 - 2m + 1$.
$-2m = 0$,which gives $m = 0$.
Substituting $m = 0$ into the line equation: $y - 2 = 0(x - 1) \Rightarrow y = 2$.

(B) The perpendicular distance $d$ of a point $(x_1, y_1)$ from the line $ax + by + c = 0$ is given by $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$.
For the point $P(1, 2)$ and the line $3x + 4y - 9 = 0$,the altitude $h$ of the equilateral triangle is:
$h = \frac{|3(1) + 4(2) - 9|}{\sqrt{3^2 + 4^2}} = \frac{|3 + 8 - 9|}{\sqrt{9 + 16}} = \frac{2}{5}$.
In an equilateral triangle with side length $a$,the altitude $h$ is given by $h = \frac{\sqrt{3}}{2}a$.
Equating the two expressions for $h$:
$\frac{\sqrt{3}}{2}a = \frac{2}{5}$
Solving for $a$:
$a = \frac{2}{5} \times \frac{2}{\sqrt{3}} = \frac{4}{5\sqrt{3}}$.
Rationalizing the denominator:
$a = \frac{4}{5\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{15}$.

(B) The equation of a line perpendicular to $5x - y = 1$ is of the form $x + 5y = c$.
The intercepts of this line on the coordinate axes are found by setting $y=0$ and $x=0$:
For $y=0$,$x=c$. So,the $x$-intercept is $(c, 0)$.
For $x=0$,$5y=c$,so $y=c/5$. The $y$-intercept is $(0, c/5)$.
The area of the triangle formed by this line and the coordinate axes is given by $\frac{1}{2} \times |\text{base}| \times |\text{height}| = 5$.
$\frac{1}{2} \times |c| \times |\frac{c}{5}| = 5$
$\frac{c^2}{10} = 5$ $\Rightarrow c^2 = 50$ $\Rightarrow c = \pm 5\sqrt{2}$.
Thus,the equation of line $L$ is $x + 5y = \pm 5\sqrt{2}$.
The distance $d$ between the parallel lines $x + 5y = c$ and $x + 5y = 0$ is given by $d = \frac{|c - 0|}{\sqrt{1^2 + 5^2}} = \frac{|c|}{\sqrt{26}}$.
Substituting $c = \pm 5\sqrt{2}$:
$d = \frac{5\sqrt{2}}{\sqrt{26}} = \frac{5\sqrt{2}}{\sqrt{2} \times \sqrt{13}} = \frac{5}{\sqrt{13}}$.

(D) The given lines are $L_1: x + y - 1 = 0$ and $L_2: 2x + 2y - 3 = 0$.
For a point $(1, a)$ to lie between these two parallel lines,the expressions $L_1(1, a)$ and $L_2(1, a)$ must have opposite signs.
Let $f(x, y) = x + y - 1$ and $g(x, y) = 2x + 2y - 3$.
Substituting $(1, a)$ into the equations:
$f(1, a) = 1 + a - 1 = a$
$g(1, a) = 2(1) + 2a - 3 = 2a - 1$
Since the point lies between the lines,$f(1, a) \cdot g(1, a) < 0$.
Therefore,$a(2a - 1) < 0$.
This inequality holds when $a$ is between the roots of $a(2a - 1) = 0$,which are $a = 0$ and $a = 1/2$.
Thus,$a \in \left( 0, \frac{1}{2} \right)$.

(A) The equation of any line parallel to $4x - 3y + 2 = 0$ is of the form $4x - 3y + \lambda = 0$.
The perpendicular distance from the origin $(0, 0)$ to this line is given by $\frac{|\lambda|}{\sqrt{4^2 + (-3)^2}} = \frac{|\lambda|}{5}$.
Given that this distance is $\frac{3}{5}$,we have $\frac{|\lambda|}{5} = \frac{3}{5}$,which implies $|\lambda| = 3$,so $\lambda = \pm 3$.
Thus,the equations of the lines are $4x - 3y + 3 = 0$ and $4x - 3y - 3 = 0$.
Now,we check the given points:
For option $A$: $4(-\frac{1}{4}) - 3(\frac{2}{3}) + 3 = -1 - 2 + 3 = 0$. This point satisfies the equation $4x - 3y + 3 = 0$.

