Equation of pair of straight lines Questions in English

(D) The given equation is $ax^2 - bxy - y^2 = 0$. Dividing by $x^2$,we get $-(\frac{y}{x})^2 - b(\frac{y}{x}) + a = 0$,which is $(\frac{y}{x})^2 + b(\frac{y}{x}) - a = 0$.
Let $m_1 = \tan \alpha$ and $m_2 = \tan \beta$ be the slopes of the lines.
These are the roots of the quadratic equation $m^2 + bm - a = 0$.
From the properties of roots,we have $\tan \alpha + \tan \beta = -b$ and $\tan \alpha \tan \beta = -a$.
Using the formula $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$,we substitute the values:
$\tan(\alpha + \beta) = \frac{-b}{1 - (-a)} = \frac{-b}{1+a}$.

(B) Let the slopes of the lines be $m_{1}$ and $m_{2}$.
Given the equation $ax^{2} + 2hxy + by^{2} = 0$,we have $m_{1} + m_{2} = \frac{-2h}{b}$ and $m_{1}m_{2} = \frac{a}{b}$.
Given the ratio of slopes is $m_{1}:m_{2} = 5:3$,let $m_{1} = 5k$ and $m_{2} = 3k$.
Then $m_{1} + m_{2} = 8k = \frac{-2h}{b} \Rightarrow k = \frac{-h}{4b}$.
Also $m_{1}m_{2} = 15k^{2} = \frac{a}{b}$.
Substituting $k$ in the second equation: $15 \left( \frac{-h}{4b} \right)^{2} = \frac{a}{b}$.
$15 \left( \frac{h^{2}}{16b^{2}} \right) = \frac{a}{b}$.
$\frac{15h^{2}}{16b} = a$.
Therefore,$\frac{h^{2}}{ab} = \frac{16}{15}$.

(C) The given equation is $K x^2 + 5 x y + y^2 = 0$.
Comparing this with the general equation $a x^2 + 2 h x y + b y^2 = 0$,we have $a = K$,$2h = 5$,and $b = 1$.
Let the slopes of the lines be $m_1$ and $m_2$.
Then,$m_1 + m_2 = -\frac{2h}{b} = -5$ and $m_1 m_2 = \frac{a}{b} = K$.
It is given that the difference between the slopes is $1$,i.e.,$|m_1 - m_2| = 1$.
Using the identity $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4 m_1 m_2$,we substitute the values:
$1^2 = (-5)^2 - 4(K)$.
$1 = 25 - 4K$.
$4K = 24$.
$K = 6$.

(B) The equation of the pair of lines is $k x^2 + x y - y^2 = 0$.
Since the lines bisect the angle between the coordinate axes,their slopes must be $m = \pm 1$.
Substituting $y = mx$ into the equation,we get $k x^2 + x(mx) - (mx)^2 = 0$.
Dividing by $x^2$ (assuming $x \neq 0$),we get $k + m - m^2 = 0$.
For $m = 1$,$k + 1 - 1^2 = 0 \Rightarrow k = 0$.
For $m = -1$,$k - 1 - (-1)^2 = 0$ $\Rightarrow k - 1 - 1 = 0$ $\Rightarrow k = 2$.
Thus,the values of $k$ are $0$ and $2$.

(C) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
For this to represent a pair of straight lines,the condition is $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Comparing the given equation $kxy + 10x + 8y + 16 = 0$ with the general form,we have:
$a = 0, b = 0, c = 16, h = k/2, g = 5, f = 4$.
Substituting these values into the condition:
$(0)(0)(16) + 2(4)(5)(k/2) - (0)(4^2) - (0)(5^2) - (16)(k/2)^2 = 0$.
$0 + 20k - 0 - 0 - 16(k^2/4) = 0$.
$20k - 4k^2 = 0$.
$4k(5 - k) = 0$.
Thus,$k = 0$ or $k = 5$.

