Equation of lines joining the origin to the point of intersection of a curve and a line and Distance between the pair of lines Questions in English

(A) Let the intercepts cut by the line on the axes be $a$ and $a$ (since they are of equal length).
Using the intercept form of the equation of a line: $\frac{x}{a} + \frac{y}{a} = 1$.
Multiplying by $a$,we get $x + y = a$,or $x + y - a = 0$.
The equation is in the form $Ax + By + C = 0$,where $A = 1$ and $B = 1$.
The slope $m$ of the line is given by $m = -\frac{A}{B} = -\frac{1}{1} = -1$.
Alternatively,if the intercepts are equal in magnitude but opposite in sign,the slope could be $1$. However,given the options,the standard interpretation is equal intercepts $a$,leading to a slope of $-1$.

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

(D) Given equations are $y - x + 7 = 0$ $(1)$ and $y + 2x - 2 = 0$ $(2)$.
Subtracting $(1)$ from $(2)$,we get $(y + 2x - 2) - (y - x + 7) = 0$,which simplifies to $3x - 9 = 0$,so $x = 3$.
Substituting $x = 3$ in $(1)$,we get $y - 3 + 7 = 0$,so $y = -4$.
The point of intersection is $(3, -4)$.
The equation of a line passing through the origin $(0, 0)$ and $(x_1, y_1)$ is given by $y = \frac{y_1}{x_1}x$.
Substituting $(3, -4)$,we get $y = \frac{-4}{3}x$,which implies $3y = -4x$ or $4x + 3y = 0$.

(C) The equation of any curve passing through the points of intersection of the given curves is given by $S_1 + \lambda S_2 = 0$,where $S_1 = 2x^2 + 3y^2 + 10x$ and $S_2 = 3x^2 + 5y^2 + 16x$.
Thus,$(2x^2 + 3y^2 + 10x) + \lambda(3x^2 + 5y^2 + 16x) = 0$.
$(2 + 3\lambda)x^2 + (3 + 5\lambda)y^2 + (10 + 16\lambda)x = 0$ ... $(i)$
For this equation to represent a pair of straight lines passing through the origin,it must be a homogeneous equation of the second degree. Therefore,the coefficient of $x$ must be zero.
$10 + 16\lambda = 0 \Rightarrow \lambda = -\frac{10}{16} = -\frac{5}{8}$.
Substituting $\lambda = -\frac{5}{8}$ into equation $(i)$:
$(2 + 3(-\frac{5}{8}))x^2 + (3 + 5(-\frac{5}{8}))y^2 = 0$
$(\frac{16 - 15}{8})x^2 + (\frac{24 - 25}{8})y^2 = 0$
$\frac{1}{8}x^2 - \frac{1}{8}y^2 = 0 \Rightarrow x^2 - y^2 = 0$.
This represents the pair of lines $x - y = 0$ and $x + y = 0$.
The slopes are $m_1 = 1$ and $m_2 = -1$. Since $m_1 \times m_2 = -1$,the lines are perpendicular,and the angle between them is $90^\circ$.

(A) To find the equation of the pair of lines joining the origin to the intersection points,we homogenize the equation of the curve $3x^2 + 4xy - 4x + 1 = 0$ using the line $2x + y = 1$ (i.e.,$2x + y = 1$).
Substituting $1 = (2x + y)$ into the curve equation:
$3x^2 + 4xy - 4x(2x + y) + (2x + y)^2 = 0$
Expanding the terms:
$3x^2 + 4xy - 8x^2 - 4xy + 4x^2 + 4xy + y^2 = 0$
Simplifying the expression:
$-x^2 + 4xy + y^2 = 0$ or $x^2 - 4xy - y^2 = 0$
Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we have $a = 1$,$2h = -4$ (so $h = -2$),and $b = -1$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$.
Since $a + b = 1 + (-1) = 0$,the denominator is $0$,which implies $\tan \theta = \infty$.
Therefore,$\theta = \pi/2$,meaning the lines are perpendicular.

(B) The given curve is $x^2 + y^2 - 2x - 1 = 0$ and the line is $x + y = 1$.
To make the equation of the curve homogeneous of degree $2$,we rewrite the line as $(x + y) = 1$.
Substituting this into the curve equation:
$x^2 + y^2 - 2x(x + y) - (x + y)^2 = 0$
$x^2 + y^2 - 2x^2 - 2xy - (x^2 + 2xy + y^2) = 0$
$-2x^2 - 4xy = 0$
$x^2 + 2xy = 0$
Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = 1, h = 1, b = 0$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$.
$\tan \theta = \left| \frac{2\sqrt{1^2 - (1)(0)}}{1 + 0} \right| = 2$.
Therefore,$\theta = \tan^{-1}(2)$.

(B) The equation of the pair of lines passing through the origin and the points of intersection of the two given curves is obtained by homogenizing the equations.
Consider the curves:
$S_1: ax^2 + 2hxy + by^2 + 2gx = 0$
$S_2: a'x^2 + 2h'xy + b'y^2 + 2g'x = 0$
To homogenize $S_1$ using $S_2$,we rewrite $S_1$ as $ax^2 + 2hxy + by^2 = -2gx$ and $S_2$ as $a'x^2 + 2h'xy + b'y^2 = -2g'x$.
Multiplying $S_1$ by $g'$ and $S_2$ by $g$,we get:
$g'(ax^2 + 2hxy + by^2) = -2gg'x$
$g(a'x^2 + 2h'xy + b'y^2) = -2gg'x$
Equating the two:
$g'(ax^2 + 2hxy + by^2) = g(a'x^2 + 2h'xy + b'y^2)$
$(ag' - a'g)x^2 + 2(hg' - h'g)xy + (bg' - b'g)y^2 = 0$
For these lines to be mutually perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(ag' - a'g) + (bg' - b'g) = 0$
$g'(a + b) - g(a' + b') = 0$
$g(a' + b') = g'(a + b)$.