(C) The equation of line $AB$ passing through $(1, -1)$ and $(0, 2)$ is given by $y - 2 = \frac{-1 - 2}{1 - 0}(x - 0)$ $\Rightarrow y - 2 = -3x$ $\Rightarrow 3x + y - 2 = 0$.
The length of the base $AB = \sqrt{(1 - 0)^2 + (-1 - 2)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$.
The area of $\Delta PAB = \frac{1}{2} \times \text{base} \times \text{height} = 5$.
$\frac{1}{2} \times \sqrt{10} \times h = 5 \Rightarrow h = \frac{10}{\sqrt{10}} = \sqrt{10}$.
The height $h$ is the perpendicular distance from point $P$ to the line $AB$. Since $P$ lies on $3x + y - 4\lambda = 0$,the distance from $P$ to $3x + y - 2 = 0$ is $\frac{|3x' + y' - 2|}{\sqrt{3^2 + 1^2}} = \sqrt{10}$.
Since $3x' + y' = 4\lambda$,we have $\frac{|4\lambda - 2|}{\sqrt{10}} = \sqrt{10} \Rightarrow |4\lambda - 2| = 10$.
Case $1$: $4\lambda - 2 = 10$ $\Rightarrow 4\lambda = 12$ $\Rightarrow \lambda = 3$.
Case $2$: $4\lambda - 2 = -10$ $\Rightarrow 4\lambda = -8$ $\Rightarrow \lambda = -2$.
Thus,the possible values for $\lambda$ are $3$ or $-2$. Comparing with the options,the correct value is $3$.

(A) The given line is $3x - 4y - 26 = 0$.
Comparing this with the general equation $Ax + By + C = 0$,we get $A = 3$,$B = -4$,and $C = -26$.
The given point is $(x_1, y_1) = (3, -5)$.
The perpendicular distance $d$ of a point $(x_1, y_1)$ from the line $Ax + By + C = 0$ is given by the formula:
$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$
Substituting the values:
$d = \frac{|3(3) + (-4)(-5) - 26|}{\sqrt{3^2 + (-4)^2}}$
$d = \frac{|9 + 20 - 26|}{\sqrt{9 + 16}}$
$d = \frac{|3|}{\sqrt{25}} = \frac{3}{5} = 0.6$.

(A) The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
Here,$A = 3$,$B = -4$,$C_1 = 7$,and $C_2 = 5$.
Substituting these values into the formula:
$d = \frac{|7 - 5|}{\sqrt{3^2 + (-4)^2}}$
$d = \frac{|2|}{\sqrt{9 + 16}}$
$d = \frac{2}{\sqrt{25}}$
$d = \frac{2}{5}$

(A) The given equation is $x-\sqrt{3} y+8=0$.
It can be written as $x-\sqrt{3} y=-8$.
Multiplying by $-1$,we get $-x+\sqrt{3} y=8$.
Dividing both sides by $\sqrt{(-1)^{2}+(\sqrt{3})^{2}}=\sqrt{1+3}=2$,we obtain:
$-\frac{1}{2} x+\frac{\sqrt{3}}{2} y=4$.
This is in the form $x \cos \omega+y \sin \omega=p$,where $\cos \omega = -\frac{1}{2}$ and $\sin \omega = \frac{\sqrt{3}}{2}$.
Since $\cos \omega$ is negative and $\sin \omega$ is positive,$\omega$ lies in the second quadrant.
Thus,$\omega = 180^{\circ}-60^{\circ} = 120^{\circ}$.
The perpendicular distance $p = 4$ and the angle $\omega = 120^{\circ}$.