(A) The given equation of the pair of lines is $ax^2 + (2a + 1)xy + 2y^2 = 0$.
Comparing this with $Ax^2 + 2Hxy + By^2 = 0$,we have $A = a$,$2H = 2a + 1$,and $B = 2$.
Let the slopes of the lines be $m_1$ and $m_2$.
Given that $m_1 = \frac{1}{m_2}$,which implies $m_1 m_2 = 1$.
For a pair of lines $ax^2 + 2hxy + by^2 = 0$,the product of slopes is $m_1 m_2 = \frac{A}{B} = \frac{a}{2}$.
Equating the two,we get $\frac{a}{2} = 1$,so $a = 2$.
The sum of the slopes is $m_1 + m_2 = -\frac{2H}{B} = -\frac{2a + 1}{2}$.
Substituting $a = 2$,we get $m_1 + m_2 = -\frac{2(2) + 1}{2} = -\frac{5}{2}$.
We need to find $m_1^2 + m_2^2$.
Using the identity $m_1^2 + m_2^2 = (m_1 + m_2)^2 - 2m_1 m_2$,we get:
$m_1^2 + m_2^2 = (-\frac{5}{2})^2 - 2(1) = \frac{25}{4} - 2 = \frac{25 - 8}{4} = \frac{17}{4}$.

(B) The slope of line $QR$ is $m = -2$. Let the slopes of lines $PQ$ and $PR$ be $m_1$ and $m_2$ respectively.
Since $\triangle PQR$ is an isosceles right-angled triangle,the angles $\angle PQR = \angle PRQ = 45^{\circ}$.
Using the formula $\tan \theta = \left| \frac{m - m_1}{1 + m \cdot m_1} \right|$,we have $\tan 45^{\circ} = \left| \frac{-2 - m_1}{1 + (-2)m_1} \right| = 1$.
This gives $\left| \frac{2 + m_1}{1 - 2m_1} \right| = 1$,so $2 + m_1 = 1 - 2m_1$ or $2 + m_1 = -(1 - 2m_1)$.
Solving $3m_1 = -1$ gives $m_1 = -1/3$. Solving $-m_1 = -3$ gives $m_2 = 3$.
The lines $PQ$ and $PR$ pass through $P(2, 1)$ with slopes $-1/3$ and $3$.
The equations are $y - 1 = -1/3(x - 2) \Rightarrow x + 3y - 5 = 0$ and $y - 1 = 3(x - 2) \Rightarrow 3x - y - 5 = 0$.
The joint equation is $(x + 3y - 5)(3x - y - 5) = 0$.
Expanding this: $3x^2 - xy - 5x + 9xy - 3y^2 - 15y - 15x + 5y + 25 = 0$.
Simplifying gives $3x^2 - 3y^2 + 8xy - 20x - 10y + 25 = 0$.

(C) The equation of the pair of lines is $ax^2 + 2hxy + by^2 = 0$. Let $m_1$ and $m_2$ be the slopes of the lines.
We know that $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$.
The difference of the slopes is given by $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1m_2$.
$(m_1 - m_2)^2 = \left(-\frac{2h}{b}\right)^2 - 4\left(\frac{a}{b}\right) = \frac{4h^2}{b^2} - \frac{4a}{b} = \frac{4h^2 - 4ab}{b^2}$.
Given $4ab = 3h^2$,we substitute this into the expression:
$(m_1 - m_2)^2 = \frac{4h^2 - 3h^2}{b^2} = \frac{h^2}{b^2}$.
Thus,$m_1 - m_2 = \pm \frac{h}{b}$.
Taking $m_1 - m_2 = \frac{h}{b}$ and $m_1 + m_2 = -\frac{2h}{b}$,we solve for $m_1$ and $m_2$:
$2m_1 = -\frac{2h}{b} + \frac{h}{b} = -\frac{h}{b} \Rightarrow m_1 = -\frac{h}{2b}$.
$2m_2 = -\frac{2h}{b} - \frac{h}{b} = -\frac{3h}{b} \Rightarrow m_2 = -\frac{3h}{2b}$.
The ratio $m_1 : m_2 = \left(-\frac{h}{2b}\right) : \left(-\frac{3h}{2b}\right) = 1 : 3$.