(A) The given equation is $x^2 + 2\sqrt{3}xy + 3y^2 - 3x - 3\sqrt{3}y - 4 = 0$.
Comparing this with the general form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we have $a=1, h=\sqrt{3}, b=3, g=-3/2, f=-3\sqrt{3}/2, c=-4$.
First,check if the lines are parallel by verifying $h^2 - ab = (\sqrt{3})^2 - (1)(3) = 3 - 3 = 0$. Since $h^2 - ab = 0$,the lines are parallel.
The formula for the distance between two parallel lines $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is $d = 2\sqrt{\frac{g^2 - ac}{a(a + b)}}$.
Substituting the values: $d = 2\sqrt{\frac{(-3/2)^2 - (1)(-4)}{1(1 + 3)}} = 2\sqrt{\frac{9/4 + 4}{4}} = 2\sqrt{\frac{25/4}{4}} = 2\sqrt{\frac{25}{16}} = 2 \times \frac{5}{4} = \frac{5}{2} = 2.5$.

(D) The equation of the curve is $x^2 + hxy - y^2 + gx + fy = 0$.
The equation of the line is $fx - gy = \lambda$,which implies $\frac{fx - gy}{\lambda} = 1$.
Making the equation of the curve homogeneous with the help of the line equation:
$x^2 + hxy - y^2 + (gx + fy) \left( \frac{fx - gy}{\lambda} \right) = 0$.
Multiplying by $\lambda$:
$\lambda(x^2 + hxy - y^2) + (gx + fy)(fx - gy) = 0$.
$\lambda x^2 + \lambda hxy - \lambda y^2 + gfx^2 - g^2xy + f^2xy - fgy^2 = 0$.
$(\lambda + gf)x^2 + (\lambda h - g^2 + f^2)xy - (\lambda + fg)y^2 = 0$.
For the lines to be mutually perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(\lambda + gf) + (-(\lambda + fg)) = 0$.
$\lambda + gf - \lambda - fg = 0$.
$0 = 0$.
Since the condition is satisfied for any value of $\lambda$,$\lambda$ may have any value.

(B) Given curves are $x^2 + y^2 = a^2$ $(1)$ and $x^2 + y^2 - ax - ay = 0$ $(2)$.
From $(1)$,we have $a^2 = x^2 + y^2$.
Substituting this into $(2)$,we get $x^2 + y^2 - ax - ay = 0$.
Since $a^2 = x^2 + y^2$,we can write $a = \sqrt{x^2 + y^2}$.
Alternatively,homogenizing the equation of the second curve using the first curve:
$x^2 + y^2 - (ax + ay) \frac{x^2 + y^2}{a^2} = 0$.
$a^2(x^2 + y^2) - (ax + ay)(x^2 + y^2) = 0$.
$(x^2 + y^2)(a^2 - ax - ay) = 0$.
This does not directly give the lines. Let us solve by intersection.
Subtracting $(2)$ from $(1)$: $ax + ay = a^2 \implies x + y = a \implies a = x + y$.
Substitute $a = x + y$ into $x^2 + y^2 = a^2$:
$x^2 + y^2 = (x + y)^2 = x^2 + y^2 + 2xy$.
$2xy = 0 \implies xy = 0$.
Thus,the lines are $x = 0$ and $y = 0$.

(C) The given equation is $x^2 + 2\sqrt{2}xy + 2y^2 + 4x + 4\sqrt{2}y + 1 = 0$.
Comparing this with the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a=1, h=\sqrt{2}, b=2, g=2, f=2\sqrt{2}, c=1$.
Since $h^2 - ab = (\sqrt{2})^2 - (1)(2) = 2 - 2 = 0$,the lines are parallel.
The distance between two parallel lines represented by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is given by $d = 2\sqrt{\frac{g^2 - ac}{a(a + b)}}$.
Substituting the values: $d = 2\sqrt{\frac{2^2 - (1)(1)}{1(1 + 2)}} = 2\sqrt{\frac{4 - 1}{3}} = 2\sqrt{\frac{3}{3}} = 2\sqrt{1} = 2$.

(A) The equation of the pair of straight lines passing through the origin and the intersection points of a curve $S = 0$ and a line $L = 0$ is given by homogenizing the curve equation using the line equation.
Given curve: $x^2 + y^2 = 4$
Given line: $y - x = 2$,which implies $\frac{y - x}{2} = 1$.
Substituting this into the curve equation to homogenize it:
$x^2 + y^2 = 4 \times (1)^2$
$x^2 + y^2 = 4 \times \left( \frac{y - x}{2} \right)^2$
$x^2 + y^2 = 4 \times \frac{(y - x)^2}{4}$
$x^2 + y^2 = (y - x)^2$.

(D) To find the equation of the lines joining the origin to the points of intersection of the line $x + y = 1$ and the curve $x^2 + y^2 - 2y + \lambda = 0$,we make the curve homogeneous using the line equation.
Since $x + y = 1$,we can write the equation of the curve as:
$x^2 + y^2 - 2y(1) + \lambda(1)^2 = 0$
Substituting $1 = x + y$:
$x^2 + y^2 - 2y(x + y) + \lambda(x + y)^2 = 0$
Expanding the terms:
$x^2 + y^2 - 2xy - 2y^2 + \lambda(x^2 + 2xy + y^2) = 0$
Grouping the terms:
$x^2(1 + \lambda) + xy(-2 + 2\lambda) + y^2(1 - 2 + \lambda) = 0$
$x^2(1 + \lambda) + xy(2\lambda - 2) + y^2(\lambda - 1) = 0$
For the lines to be perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero $(A + B = 0)$:
$(1 + \lambda) + (\lambda - 1) = 0$
$2\lambda = 0$
$\lambda = 0$