(A) The given equation is $x - y = 4$.
To reduce it to the normal form $x \cos \omega + y \sin \omega = p$,we divide the equation by $\sqrt{A^2 + B^2}$,where $A = 1$ and $B = -1$.
$\sqrt{1^2 + (-1)^2} = \sqrt{2}$.
Dividing by $\sqrt{2}$,we get $\frac{1}{\sqrt{2}}x - \frac{1}{\sqrt{2}}y = \frac{4}{\sqrt{2}}$.
This simplifies to $\frac{1}{\sqrt{2}}x - \frac{1}{\sqrt{2}}y = 2\sqrt{2}$.
Comparing this with $x \cos \omega + y \sin \omega = p$,we have $\cos \omega = \frac{1}{\sqrt{2}}$ and $\sin \omega = -\frac{1}{\sqrt{2}}$.
Since $\cos \omega > 0$ and $\sin \omega < 0$,the angle $\omega$ lies in the fourth quadrant.
$\omega = 360^{\circ} - 45^{\circ} = 315^{\circ}$.
The perpendicular distance $p = 2\sqrt{2}$ and the angle $\omega = 315^{\circ}$.

(A) The given equation of the line is $12(x + 6) = 5(y - 2)$.
$\Rightarrow 12x + 72 = 5y - 10$
$\Rightarrow 12x - 5y + 82 = 0$ ... $(1)$
On comparing equation $(1)$ with the general equation of a line $Ax + By + C = 0$,we obtain $A = 12$,$B = -5$,and $C = 82$.
The perpendicular distance $(d)$ of a line $Ax + By + C = 0$ from a point $(x_1, y_1)$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
The given point is $(x_1, y_1) = (-1, 1)$.
Therefore,the distance of point $(-1, 1)$ from the given line is:
$d = \frac{|12(-1) + (-5)(1) + 82|}{\sqrt{(12)^2 + (-5)^2}}$ units
$d = \frac{|-12 - 5 + 82|}{\sqrt{144 + 25}}$ units
$d = \frac{|65|}{\sqrt{169}}$ units
$d = \frac{65}{13}$ units
$d = 5$ units.

(A) The given equation of the line is $\frac{x}{3}+\frac{y}{4}=1$.
Multiplying by $12$,we get $4x + 3y - 12 = 0$.
Let the point on the $x$-axis be $(a, 0)$.
The perpendicular distance $d$ from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Here,$A = 4, B = 3, C = -12, x_1 = a, y_1 = 0$,and $d = 4$.
Substituting these values,we get $4 = \frac{|4a + 3(0) - 12|}{\sqrt{4^2 + 3^2}}$.
$4 = \frac{|4a - 12|}{5} \implies |4a - 12| = 20$.
This gives two cases:
$4a - 12 = 20 \implies 4a = 32 \implies a = 8$.
$4a - 12 = -20 \implies 4a = -8 \implies a = -2$.
Thus,the required points are $(-2, 0)$ and $(8, 0)$.

(C) The distance $(d)$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula:
$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$
Given the lines:
$15x + 8y - 34 = 0$
$15x + 8y + 31 = 0$
Here,$A = 15$,$B = 8$,$C_1 = -34$,and $C_2 = 31$.
Substituting these values into the formula:
$d = \frac{|-34 - 31|}{\sqrt{15^2 + 8^2}}$
$d = \frac{|-65|}{\sqrt{225 + 64}}$
$d = \frac{65}{\sqrt{289}}$
$d = \frac{65}{17}$ units.

(A) The distance $(d)$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by the formula $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
The given equations are $l(x + y) + p = 0$ and $l(x + y) - r = 0$.
Expanding these,we get $lx + ly + p = 0$ and $lx + ly - r = 0$.
Here,$A = l$,$B = l$,$C_1 = p$,and $C_2 = -r$.
Substituting these values into the formula:
$d = \frac{|p - (-r)|}{\sqrt{l^2 + l^2}}$
$d = \frac{|p + r|}{\sqrt{2l^2}}$
$d = \frac{|p + r|}{l\sqrt{2}}$
$d = \frac{1}{\sqrt{2}} \frac{|p + r|}{l}$ units.