(A) The given equation is $x^{2}-y^{2}=0$,which can be written as $(x-y)(x+y)=0$.
Thus,the slopes of the given lines are $m_{1}=1$ and $m_{2}=-1$.
The equations of the lines passing through $(2,3)$ and parallel to these lines are:
$(y-3)=1(x-2) \Rightarrow x-y+1=0$
$(y-3)=-1(x-2) \Rightarrow x+y-5=0$
The joint equation is the product of these two lines:
$(x-y+1)(x+y-5)=0$
Expanding this:
$x(x+y-5) - y(x+y-5) + 1(x+y-5) = 0$
$x^{2}+xy-5x-xy-y^{2}+5y+x+y-5=0$
$x^{2}-y^{2}-4x+6y-5=0$

(C) The general equation of a second-degree curve is $Ax^{2} + 2Hxy + By^{2} + 2Gx + 2Fy + C = 0$.
For this to represent a pair of lines,the condition is $\Delta = ABC + 2FGH - AF^{2} - BG^{2} - CH^{2} = 0$.
Comparing the given equation $kxy + 5x + 3y + 2 = 0$ with the general form,we have:
$A = 0, B = 0, C = 2, 2H = k$ $\Rightarrow H = \frac{k}{2}, 2G = 5$ $\Rightarrow G = \frac{5}{2}, 2F = 3$ $\Rightarrow F = \frac{3}{2}$.
Substituting these values into the condition $\Delta = 0$:
$0 + 2(\frac{3}{2})(\frac{5}{2})(\frac{k}{2}) - 0 - 0 - 2(\frac{k}{2})^{2} = 0$.
$\frac{15k}{4} - \frac{2k^{2}}{4} = 0$.
$15k - 2k^{2} = 0$.
$k(15 - 2k) = 0$.
Thus,$k = 0$ or $k = \frac{15}{2}$.

(D) Given equation: $4x^{2}-y^{2}+2x+y=0$
We can rewrite the expression as: $(2x)^{2} - y^{2} + (2x+y) = 0$
Using the identity $a^{2}-b^{2} = (a-b)(a+b)$,we get: $(2x-y)(2x+y) + (2x+y) = 0$
Factoring out $(2x+y)$: $(2x+y)(2x-y+1) = 0$
Thus,the separate equations are $2x+y=0$ and $2x-y+1=0$.

(A) The given equation of the pair of lines is $x^2-4pxy+8y^2=0$.
This is in the form $ax^2+2hxy+by^2=0$,where $a=1$,$2h=-4p$,and $b=8$.
Let $m_1$ and $m_2$ be the slopes of the lines.
We know that $m_1+m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$.
Substituting the values,$m_1+m_2 = -\frac{-4p}{8} = \frac{4p}{8} = \frac{p}{2}$ and $m_1m_2 = \frac{1}{8}$.
According to the problem,the sum of the slopes is three times their product:
$m_1+m_2 = 3(m_1m_2)$
$\frac{p}{2} = 3 \times \frac{1}{8}$
$\frac{p}{2} = \frac{3}{8}$
$p = \frac{3}{8} \times 2 = \frac{3}{4}$.

(D) Let $O(0,0), A(1,2), B(3,4)$.
$1$. Median from $O$ to $AB$:
The midpoint $D$ of $AB$ is $(\frac{1+3}{2}, \frac{2+4}{2}) = (2,3)$.
The equation of the median $OD$ passing through $(0,0)$ and $(2,3)$ is $y = \frac{3}{2}x$,which simplifies to $3x - 2y = 0$.
$2$. Altitude from $O$ to $AB$:
The slope of $AB$ is $m_{AB} = \frac{4-2}{3-1} = \frac{2}{2} = 1$.
The slope of the altitude $OP$ is $m_{OP} = -\frac{1}{m_{AB}} = -1$.
The equation of the altitude $OP$ passing through $(0,0)$ with slope $-1$ is $y = -x$,which simplifies to $x + y = 0$.
$3$. Joint equation:
The joint equation of the altitude and median is $(3x - 2y)(x + y) = 0$.
Expanding this,we get $3x^2 + 3xy - 2xy - 2y^2 = 0$,which is $3x^2 + xy - 2y^2 = 0$.

(B) The line that bisects the angle between the positive $x$-axis and positive $y$-axis is $y=x$.
Since this line is a part of the pair $ax^{2}+2hxy+by^{2}=0$,it must satisfy the equation.
Substituting $y=x$ into the equation:
$ax^{2}+2hx(x)+b(x)^{2}=0$
$ax^{2}+2hx^{2}+bx^{2}=0$
$(a+2h+b)x^{2}=0$
For this to hold for all $x$,we must have $a+b+2h=0$,which implies $a+b=-2h$.