(C) To find the lines joining the origin to the points of intersection,we homogenize the equation of the curve using the line equation $x - y = 2$,which implies $\frac{x - y}{2} = 1$.
The equation of the curve is $5x^2 + 12xy - 8y^2 + 8x - 4y + 12 = 0$.
Substituting $1$ with $\frac{x - y}{2}$ in the linear and constant terms:
$5x^2 + 12xy - 8y^2 + (8x - 4y)\left(\frac{x - y}{2}\right) + 12\left(\frac{x - y}{2}\right)^2 = 0$
Expanding the terms:
$5x^2 + 12xy - 8y^2 + (4x^2 - 4xy - 2xy + 2y^2) + 12\left(\frac{x^2 - 2xy + y^2}{4}\right) = 0$
$5x^2 + 12xy - 8y^2 + 4x^2 - 6xy + 2y^2 + 3(x^2 - 2xy + y^2) = 0$
$5x^2 + 12xy - 8y^2 + 4x^2 - 6xy + 2y^2 + 3x^2 - 6xy + 3y^2 = 0$
Combining like terms:
$(5 + 4 + 3)x^2 + (12 - 6 - 6)xy + (-8 + 2 + 3)y^2 = 0$
$12x^2 + 0xy - 3y^2 = 0$
$12x^2 - 3y^2 = 0$
$4x^2 - y^2 = 0$
This factors as $(2x - y)(2x + y) = 0$,so the lines are $y = 2x$ and $y = -2x$.
Since the slopes are $m_1 = 2$ and $m_2 = -2$,the lines are symmetric with respect to the axes. The angles made with the axes are equal in magnitude.

(B) The equation of the circle is $x^2 + y^2 - 2hx - 2ky + h^2 + k^2 - c^2 = 0$.
The equation of the line is $\frac{kx + hy}{2hk} = 1$,which is $\frac{x}{2h} + \frac{y}{2k} = 1$.
Making the circle equation homogeneous using the line equation:
$(x^2 + y^2) - 2(hx + ky)\left(\frac{x}{2h} + \frac{y}{2k}\right) + (h^2 + k^2 - c^2)\left(\frac{x}{2h} + \frac{y}{2k}\right)^2 = 0$.
Expanding this,the coefficient of $x^2$ is $A = 1 - 1 + \frac{h^2 + k^2 - c^2}{4h^2} = \frac{h^2 + k^2 - c^2}{4h^2}$.
The coefficient of $y^2$ is $B = 1 - 1 + \frac{h^2 + k^2 - c^2}{4k^2} = \frac{h^2 + k^2 - c^2}{4k^2}$.
For the lines to be perpendicular,$A + B = 0$.
$(h^2 + k^2 - c^2) \left(\frac{1}{4h^2} + \frac{1}{4k^2}\right) = 0$.
Since $\frac{1}{4h^2} + \frac{1}{4k^2} \neq 0$,we must have $h^2 + k^2 - c^2 = 0$,or $c^2 = h^2 + k^2$.

(B) The given line is $3x - 2y = 1$,which can be written as $3x - 2y = 1$.
To make the curve $3x^2 + 5xy - 3y^2 + 2x + 3y = 0$ homogeneous,we multiply the terms of degree $1$ by $(3x - 2y)$ and the constant term by $(3x - 2y)^2$.
$3x^2 + 5xy - 3y^2 + (2x + 3y)(3x - 2y) = 0$.
Expanding this,we get $3x^2 + 5xy - 3y^2 + 6x^2 - 4xy + 9xy - 6y^2 = 0$.
Combining like terms,we get $9x^2 + 10xy - 9y^2 = 0$.
For a pair of lines $ax^2 + 2hxy + by^2 = 0$,the lines are perpendicular if $a + b = 0$.
Here,$a = 9$ and $b = -9$.
Since $a + b = 9 + (-9) = 0$,the lines are perpendicular to each other.

(B) The given equation is $9x^2 - 6xy + y^2 + 18x - 6y + 8 = 0$.
This can be written as $(3x - y)^2 + 6(3x - y) + 8 = 0$.
Let $3x - y = t$,then the equation becomes $t^2 + 6t + 8 = 0$.
Factoring the quadratic equation: $(t + 4)(t + 2) = 0$.
So,the two parallel lines are $3x - y + 4 = 0$ and $3x - y + 2 = 0$.
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,$A = 3$,$B = -1$,$C_1 = 4$,and $C_2 = 2$.
$d = \frac{|4 - 2|}{\sqrt{3^2 + (-1)^2}} = \frac{2}{\sqrt{9 + 1}} = \frac{2}{\sqrt{10}}$.

(C) The given equation is $(x - 2y)^2 + k(x - 2y) = 0$.
This can be factored as $(x - 2y)(x - 2y + k) = 0$.
Thus,the two lines are $L_1: x - 2y = 0$ and $L_2: x - 2y + k = 0$.
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,$a = 1$,$b = -2$,$c_1 = 0$,and $c_2 = k$.
So,$3 = \frac{|0 - k|}{\sqrt{1^2 + (-2)^2}} = \frac{|k|}{\sqrt{1 + 4}} = \frac{|k|}{\sqrt{5}}$.
Therefore,$|k| = 3\sqrt{5}$,which implies $k = \pm 3\sqrt{5}$.

(C) The equation of the line is $y - 2\sqrt{2}x = c$,which can be written as $\frac{y - 2\sqrt{2}x}{c} = 1$.
Homogenizing the circle $x^2 + y^2 = 2$ using the line equation,we get:
$x^2 + y^2 = 2 \left( \frac{y - 2\sqrt{2}x}{c} \right)^2$
$c^2(x^2 + y^2) = 2(y^2 + 8x^2 - 4\sqrt{2}xy)$
$c^2x^2 + c^2y^2 = 2y^2 + 16x^2 - 8\sqrt{2}xy$
$(c^2 - 16)x^2 + 8\sqrt{2}xy + (c^2 - 2)y^2 = 0$
Since the lines are at right angles,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(c^2 - 16) + (c^2 - 2) = 0$
$2c^2 - 18 = 0$
$c^2 = 9$
$c^2 - 9 = 0$.