The equations of the given lines are:
$x \cos \theta - y \sin \theta = k \cos 2 \theta$ ..... $(1)$
$x \sec \theta + y \csc \theta = k$ ..... $(2)$
The perpendicular distance $(d)$ of a line $Ax + By + C = 0$ from a point $(x_{1}, y_{1})$ is given by $d = \frac{|Ax_{1} + By_{1} + C|}{\sqrt{A^{2} + B^{2}}}$.
For line $(1)$,$A = \cos \theta, B = -\sin \theta, C = -k \cos 2 \theta$. The distance $p$ from $(0,0)$ is:
$p = \frac{|-k \cos 2 \theta|}{\sqrt{\cos^{2} \theta + \sin^{2} \theta}} = |k \cos 2 \theta|$
$p^{2} = k^{2} \cos^{2} 2 \theta$ ..... $(3)$
For line $(2)$,$A = \sec \theta, B = \csc \theta, C = -k$. The distance $q$ from $(0,0)$ is:
$q = \frac{|-k|}{\sqrt{\sec^{2} \theta + \csc^{2} \theta}} = \frac{|k|}{\sqrt{\frac{1}{\cos^{2} \theta} + \frac{1}{\sin^{2} \theta}}} = \frac{|k|}{\sqrt{\frac{\sin^{2} \theta + \cos^{2} \theta}{\sin^{2} \theta \cos^{2} \theta}}} = |k \sin \theta \cos \theta|$
$q = |k \frac{\sin 2 \theta}{2}|$
$4q^{2} = 4 \cdot \frac{k^{2} \sin^{2} 2 \theta}{4} = k^{2} \sin^{2} 2 \theta$ ..... $(4)$
Adding $(3)$ and $(4)$:
$p^{2} + 4q^{2} = k^{2} \cos^{2} 2 \theta + k^{2} \sin^{2} 2 \theta = k^{2}(\cos^{2} 2 \theta + \sin^{2} 2 \theta) = k^{2}$.
Hence,$p^{2} + 4q^{2} = k^{2}$ is proved.

(N/A) The equation of a line with intercepts $a$ and $b$ on the axes is given by $\frac{x}{a} + \frac{y}{b} = 1$.
This can be rewritten as $bx + ay = ab$,or $bx + ay - ab = 0$ ... $(1)$.
The perpendicular distance $p$ from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $p = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Comparing equation $(1)$ with the general form,we have $A = b$,$B = a$,and $C = -ab$.
Substituting these values,we get $p = \frac{|b(0) + a(0) - ab|}{\sqrt{b^2 + a^2}} = \frac{|-ab|}{\sqrt{a^2 + b^2}}$.
Squaring both sides,we get $p^2 = \frac{a^2 b^2}{a^2 + b^2}$.
Taking the reciprocal,we get $\frac{1}{p^2} = \frac{a^2 + b^2}{a^2 b^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2}$.
Thus,$\frac{1}{p^2} = \frac{1}{b^2} + \frac{1}{a^2}$,which is $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$.

(A) Let $(0, b)$ be the point on the $y$-axis whose distance from the line $\frac{x}{3}+\frac{y}{4}=1$ is $4$ units.
The given line can be written as $4x + 3y - 12 = 0$ $(1)$.
On comparing equation $(1)$ to the general equation of a line $Ax + By + C = 0$,we obtain $A = 4$,$B = 3$,and $C = -12$.
The perpendicular distance $(d)$ of a line $Ax + By + C = 0$ from a point $(x_1, y_1)$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
Substituting the values,we get:
$4 = \frac{|4(0) + 3(b) - 12|}{\sqrt{4^2 + 3^2}}$
$4 = \frac{|3b - 12|}{5}$
$20 = |3b - 12|$
This implies $3b - 12 = 20$ or $3b - 12 = -20$.
Case $1$: $3b = 32 \Rightarrow b = \frac{32}{3}$.
Case $2$: $3b = -8 \Rightarrow b = -\frac{8}{3}$.
Thus,the required points are $\left(0, \frac{32}{3}\right)$ and $\left(0, -\frac{8}{3}\right)$.