(B) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
Comparing $hxy + 10x + 6y + 4 = 0$ with the general form,we get $a = 0, b = 0, 2h' = h \Rightarrow h' = h/2, g = 5, f = 3, c = 4$.
For the equation to represent a pair of lines,the determinant must be zero:
$\Delta = \begin{vmatrix} a & h' & g \\ h' & b & f \\ g & f & c \end{vmatrix} = 0$
$\begin{vmatrix} 0 & h/2 & 5 \\ h/2 & 0 & 3 \\ 5 & 3 & 4 \end{vmatrix} = 0$
$0(0 - 9) - \frac{h}{2}(2h/2 - 15) + 5(3h/2 - 0) = 0$
$-\frac{h}{2}(h - 15) + \frac{15h}{2} = 0$
$-\frac{h^2}{2} + \frac{15h}{2} + \frac{15h}{2} = 0$
$-\frac{h^2}{2} + 15h = 0$
$-h^2 + 30h = 0 \Rightarrow h(30 - h) = 0$
Therefore,$h = 0$ or $h = 30$.

(A) The given equation is $x^{2} - 4xy + 3y^{2} = 0$.
Factoring the equation: $x^{2} - 3xy - xy + 3y^{2} = 0$ $\Rightarrow x(x - 3y) - y(x - 3y) = 0$ $\Rightarrow (x - y)(x - 3y) = 0$.
The lines are $x - y = 0$ and $x - 3y = 0$.
Lines parallel to these passing through $(3, -2)$ are of the form $(x - y + k_{1}) = 0$ and $(x - 3y + k_{2}) = 0$.
For $(x - y + k_{1}) = 0$ passing through $(3, -2)$: $3 - (-2) + k_{1} = 0$ $\Rightarrow 5 + k_{1} = 0$ $\Rightarrow k_{1} = -5$.
For $(x - 3y + k_{2}) = 0$ passing through $(3, -2)$: $3 - 3(-2) + k_{2} = 0$ $\Rightarrow 3 + 6 + k_{2} = 0$ $\Rightarrow k_{2} = -9$.
The joint equation is $(x - y - 5)(x - 3y - 9) = 0$.
Expanding: $x(x - 3y - 9) - y(x - 3y - 9) - 5(x - 3y - 9) = 0$.
$x^{2} - 3xy - 9x - xy + 3y^{2} + 9y - 5x + 15y + 45 = 0$.
$x^{2} + 3y^{2} - 4xy - 14x + 24y + 45 = 0$.

(C) The given equation of the pair of lines is $7x^2 - 14xy + py^2 - 12x + qy - 4 = 0$.
Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = 7$,$h = -7$,$b = p$,$g = -6$,$f = \frac{q}{2}$,and $c = -4$.
Since the lines are parallel,$h^2 = ab$.
Substituting the values,$(-7)^2 = 7p$ $\Rightarrow 49 = 7p$ $\Rightarrow p = 7$.
For the equation to represent a pair of lines,the condition $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ must be satisfied.
Substituting $p = 7$,$a = 7$,$h = -7$,$g = -6$,$f = \frac{q}{2}$,and $c = -4$:
$7(7)(-4) + 2(\frac{q}{2})(-6)(-7) - 7(\frac{q}{2})^2 - 7(-6)^2 - (-4)(-7)^2 = 0$.
$-196 + 42q - \frac{7q^2}{4} - 252 + 196 = 0$.
$42q - \frac{7q^2}{4} - 252 = 0$.
Multiplying by $-4/7$,we get $q^2 - 24q + 144 = 0$.
$(q - 12)^2 = 0 \Rightarrow q = 12$.
Finally,$\sqrt{p^2 + q^2 - pq} = \sqrt{7^2 + 12^2 - (7)(12)} = \sqrt{49 + 144 - 84} = \sqrt{109}$.