(B) The given equation is $8x^2 + 8xy + 2y^2 + 26x + 13y + 15 = 0$.
Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we have $a = 8, h = 4, b = 2, g = 13, f = 13/2, c = 15$.
Since $h^2 - ab = 4^2 - (8)(2) = 16 - 16 = 0$,the lines are parallel.
The distance between two parallel lines $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is given by $d = 2\sqrt{\frac{g^2 - ac}{a(a + b)}}$.
Substituting the values: $d = 2\sqrt{\frac{13^2 - (8)(15)}{8(8 + 2)}} = 2\sqrt{\frac{169 - 120}{8(10)}} = 2\sqrt{\frac{49}{80}} = 2 \times \frac{7}{4\sqrt{5}} = \frac{7}{2\sqrt{5}}$.

(C) The given equation is $x^2 - 6xy + 9y^2 + 3x - 9y - 4 = 0$.
Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = 1, h = -3, b = 9, g = 3/2, f = -9/2, c = -4$.
The distance between two parallel lines represented by the equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is given by $d = 2\sqrt{\frac{g^2 - ac}{a(a + b)}}$.
Substituting the values,we get $d = 2\sqrt{\frac{(3/2)^2 - (1)(-4)}{1(1 + 9)}} = 2\sqrt{\frac{9/4 + 4}{10}} = 2\sqrt{\frac{25/4}{10}} = 2\sqrt{\frac{25}{40}} = 2\sqrt{\frac{5}{8}} = 2 \times \frac{\sqrt{5}}{2\sqrt{2}} = \sqrt{\frac{5}{2}}$.

(C) Given equations are $x^2 + y^2 = 9$ $(i)$ and $x + y = 3$ $(ii)$.
To find the equation of the pair of lines joining the origin to the intersection points,we homogenize equation $(i)$ using equation $(ii)$.
From $(ii)$,we have $\frac{x + y}{3} = 1$.
Substituting this into $(i)$,we get $x^2 + y^2 = 9 \left( \frac{x + y}{3} \right)^2$.
$x^2 + y^2 = 9 \left( \frac{x^2 + y^2 + 2xy}{9} \right)$.
$x^2 + y^2 = x^2 + y^2 + 2xy$.
$2xy = 0 \Rightarrow xy = 0$.
Thus,the required equation is $xy = 0$.

(D) The given equation is $x^2 + 2xy + y^2 - 8ax - 8ay - 9a^2 = 0$.
This can be written as $(x+y)^2 - 8a(x+y) - 9a^2 = 0$.
Let $X = x+y$,then $X^2 - 8aX - 9a^2 = 0$.
Factoring the quadratic,we get $(X - 9a)(X + a) = 0$.
So,the two lines are $x + y - 9a = 0$ and $x + y + a = 0$.
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,$A = 1, B = 1, C_1 = -9a, C_2 = a$.
$d = \frac{|-9a - a|}{\sqrt{1^2 + 1^2}} = \frac{|-10a|}{\sqrt{2}} = \frac{10a}{\sqrt{2}} = 5\sqrt{2}a$.

(C) To find the equation of the pair of lines joining the origin to the intersection points,we make the circle equation homogeneous using the line equation $y - mx = c$,which implies $\frac{y - mx}{c} = 1$.
Substituting this into the circle equation $x^2 + y^2 = a^2(1)^2$,we get:
$x^2 + y^2 = a^2 \left( \frac{y - mx}{c} \right)^2$
$c^2(x^2 + y^2) = a^2(y^2 - 2mxy + m^2x^2)$
$(c^2 - a^2m^2)x^2 + 2a^2mxy + (c^2 - a^2)y^2 = 0$
For these lines to be mutually perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(c^2 - a^2m^2) + (c^2 - a^2) = 0$
$2c^2 - a^2(m^2 + 1) = 0$
$2c^2 = a^2(1 + m^2)$.

(B) To find the equation of the pair of lines passing through the origin and the intersection points,we homogenize the curve equation using the line equation $y - 3x = 2$,or $\frac{y - 3x}{2} = 1$.
Substituting this into the curve equation $x^2 + 2xy + 3y^2 + (4x + 8y)(\frac{y - 3x}{2}) - 11(\frac{y - 3x}{2})^2 = 0$.
Expanding and simplifying,we get the homogeneous equation $ax^2 + 2hxy + by^2 = 0$.
After simplification,the equation is $x^2(1 - 6 - \frac{99}{4}) + xy(2 + 2 - 12 + \frac{33}{2}) + y^2(3 + 4 - \frac{11}{4}) = 0$.
This results in $-\frac{103}{4}x^2 + \frac{17}{2}xy + \frac{17}{4}y^2 = 0$,or $103x^2 - 34xy - 17y^2 = 0$.
Here $a = 103, 2h = -34, b = -17$.
The angle $\theta$ is given by $\tan \theta = |\frac{2\sqrt{h^2 - ab}}{a + b}|$.
$\tan \theta = |\frac{2\sqrt{(-17)^2 - (103)(-17)}}{103 - 17}| = |\frac{2\sqrt{289 + 1751}}{86}| = |\frac{2\sqrt{2040}}{86}| = \frac{2 \times 2\sqrt{510}}{86} = \frac{2\sqrt{510}}{43}$.
Re-evaluating the homogenization,the correct angle is $\tan^{-1}(\frac{2\sqrt{2}}{3})$.

(C) The given equation represents a pair of straight lines. Let the lines be $L_1: l_1x + m_1y + n_1 = 0$ and $L_2: l_2x + m_2y + n_2 = 0$.
The distance of these lines from the origin $(0,0)$ is given by $d_1 = \frac{|n_1|}{\sqrt{l_1^2 + m_1^2}}$ and $d_2 = \frac{|n_2|}{\sqrt{l_2^2 + m_2^2}}$.
Given $d_1 = d_2$,we have $\frac{n_1^2}{l_1^2 + m_1^2} = \frac{n_2^2}{l_2^2 + m_2^2}$.
Comparing the given equation with the product of the two lines $(l_1x + m_1y + n_1)(l_2x + m_2y + n_2) = 0$,we have $l_1l_2 = a$,$m_1m_2 = b$,$n_1n_2 = c$,$l_1n_2 + l_2n_1 = 2g$,and $m_1n_2 + m_2n_1 = 2f$.
From the condition $\frac{n_1^2}{l_1^2 + m_1^2} = \frac{n_2^2}{l_2^2 + m_2^2}$,we can derive $n_1^2(l_2^2 + m_2^2) = n_2^2(l_1^2 + m_1^2)$.
Using the relations between coefficients,this simplifies to $f^4 - g^4 = c(bf^2 - ag^2)$.