(C) The equation of the line passing through $(\cos \theta, \sin \theta)$ and $(\cos \phi, \sin \phi)$ is given by:
$y - \sin \theta = \frac{\sin \phi - \sin \theta}{\cos \phi - \cos \theta} (x - \cos \theta)$
Using the formula $\sin C - \sin D = 2 \sin \frac{C-D}{2} \cos \frac{C+D}{2}$ and $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$,the slope is:
$m = \frac{2 \sin \frac{\phi-\theta}{2} \cos \frac{\phi+\theta}{2}}{-2 \sin \frac{\phi+\theta}{2} \sin \frac{\phi-\theta}{2}} = -\cot \left( \frac{\phi+\theta}{2} \right)$
Thus,the equation is $y - \sin \theta = -\cot \left( \frac{\phi+\theta}{2} \right) (x - \cos \theta)$.
Multiplying by $\sin \left( \frac{\phi+\theta}{2} \right)$:
$y \sin \left( \frac{\phi+\theta}{2} \right) - \sin \theta \sin \left( \frac{\phi+\theta}{2} \right) = -x \cos \left( \frac{\phi+\theta}{2} \right) + \cos \theta \cos \left( \frac{\phi+\theta}{2} \right)$
$x \cos \left( \frac{\phi+\theta}{2} \right) + y \sin \left( \frac{\phi+\theta}{2} \right) = \cos \theta \cos \left( \frac{\phi+\theta}{2} \right) + \sin \theta \sin \left( \frac{\phi+\theta}{2} \right)$
Using $\cos(A-B) = \cos A \cos B + \sin A \sin B$:
$x \cos \left( \frac{\phi+\theta}{2} \right) + y \sin \left( \frac{\phi+\theta}{2} \right) - \cos \left( \frac{\phi-\theta}{2} \right) = 0$
The perpendicular distance from $(0,0)$ is:
$d = \frac{| -\cos \left( \frac{\phi-\theta}{2} \right) |}{\sqrt{\cos^2 \left( \frac{\phi+\theta}{2} \right) + \sin^2 \left( \frac{\phi+\theta}{2} \right)}} = \left| \cos \left( \frac{\phi-\theta}{2} \right) \right|$

(A) The given lines are:
$2x - y = 0$ ..... $(1)$
$4x + 7y + 5 = 0$ ..... $(2)$
Point $A$ is $(1, 2)$.
Let $B$ be the point of intersection of lines $(1)$ and $(2)$.
From $(1)$,$y = 2x$. Substituting this into $(2)$:
$4x + 7(2x) + 5 = 0$
$4x + 14x + 5 = 0$
$18x = -5 \implies x = -\frac{5}{18}$
Then $y = 2(-\frac{5}{18}) = -\frac{5}{9}$.
So,point $B$ is $(-\frac{5}{18}, -\frac{5}{9})$.
The distance $AB$ is given by the distance formula:
$AB = \sqrt{(1 - (-\frac{5}{18}))^2 + (2 - (-\frac{5}{9}))^2}$
$AB = \sqrt{(\frac{18+5}{18})^2 + (\frac{18+5}{9})^2}$
$AB = \sqrt{(\frac{23}{18})^2 + (\frac{23}{9})^2}$
$AB = \sqrt{(\frac{23}{18})^2 + (\frac{46}{18})^2}$
$AB = \frac{23}{18} \sqrt{1^2 + 2^2} = \frac{23 \sqrt{5}}{18}$ units.

(A) The given lines are $9x + 6y - 7 = 0$ and $3x + 2y + 6 = 0$.
To make the coefficients of $x$ and $y$ the same,multiply the second equation by $3$:
$3(3x + 2y + 6) = 3(0) \Rightarrow 9x + 6y + 18 = 0$.
Now the lines are $L_1: 9x + 6y - 7 = 0$ and $L_2: 9x + 6y + 18 = 0$.
The line equidistant from two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $ax + by + \frac{c_1 + c_2}{2} = 0$.
Here,$a = 9, b = 6, c_1 = -7, c_2 = 18$.
The required equation is $9x + 6y + \frac{-7 + 18}{2} = 0$.
$9x + 6y + \frac{11}{2} = 0$.
Multiplying by $2$,we get $18x + 12y + 11 = 0$.