(A) The general equation of a second-degree curve is $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$.
Comparing this with the given equation $hxy + gx + fy + c = 0$,we get:
$A = 0, B = 0, C = c, H = \frac{h}{2}, G = \frac{g}{2}, F = \frac{f}{2}$.
For the equation to represent a pair of lines,the condition is $\Delta = \begin{vmatrix} A & H & G \\ H & B & F \\ G & F & C \end{vmatrix} = 0$.
Substituting the values:
$\begin{vmatrix} 0 & \frac{h}{2} & \frac{g}{2} \\ \frac{h}{2} & 0 & \frac{f}{2} \\ \frac{g}{2} & \frac{f}{2} & c \end{vmatrix} = 0$.
Expanding along the first row:
$0 - \frac{h}{2} \left( \frac{ch}{2} - \frac{gf}{4} \right) + \frac{g}{2} \left( \frac{hf}{4} - 0 \right) = 0$.
$-\frac{ch^2}{4} + \frac{hgf}{8} + \frac{ghf}{8} = 0$.
$-\frac{ch^2}{4} + \frac{hgf}{4} = 0$.
Multiplying by $4$,we get $-ch^2 + hgf = 0$,which implies $hgf = ch^2$.
Dividing by $h$ (assuming $h \neq 0$),we get $gf = ch$.

(A) The given equation is $ax^2+8xy+5y^2=0$.
Comparing this with the general equation $Ax^2+2Hxy+By^2=0$,we have $A=a$,$2H=8$ (so $H=4$),and $B=5$.
Let the slopes of the lines be $m_1$ and $m_2$.
The sum of slopes is $m_1+m_2 = -\frac{2H}{B} = -\frac{8}{5}$.
The product of slopes is $m_1m_2 = \frac{A}{B} = \frac{a}{5}$.
According to the given condition,the sum of slopes is twice their product:
$m_1+m_2 = 2(m_1m_2)$
$-\frac{8}{5} = 2\left(\frac{a}{5}\right)$
$-\frac{8}{5} = \frac{2a}{5}$
$2a = -8$
$a = -4$.

(B) The general equation of a second-degree curve $ax^{2} + 2hxy + by^{2} + 2gx + 2fy + k = 0$ represents a pair of lines if the determinant of the matrix $\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & k \end{vmatrix} = 0$.
Comparing the given equation $ax^{2} + 0xy + by^{2} + cx + cy + 0 = 0$ with the general form,we have $h = 0$,$g = c/2$,$f = c/2$,and $k = 0$.
Substituting these values into the determinant condition:
$\begin{vmatrix} a & 0 & c/2 \\ 0 & b & c/2 \\ c/2 & c/2 & 0 \end{vmatrix} = 0$.
Expanding along the first row:
$a(0 - c^{2}/4) - 0 + (c/2)(0 - bc/2) = 0$.
$-ac^{2}/4 - bc^{2}/4 = 0$.
Multiplying by $-4$:
$ac^{2} + bc^{2} = 0$.
$c^{2}(a + b) = 0$.
Since $c \neq 0$,we must have $a + b = 0$.

(D) Comparing the given equation with $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$,we get $a=3, h=5, b=3, g=0, f=8, c=k$.
Since the equation represents a pair of lines,the condition is $abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$.
Substituting the values: $(3)(3)(k)+2(8)(0)(5)-3(8)^{2}-3(0)^{2}-k(5)^{2}=0$.
$9k+0-192-0-25k=0$.
$-16k=192$.
$k = -12$.

(C) general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if the determinant $\Delta = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0$.
For option $A$: $x^2 - x = 0 \implies x(x-1) = 0$,which represents lines $x=0$ and $x=1$.
For option $B$: $xy - x = 0 \implies x(y-1) = 0$,which represents lines $x=0$ and $y=1$.
For option $C$: $y^2 - x + 1 = 0$. This is a parabola,not a pair of lines,as it cannot be factored into two linear factors.
For option $D$: $xy + x + y + 1 = 0 \implies x(y+1) + 1(y+1) = 0 \implies (x+1)(y+1) = 0$,which represents lines $x=-1$ and $y=-1$.
Thus,the equation that does not represent a pair of lines is $y^2 - x + 1 = 0$.