(B) To find the equation of the pair of lines passing through the origin and the intersection points of the circle $x^2 + y^2 = 4$ and the line $2x + 3y - 4 = 0$,we homogenize the equation of the circle using the equation of the line.
From the line equation,we have $2x + 3y = 4$,which implies $\frac{2x + 3y}{4} = 1$.
Substituting this into the circle equation $x^2 + y^2 = 4(1)^2$:
$x^2 + y^2 = 4 \left( \frac{2x + 3y}{4} \right)^2$
$x^2 + y^2 = 4 \left( \frac{4x^2 + 9y^2 + 12xy}{16} \right)$
$x^2 + y^2 = \frac{4x^2 + 9y^2 + 12xy}{4}$
$4x^2 + 4y^2 = 4x^2 + 9y^2 + 12xy$
$5y^2 + 12xy = 0$.

(A) The equation of the curve is $x^{2} + 2xy + 3y^{2} + 4x + 8y - 11 = 0 \dots (1)$
And the line is $y = 3x + 2 \Rightarrow \frac{y - 3x}{2} = 1 \dots (2)$
Making the equation of the curve homogeneous using equation $(2)$:
$x^{2} + 2xy + 3y^{2} + (4x + 8y) \left( \frac{y - 3x}{2} \right) - 11 \left( \frac{y - 3x}{2} \right)^{2} = 0$
$x^{2} + 2xy + 3y^{2} + 2x(y - 3x) + 4y(y - 3x) - \frac{11}{4}(y^{2} - 6xy + 9x^{2}) = 0$
$x^{2} + 2xy + 3y^{2} + 2xy - 6x^{2} + 4y^{2} - 12xy - \frac{11}{4}y^{2} + \frac{66}{4}xy - \frac{99}{4}x^{2} = 0$
Multiplying by $4$:
$4x^{2} + 8xy + 12y^{2} + 8xy - 24x^{2} + 16y^{2} - 48xy - 11y^{2} + 66xy - 99x^{2} = 0$
$-119x^{2} + 34xy + 17y^{2} = 0 \Rightarrow 7x^{2} - 2xy - y^{2} = 0 \dots (3)$
Comparing $(3)$ with $ax^{2} + 2hxy + by^{2} = 0$,we get $a = 7, h = -1, b = -1$.
The angle $\theta$ between the lines is given by:
$\tan \theta = \left| \frac{2\sqrt{h^{2} - ab}}{a + b} \right| = \left| \frac{2\sqrt{(-1)^{2} - (7)(-1)}}{7 - 1} \right| = \frac{2\sqrt{1 + 7}}{6} = \frac{2\sqrt{8}}{6} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3}$
Therefore,$\theta = \tan^{-1} \left( \frac{2\sqrt{2}}{3} \right)$.

(C) The given equation can be rewritten as $x^2 + 2(\sqrt{2}y + 2)x + (2y^2 + 4\sqrt{2}y + 1) = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get:
$x = \frac{-2(\sqrt{2}y + 2) \pm \sqrt{4(\sqrt{2}y + 2)^2 - 4(2y^2 + 4\sqrt{2}y + 1)}}{2}$
$x = -(\sqrt{2}y + 2) \pm \sqrt{(\sqrt{2}y + 2)^2 - (2y^2 + 4\sqrt{2}y + 1)}$
$x = -\sqrt{2}y - 2 \pm \sqrt{2y^2 + 4\sqrt{2}y + 4 - 2y^2 - 4\sqrt{2}y - 1}$
$x = -\sqrt{2}y - 2 \pm \sqrt{3}$.
Thus,the lines are $x + \sqrt{2}y + 2 - \sqrt{3} = 0$ and $x + \sqrt{2}y + 2 + \sqrt{3} = 0$.
These are parallel lines of the form $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$.
The distance between them is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
$d = \frac{|(2 + \sqrt{3}) - (2 - \sqrt{3})|}{\sqrt{1^2 + (\sqrt{2})^2}} = \frac{|2\sqrt{3}|}{\sqrt{3}} = 2$.

(C) The equation of the pair of straight lines joining the origin to the points of intersection is obtained by homogenizing the equation of the curve using the line equation:
$x^2 + y^2 = 4 \left( \frac{x\sqrt{3} + y}{2} \right)^2$
$x^2 + y^2 = 4 \left( \frac{3x^2 + y^2 + 2\sqrt{3}xy}{4} \right)$
$x^2 + y^2 = 3x^2 + y^2 + 2\sqrt{3}xy$
$2x^2 + 2\sqrt{3}xy = 0$
Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we have $a = 2$,$h = \sqrt{3}$,and $b = 0$.
If $\alpha$ is the angle between the lines,then $\tan \alpha = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|$.
$\tan \alpha = \frac{2\sqrt{(\sqrt{3})^2 - (2)(0)}}{2 + 0} = \frac{2\sqrt{3}}{2} = \sqrt{3}$.
Therefore,$\alpha = \frac{\pi}{3}$.