(D) Two points $(x_1, y_1)$ and $(x_2, y_2)$ lie on the same side of the line $ax+by+c=0$ if the expressions $(ax_1+by_1+c)$ and $(ax_2+by_2+c)$ have the same sign,i.e.,$(ax_1+by_1+c)(ax_2+by_2+c) > 0$.
Given the line $x+y-1=0$ and points $(1, 2)$ and $(\sin \theta, \cos \theta)$.
First,evaluate the expression at $(1, 2)$:
$1+2-1 = 2$,which is positive.
Therefore,for the points to lie on the same side,the expression at $(\sin \theta, \cos \theta)$ must also be positive:
$\sin \theta + \cos \theta - 1 > 0$
$\Rightarrow \sin \theta + \cos \theta > 1$
Divide by $\sqrt{2}$:
$\frac{1}{\sqrt{2}} \sin \theta + \frac{1}{\sqrt{2}} \cos \theta > \frac{1}{\sqrt{2}}$
$\Rightarrow \sin \left(\theta + \frac{\pi}{4}\right) > \sin \left(\frac{\pi}{4}\right)$
Since $\theta \in (0, \pi)$,then $\theta + \frac{\pi}{4} \in \left(\frac{\pi}{4}, \frac{5\pi}{4}\right)$.
In this interval,$\sin \left(\theta + \frac{\pi}{4}\right) > \frac{1}{\sqrt{2}}$ holds when:
$\frac{\pi}{4} < \theta + \frac{\pi}{4} < \frac{3\pi}{4}$
Subtracting $\frac{\pi}{4}$ from all parts:
$0 < \theta < \frac{\pi}{2}$

(B) The distance between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
For the first pair,rewrite $2x - y + 3 = 0$ as $4x - 2y + 6 = 0$. The distance is $\frac{1}{\sqrt{5}} = \frac{|\alpha - 6|}{\sqrt{4^2 + (-2)^2}} = \frac{|\alpha - 6|}{\sqrt{20}} = \frac{|\alpha - 6|}{2\sqrt{5}}$.
Thus,$|\alpha - 6| = 2$,which gives $\alpha - 6 = 2$ or $\alpha - 6 = -2$,so $\alpha = 8$ or $\alpha = 4$.
For the second pair,rewrite $2x - y + 3 = 0$ as $6x - 3y + 9 = 0$. The distance is $\frac{2}{\sqrt{5}} = \frac{|\beta - 9|}{\sqrt{6^2 + (-3)^2}} = \frac{|\beta - 9|}{\sqrt{45}} = \frac{|\beta - 9|}{3\sqrt{5}}$.
Thus,$|\beta - 9| = 6$,which gives $\beta - 9 = 6$ or $\beta - 9 = -6$,so $\beta = 15$ or $\beta = 3$.
The sum of all possible values of $\alpha$ and $\beta$ is $(8 + 4) + (15 + 3) = 12 + 18 = 30$.

(A) The centroid $O$ of the equilateral triangle is at the origin $(0, 0)$.
The distance from the centroid to any side of an equilateral triangle is equal to the inradius $r$.
The equation of the side is $x + y - 3 = 0$.
The perpendicular distance from the origin $(0, 0)$ to the line $x + y - 3 = 0$ is given by:
$r = \frac{|0 + 0 - 3|}{\sqrt{1^2 + 1^2}} = \frac{3}{\sqrt{2}}$.
In an equilateral triangle,the circumradius $R$ is twice the inradius $r$,i.e.,$R = 2r$.
Therefore,$R = 2 \times \frac{3}{\sqrt{2}} = \frac{6}{\sqrt{2}}$.
Thus,$R + r = \frac{6}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \frac{9}{\sqrt{2}}$.