(B) The given equation is of the form $Ax^{2} + 2Hxy + By^{2} = 0$,where $A = \sec^{2} \theta - \sin^{2} \theta$,$2H = -2 \tan \theta$,and $B = \sin^{2} \theta$.
For a pair of lines $Ax^{2} + 2Hxy + By^{2} = 0$,the sum of slopes is $m_{1} + m_{2} = -\frac{2H}{B}$ and the product of slopes is $m_{1}m_{2} = \frac{A}{B}$.
Here,$m_{1} + m_{2} = \frac{2 \tan \theta}{\sin^{2} \theta}$ and $m_{1}m_{2} = \frac{\sec^{2} \theta - \sin^{2} \theta}{\sin^{2} \theta}$.
We know that $|m_{1} - m_{2}| = \sqrt{(m_{1} + m_{2})^{2} - 4m_{1}m_{2}}$.
Substituting the values:
$|m_{1} - m_{2}| = \sqrt{\left(\frac{2 \tan \theta}{\sin^{2} \theta}\right)^{2} - 4\left(\frac{\sec^{2} \theta - \sin^{2} \theta}{\sin^{2} \theta}\right)}$
$|m_{1} - m_{2}| = \sqrt{\frac{4 \tan^{2} \theta - 4(\sec^{2} \theta - \sin^{2} \theta) \sin^{2} \theta}{\sin^{4} \theta}}$
Using $\tan^{2} \theta = \frac{\sin^{2} \theta}{\cos^{2} \theta}$ and $\sec^{2} \theta = \frac{1}{\cos^{2} \theta}$:
$|m_{1} - m_{2}| = \sqrt{\frac{4 \frac{\sin^{2} \theta}{\cos^{2} \theta} - 4(\frac{1}{\cos^{2} \theta} - \sin^{2} \theta) \sin^{2} \theta}{\sin^{4} \theta}} = \sqrt{\frac{4 \frac{\sin^{2} \theta}{\cos^{2} \theta} - 4 \frac{\sin^{2} \theta}{\cos^{2} \theta} + 4 \sin^{4} \theta}{\sin^{4} \theta}} = \sqrt{\frac{4 \sin^{4} \theta}{\sin^{4} \theta}} = \sqrt{4} = 2$.

(D) Let $D$ be the midpoint of $AB$. The coordinates of $D$ are $\left(\frac{1+3}{2}, \frac{2+4}{2}\right) = (2,3)$.
The median $OD$ passes through $(0,0)$ and $(2,3)$. Its equation is $y = \frac{3}{2}x$,which simplifies to $3x-2y=0$.
The slope of line $AB$ is $m_{AB} = \frac{4-2}{3-1} = \frac{2}{2} = 1$.
The altitude $OE$ is perpendicular to $AB$,so its slope is $m_{OE} = -\frac{1}{m_{AB}} = -1$.
The equation of the altitude $OE$ is $y = -x$,which simplifies to $x+y=0$.
The joint equation of the median and altitude is $(3x-2y)(x+y) = 0$.
Expanding this,we get $3x^2 + 3xy - 2xy - 2y^2 = 0$,which is $3x^2 + xy - 2y^2 = 0$.

(A) The given equation is $x^2+2hxy+2y^2=0$.
Comparing with $ax^2+2hxy+by^2=0$,we have $a=1, h=h, b=2$.
Let the slopes of the lines be $m_1$ and $m_2$.
We know that $m_1+m_2 = -\frac{2h}{b} = -\frac{2h}{2} = -h$ and $m_1m_2 = \frac{a}{b} = \frac{1}{2}$.
Given the ratio of slopes is $m_1:m_2 = 1:2$,so $m_2 = 2m_1$.
Substituting $m_2$ in the product equation: $m_1(2m_1) = \frac{1}{2}$ $\Rightarrow 2m_1^2 = \frac{1}{2}$ $\Rightarrow m_1^2 = \frac{1}{4}$ $\Rightarrow m_1 = \pm \frac{1}{2}$.
If $m_1 = \frac{1}{2}$,then $m_2 = 1$.
Then $m_1+m_2 = \frac{1}{2} + 1 = \frac{3}{2}$.
Since $m_1+m_2 = -h$,we get $-h = \frac{3}{2} \Rightarrow h = -\frac{3}{2}$.
If $m_1 = -\frac{1}{2}$,then $m_2 = -1$.
Then $m_1+m_2 = -\frac{1}{2} - 1 = -\frac{3}{2}$.
Since $m_1+m_2 = -h$,we get $-h = -\frac{3}{2} \Rightarrow h = \frac{3}{2}$.
Thus,$h = \pm \frac{3}{2}$. Given the options,the correct value is $\frac{3}{2}$.