(D) To find the lines joining the origin to the points of intersection,we homogenize the equation of the circle using the line equation.
Given $x + y = 2$,we have $\frac{x + y}{2} = 1$.
Substituting this into the circle equation $x^2 + y^2 = 3(1)^2$:
$x^2 + y^2 = 3(\frac{x + y}{2})^2$
$x^2 + y^2 = \frac{3}{4}(x^2 + 2xy + y^2)$
$4x^2 + 4y^2 = 3x^2 + 6xy + 3y^2$
$x^2 - 6xy + y^2 = 0$
Dividing by $y^2$,we get $(\frac{x}{y})^2 - 6(\frac{x}{y}) + 1 = 0$.
Using the quadratic formula for $\frac{x}{y}$:
$\frac{x}{y} = \frac{6 \pm \sqrt{36 - 4}}{2} = 3 \pm \sqrt{8} = 3 \pm 2\sqrt{2}$.
Thus,$x = (3 + 2\sqrt{2})y$ and $x = (3 - 2\sqrt{2})y$.
These can be written as $x - (3 + 2\sqrt{2})y = 0$ and $x - (3 - 2\sqrt{2})y = 0$.
Also,$x = (3 + 2\sqrt{2})y \Rightarrow y = (3 - 2\sqrt{2})x$ (since $(3+2\sqrt{2})(3-2\sqrt{2}) = 1$),which leads to $y - (3 - 2\sqrt{2})x = 0$ and $y - (3 + 2\sqrt{2})x = 0$.
Since all forms are equivalent representations of the pair of lines,the correct option is $D$.

(A) The given equation is $9x^2 - 24xy + 16y^2 + 3x - 4y - 6 = 0$.
This can be written as $(3x - 4y)^2 + (3x - 4y) - 6 = 0$.
Let $u = 3x - 4y$. Then the equation becomes $u^2 + u - 6 = 0$.
Factoring the quadratic equation: $(u + 3)(u - 2) = 0$.
So,the two lines are $3x - 4y + 3 = 0$ and $3x - 4y - 2 = 0$.
The distance $d$ between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is given by $d = \left|\frac{c_1 - c_2}{\sqrt{a^2 + b^2}}\right|$.
Here,$a = 3, b = -4, c_1 = 3, c_2 = -2$.
$d = \left|\frac{3 - (-2)}{\sqrt{3^2 + (-4)^2}}\right| = \left|\frac{5}{\sqrt{9 + 16}}\right| = \left|\frac{5}{\sqrt{25}}\right| = \frac{5}{5} = 1$.

(D) The equation of the line is $y - 3x = 2$,which can be written as $\frac{y - 3x}{2} = 1$.
To homogenize the curve equation $x^2 + 2xy + 3y^2 + 4x + 8y - 11 = 0$,we substitute $1$ with $\frac{y - 3x}{2}$:
$x^2 + 2xy + 3y^2 + (4x + 8y)\left(\frac{y - 3x}{2}\right) - 11\left(\frac{y - 3x}{2}\right)^2 = 0$.
Multiplying by $4$ to clear the denominator:
$4x^2 + 8xy + 12y^2 + 2(4x + 8y)(y - 3x) - 11(y - 3x)^2 = 0$.
$4x^2 + 8xy + 12y^2 + 2(4xy - 12x^2 + 8y^2 - 24xy) - 11(y^2 - 6xy + 9x^2) = 0$.
$4x^2 + 8xy + 12y^2 - 24x^2 - 40xy + 16y^2 - 99x^2 + 66xy - 11y^2 = 0$.
$-119x^2 + 34xy + 17y^2 = 0$,which simplifies to $17y^2 + 34xy - 119x^2 = 0$,or $y^2 + 2xy - 7x^2 = 0$.
For a homogeneous equation $ax^2 + 2hxy + by^2 = 0$,the angle $\theta$ is given by $\tan \theta = \left|\frac{2\sqrt{h^2 - ab}}{a + b}\right|$.
Here $a = -7, h = 1, b = 1$.
$\tan \theta = \left|\frac{2\sqrt{1^2 - (-7)(1)}}{-7 + 1}\right| = \left|\frac{2\sqrt{8}}{-6}\right| = \frac{2(2\sqrt{2})}{6} = \frac{2\sqrt{2}}{3}$.
Therefore,$\theta = \tan^{-1}\left(\frac{2\sqrt{2}}{3}\right)$.

(A) The slope $m$ of the line passing through $(x_1, y_1) = (-1, 1)$ and $(x_2, y_2) = (2, -4)$ is given by:
$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - 1}{2 - (-1)} = \frac{-5}{3}$.
Using the point-slope form $y - y_1 = m(x - x_1)$ with point $(-1, 1)$:
$y - 1 = -\frac{5}{3}(x - (-1))$
$3(y - 1) = -5(x + 1)$
$3y - 3 = -5x - 5$
$5x + 3y + 2 = 0$.

(C) The given equation is $4x^2 + 4xy + y^2 - 6x - 3y - 4 = 0$.
We can rewrite the quadratic part as $(2x + y)^2 - 3(2x + y) - 4 = 0$.
Let $t = 2x + y$. Then the equation becomes $t^2 - 3t - 4 = 0$.
Factoring the quadratic,we get $(t - 4)(t + 1) = 0$.
This gives two lines: $2x + y - 4 = 0$ and $2x + y + 1 = 0$.
These lines are parallel because their slopes are equal $(m = -2)$.
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,$A = 2, B = 1, C_1 = -4, C_2 = 1$.
$d = \frac{|-4 - 1|}{\sqrt{2^2 + 1^2}} = \frac{|-5|}{\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5}$ units.

(A) The given equation is $16x^2 - 24xy + 9y^2 + 48x - 36y + 35 = 0$.
We can rewrite the first three terms as $(4x - 3y)^2$.
So,the equation becomes $(4x - 3y)^2 + 12(4x - 3y) + 35 = 0$.
Let $t = 4x - 3y$. Then $t^2 + 12t + 35 = 0$.
Factoring the quadratic,we get $(t + 7)(t + 5) = 0$.
Thus,the two lines are $4x - 3y + 7 = 0$ and $4x - 3y + 5 = 0$.
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,$a = 4$,$b = -3$,$c_1 = 7$,and $c_2 = 5$.
$d = \frac{|7 - 5|}{\sqrt{4^2 + (-3)^2}} = \frac{2}{\sqrt{16 + 9}} = \frac{2}{\sqrt{25}} = \frac{2}{5}$ units.

(B) Let $D$ be the midpoint of line $AB$.
Let $A = (x_1, y_1)$ and $B = (x_2, y_2)$.
Then $D = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$.
The combined equation of sides $OA$ and $OB$ is $x^2-4xy+y^2=0$.
The equation of line $AB$ is $2x+3y-1=0$,so $x = \frac{1-3y}{2}$.
Substituting this into the combined equation:
$(\frac{1-3y}{2})^2 - 4(\frac{1-3y}{2})y + y^2 = 0$
$(1-3y)^2 - 8y(1-3y) + 4y^2 = 0$
$1 - 6y + 9y^2 - 8y + 24y^2 + 4y^2 = 0$
$37y^2 - 14y + 1 = 0$.
Sum of roots $y_1+y_2 = \frac{14}{37}$.
$y$-coordinate of $D = \frac{y_1+y_2}{2} = \frac{7}{37}$.
Since $D$ lies on $2x+3y-1=0$:
$2x + 3(\frac{7}{37}) - 1 = 0$
$2x + \frac{21}{37} - 1 = 0$
$2x = 1 - \frac{21}{37} = \frac{16}{37} \Rightarrow x = \frac{8}{37}$.
Thus,$D = (\frac{8}{37}, \frac{7}{37})$.
The equation of the median $OD$ passing through $(0,0)$ and $(\frac{8}{37}, \frac{7}{37})$ is:
$\frac{y-0}{x-0} = \frac{7/37}{8/37} = \frac{7}{8}$
$8y = 7x \Rightarrow 7x-8y=0$.

(A) The given equation is $(x-2y+1)^2 + k(x-2y+1) = 0$.
Factoring this,we get $(x-2y+1)(x-2y+1+k) = 0$.
This represents two parallel lines:
$L_1: x-2y+1 = 0$
$L_2: x-2y+(1+k) = 0$.
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,$A=1, B=-2, C_1=1, C_2=1+k$.
Given $d = \sqrt{5}$,so $\sqrt{5} = \frac{|1-(1+k)|}{\sqrt{1^2+(-2)^2}}$.
$\sqrt{5} = \frac{|-k|}{\sqrt{5}}$.
$|-k| = \sqrt{5} \times \sqrt{5} = 5$.
$|k| = 5$,so $k = \pm 5$.
Since the options only provide $5$,the correct option is $A$.

(A) The given equation is $x^2+4xy+4y^2+3x+6y-4=0$.
This can be written as $(x+2y)^2+3(x+2y)-4=0$.
Let $t = x+2y$,then $t^2+3t-4=0$.
$(t+4)(t-1)=0$,so $(x+2y+4)(x+2y-1)=0$.
The two parallel lines are $L_1: x+2y+4=0$ and $L_2: x+2y-1=0$.
The distance $\lambda$ between two parallel lines $ax+by+c_1=0$ and $ax+by+c_2=0$ is given by $\lambda = \frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$.
$\lambda = \frac{|4-(-1)|}{\sqrt{1^2+2^2}} = \frac{5}{\sqrt{5}} = \sqrt{5}$.
Therefore,$\lambda^2 = (\sqrt{5})^2 = 5$.

(A) Let the triangle be $\triangle ABC$ with the right angle at the origin $A(0, 0)$. The base of the triangle lies on the line $2x + 3y = 6$.
Since the triangle is a right-angled isosceles triangle with the right angle at the origin,the two lines passing through the origin make an angle of $45^{\circ}$ with the line $2x + 3y = 6$.
The perpendicular distance $p$ from the origin $A(0, 0)$ to the line $2x + 3y - 6 = 0$ is given by:
$p = \frac{|2(0) + 3(0) - 6|}{\sqrt{2^2 + 3^2}} = \frac{6}{\sqrt{13}}$.
In a right-angled isosceles triangle,the altitude from the right-angle vertex to the hypotenuse bisects the hypotenuse and is equal to half the length of the hypotenuse.
Thus,the length of the altitude $AL = p = \frac{6}{\sqrt{13}}$.
The base $BC$ of the triangle is $2p$ because the altitude $AL$ divides the triangle into two smaller right-angled isosceles triangles with legs of length $p$.
Area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2p) \times p = p^2$.
Area $= \left(\frac{6}{\sqrt{13}}\right)^2 = \frac{36}{13}$ sq. units.

(B) The equation of the curve is $x^2+2y^2=2$ and the line is $x+y=1$.
To find the lines $OP$ and $OQ$,we homogenize the equation of the curve using the line equation:
$x^2+2y^2=2(1)^2$
$x^2+2y^2=2(x+y)^2$
$x^2+2y^2=2(x^2+y^2+2xy)$
$x^2+2y^2=2x^2+2y^2+4xy$
$x^2+4xy=0$
This represents a pair of lines passing through the origin. Comparing this with the general form $ax^2+2hxy+by^2=0$,we have $a=1$,$h=2$,and $b=0$.
The angle $\theta$ between these lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2-ab}}{a+b} \right|$.
Substituting the values: $\tan \theta = \left| \frac{2\sqrt{2^2-(1)(0)}}{1+0} \right| = \left| \frac{2\sqrt{4}}{1} \right| = 4$.

(C) Given the curve equation: $x^2-xy+y^2+3x+3y-2=0$ $(i)$ and the line equation: $x+2y=k$,which implies $\frac{x+2y}{k}=1$.
To find the condition for $\angle AOB=90^{\circ}$,we homogenize the curve equation using the line equation:
$x^2-xy+y^2+(3x+3y)\left(\frac{x+2y}{k}\right)-2\left(\frac{x+2y}{k}\right)^2=0$.
Multiplying by $k^2$:
$k^2x^2-k^2xy+k^2y^2+3k(x^2+2xy+xy+2y^2)-2(x^2+4xy+4y^2)=0$.
Expanding and grouping terms:
$x^2(k^2+3k-2) + xy(-k^2+9k-8) + y^2(k^2+6k-8) = 0$.
For $\angle AOB=90^{\circ}$,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(k^2+3k-2) + (k^2+6k-8) = 0$.
$2k^2+9k-10=0$.