(C) Let the slopes of the lines be $m_1$ and $m_2$. For the equation $ax^2 + 2hxy + by^2 = 0$,we have:
$m_1 + m_2 = \frac{-2h}{b}$ $(i)$
$m_1 m_2 = \frac{a}{b}$ $(ii)$
Given $4ab = 3h^2$,so $ab = \frac{3h^2}{4}$.
Substituting this into $(ii)$,we get $m_1 m_2 = \frac{3h^2}{4b^2}$.
Now,$(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2 = \frac{4h^2}{b^2} - 4(\frac{3h^2}{4b^2}) = \frac{4h^2 - 3h^2}{b^2} = \frac{h^2}{b^2}$.
Thus,$m_1 - m_2 = \frac{h}{b}$ $(iii)$.
Adding $(i)$ and $(iii)$: $2m_1 = \frac{-2h}{b} + \frac{h}{b} = \frac{-h}{b} \implies m_1 = \frac{-h}{2b}$.
Subtracting $(iii)$ from $(i)$: $2m_2 = \frac{-2h}{b} - \frac{h}{b} = \frac{-3h}{b} \implies m_2 = \frac{-3h}{2b}$.
The ratio $m_1 : m_2 = \frac{-h}{2b} : \frac{-3h}{2b} = 1 : 3$.

(C) The given equation of the pair of lines is $K x^2 + 6 x y + y^2 = 0$.
Comparing this with the general form $A x^2 + 2 H x y + B y^2 = 0$,we get $A = K$,$H = 3$,and $B = 1$.
Let the slopes of the two lines be $m_1$ and $m_2$.
We know that $m_1 + m_2 = \frac{-2 H}{B} = -6$ and $m_1 m_2 = \frac{A}{B} = K$.
Given that one slope is three times the other,let $m_2 = 3 m_1$.
Substituting this into the sum of slopes: $m_1 + 3 m_1 = -6$ $\Rightarrow 4 m_1 = -6$ $\Rightarrow m_1 = -\frac{3}{2}$.
Now,using the product of slopes: $m_1 \times (3 m_1) = K \Rightarrow 3 m_1^2 = K$.
Substituting $m_1 = -\frac{3}{2}$: $K = 3 \times (-\frac{3}{2})^2 = 3 \times \frac{9}{4} = \frac{27}{4}$.

(B) The given pair of lines is $ax^2+2hxy+by^2=0$. Dividing by $x^2$,we get $a+2h(\frac{y}{x})+b(\frac{y}{x})^2=0$. Let $k = \frac{y}{x}$ be the slope of one of the lines. Then $bk^2+2hk+a=0$.
The slope of the line $mx+ny=18$ is $-\frac{m}{n}$.
Since the line is perpendicular to $mx+ny=18$,its slope $k$ must be the negative reciprocal of $-\frac{m}{n}$,so $k = \frac{n}{m}$.
Substituting $k = \frac{n}{m}$ into the equation $bk^2+2hk+a=0$,we get $b(\frac{n}{m})^2+2h(\frac{n}{m})+a=0$.
Multiplying by $m^2$,we obtain $bn^2+2hmn+am^2=0$.

(B) The given equation is $\frac{x^2}{a} + \frac{2xy}{h} + \frac{y^2}{b} = 0$. Multiplying by $abh$,we get $bhx^2 + 2abyx + ahy^2 = 0$.
Comparing this with the standard form $Ax^2 + 2Hxy + By^2 = 0$,we have $A = bh$,$2H = 2ab \Rightarrow H = ab$,and $B = ah$.
Let the slopes be $m_1$ and $m_2$. We are given $m_2 = 2m_1$.
For the equation $Ax^2 + 2Hxy + By^2 = 0$,the sum of slopes $m_1 + m_2 = -\frac{2H}{B}$ and the product of slopes $m_1 m_2 = \frac{A}{B}$.
$m_1 + 2m_1 = 3m_1 = -\frac{2ab}{ah} = -\frac{2b}{h} \Rightarrow m_1 = -\frac{2b}{3h}$.
$m_1(2m_1) = 2m_1^2 = \frac{bh}{ah} = \frac{b}{a} \Rightarrow 2\left(-\frac{2b}{3h}\right)^2 = \frac{b}{a}$.
$2 \times \frac{4b^2}{9h^2} = \frac{b}{a} \Rightarrow \frac{8b^2}{9h^2} = \frac{b}{a}$.
Dividing both sides by $b$ (assuming $b \neq 0$),we get $\frac{8b}{9h^2} = \frac{1}{a} \Rightarrow \frac{ab}{h^2} = \frac{9}{8}$.