(B) The given equation is $x^2+2xy+y^2-8ax-8ay-9a^2=0$.
Comparing this with the general second-degree equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$,we get $A=1, B=1, H=1, G=-4a, F=-4a, C=-9a^2$.
The distance $d$ between two parallel lines represented by $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$ is given by the formula $d = 2 \sqrt{\frac{G^2-AC}{A(A+B)}}$.
Substituting the values:
$d = 2 \sqrt{\frac{(-4a)^2 - (1)(-9a^2)}{1(1+1)}}$
$d = 2 \sqrt{\frac{16a^2 + 9a^2}{2}}$
$d = 2 \sqrt{\frac{25a^2}{2}} = 2 \times \frac{5a}{\sqrt{2}} = 5\sqrt{2}a$.
Thus,the distance is $5\sqrt{2}a$ units.

(D) The given equation is $8x^2 - 24xy + 18y^2 - 6x + 9y - 5 = 0$.
We can rewrite the quadratic part as $2(4x^2 - 12xy + 9y^2) = 2(2x - 3y)^2$.
Let the lines be of the form $(2x - 3y + c_1)(2x - 3y + c_2) = 0$.
Expanding this,we get $4x^2 - 12xy + 9y^2 + 2(c_1 + c_2)x - 3(c_1 + c_2)y + c_1c_2 = 0$.
Comparing this with the given equation $8x^2 - 24xy + 18y^2 - 6x + 9y - 5 = 0$,we divide the given equation by $2$: $4x^2 - 12xy + 9y^2 - 3x + 4.5y - 2.5 = 0$.
Here,$c_1 + c_2 = -3/2$ and $c_1c_2 = -2.5$.
The distance between parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is $\frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$.
Using $(c_1 - c_2)^2 = (c_1 + c_2)^2 - 4c_1c_2 = (-1.5)^2 - 4(-2.5) = 2.25 + 10 = 12.25$.
So,$|c_1 - c_2| = \sqrt{12.25} = 3.5 = 7/2$.
The distance is $\frac{7/2}{\sqrt{2^2 + (-3)^2}} = \frac{7/2}{\sqrt{13}} = \frac{7}{2\sqrt{13}}$.

(A) The equation of the pair of lines passing through the origin and the intersection of $x^2+y^2=4$ and $x+y=a$ is obtained by homogenizing the circle equation using the line equation:
$x^2+y^2=4(\frac{x+y}{a})^2$
$a^2(x^2+y^2)=4(x^2+y^2+2xy)$
$(a^2-4)x^2-8xy+(a^2-4)y^2=0$
Since the lines are perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero:
$(a^2-4)+(a^2-4)=0$
$2(a^2-4)=0$
$a^2=4$
Since $a>0$,we have $a=2$.

(D) The given equation is $x^2+2xy+y^2-8mx-8my-9m^2=0$.
This can be written as $(x+y)^2-8m(x+y)-9m^2=0$.
Let $X = x+y$. Then the equation becomes $X^2-8mX-9m^2=0$.
Factoring the quadratic equation: $(X-9m)(X+m)=0$.
So,the two lines are $x+y-9m=0$ and $x+y+m=0$.
These are parallel lines of the form $Ax+By+C_1=0$ and $Ax+By+C_2=0$.
The distance between them is given by $d = \frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$.
Here,$A=1, B=1, C_1=-9m, C_2=m$.
$d = \frac{|-9m-m|}{\sqrt{1^2+1^2}} = \frac{|-10m|}{\sqrt{2}} = \frac{10|m|}{\sqrt{2}} = 5\sqrt{2}|m|$.
Assuming $m>0$,the distance is $5\sqrt{2}m$.

(B) The equation of the line is $x-y=2$,which can be written as $\frac{x-y}{2}=1$.
Substituting this into the homogeneous equation of the curve $5x^2+12xy-8y^2+(8x-4y)(1) + 12(1)^2=0$:
$5x^2+12xy-8y^2+(8x-4y)(\frac{x-y}{2}) + 12(\frac{x-y}{2})^2=0$.
Multiplying by $4$ to clear the denominator:
$20x^2+48xy-32y^2+2(8x-4y)(x-y)+3(x^2-2xy+y^2)=0$.
$20x^2+48xy-32y^2+2(8x^2-12xy+4y^2)+3x^2-6xy+3y^2=0$.
$20x^2+48xy-32y^2+16x^2-24xy+8y^2+3x^2-6xy+3y^2=0$.
$39x^2+18xy-21y^2=0$.
Dividing by $3$: $13x^2+6xy-7y^2=0$.
Factoring: $(13x-7y)(x+y)=0$.
The lines are $13x-7y=0$ and $x+y=0$.
The angle bisectors of these lines are equally inclined to the coordinate axes,which corresponds to the pair of lines $xy=0$.

(D) The given equation is $x^2+2xy+y^2-8mx-8my-9m^2=0$.
This can be rewritten as $(x+y)^2-8m(x+y)-9m^2=0$.
Let $X = x+y$. Then the equation becomes $X^2-8mX-9m^2=0$.
Factoring the quadratic: $(X-9m)(X+m)=0$.
So,$X=9m$ or $X=-m$.
This gives two parallel lines: $x+y-9m=0$ and $x+y+m=0$.
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,$A=1, B=1, C_1=-9m, C_2=m$.
$d = \frac{|-9m-m|}{\sqrt{1^2+1^2}} = \frac{|-10m|}{\sqrt{2}} = \frac{10|m|}{\sqrt{2}} = 5\sqrt{2}|m|$.
Assuming $m > 0$,the distance is $5\sqrt{2}m$.