(A) The slope of the line $3x + 2y - 8 = 0$ is $m_1 = -\frac{3}{2}$.
Let $m$ be the slope of the lines passing through the origin that make an angle of $\frac{\pi}{4}$ with the given line.
Using the formula $\tan \theta = \left| \frac{m - m_1}{1 + mm_1} \right|$,we have:
$\tan \frac{\pi}{4} = \left| \frac{m - (-3/2)}{1 + m(-3/2)} \right|$
$1 = \left| \frac{2m + 3}{2 - 3m} \right|$
Squaring both sides:
$(2 - 3m)^2 = (2m + 3)^2$
$4 - 12m + 9m^2 = 4m^2 + 12m + 9$
$5m^2 - 24m - 5 = 0$
Since the lines pass through the origin,their equations are $y = mx$,so $m = \frac{y}{x}$.
Substituting $m = \frac{y}{x}$ into the auxiliary equation:
$5(\frac{y}{x})^2 - 24(\frac{y}{x}) - 5 = 0$
Multiplying by $x^2$,we get:
$5y^2 - 24xy - 5x^2 = 0$
Rearranging,we get $5x^2 + 24xy - 5y^2 = 0$.

(D) Given the homogeneous equation $ax^{2}+2hxy+by^{2}=0 \quad \dots(i)$
This represents a pair of straight lines passing through the origin. Let the slopes of these lines be $m$ and $m_{1}$.
Thus,the lines are $y=mx$ and $y=m_{1}x$.
Their product is $(y-mx)(y-m_{1}x) = 0$,which simplifies to $y^{2}-(m+m_{1})xy+mm_{1}x^{2}=0$.
Dividing the original equation by $b$,we get $y^{2}+\frac{2h}{b}xy+\frac{a}{b}x^{2}=0 \quad \dots(ii)$.
Comparing $(i)$ and $(ii)$,we have:
$m+m_{1} = -\frac{2h}{b}$ and $mm_{1} = \frac{a}{b}$.
Since $m$ is a root of the quadratic equation $bm^{2}+2hm+a=0$ (obtained by substituting $y=mx$ in the original equation):
$bm^{2}+2hm+a=0$
$bm^{2}+hm = -a-hm$
$m(bm+h) = -(a+hm)$
This does not lead directly to the target. Let us use the relation $m_{1} = -\frac{2h}{b}-m$.
Substituting into $mm_{1} = \frac{a}{b}$:
$m(-\frac{2h}{b}-m) = \frac{a}{b}$
$-2hm - bm^{2} = a$
$bm^{2} + 2hm + a = 0$
Adding $h^{2}$ to both sides:
$bm^{2} + 2hm + h^{2} = h^{2} - a$
This is not quite right. Let's re-evaluate: $bm^{2}+2hm+a=0 \implies bm^{2}+hm = -a-hm$. Actually,from $bm^{2}+2hm+a=0$,we have $bm^{2}+hm = -a-hm$.
Wait,the expression is $(h+bm)^{2} = h^{2} + 2hbm + b^{2}m^{2}$.
Since $bm^{2}+2hm+a=0$,then $b^{2}m^{2}+2hbm+ab=0$.
Thus $b^{2}m^{2}+2hbm = -ab$.
Adding $h^{2}$ to both sides: $h^{2}+2hbm+b^{2}m^{2} = h^{2}-ab$.
Therefore,$(h+bm)^{2} = h^{2}-ab$.

(D) The general equation of a second-degree curve is $ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0$.
Comparing this with the given equation $3x^{2} + xy - y^{2} - 3x + 6y + k = 0$,we get:
$a = 3, b = -1, h = \frac{1}{2}, g = -\frac{3}{2}, f = 3, c = k$.
For the equation to represent a pair of lines,the condition is $abc + 2fgh - af^{2} - bg^{2} - ch^{2} = 0$.
Substituting the values:
$3(-1)(k) + 2(3)(-\frac{3}{2})(\frac{1}{2}) - 3(3)^{2} - (-1)(-\frac{3}{2})^{2} - k(\frac{1}{2})^{2} = 0$.
$-3k - \frac{9}{2} - 27 + \frac{9}{4} - \frac{k}{4} = 0$.
Multiplying by $4$ to clear the denominators:
$-12k - 18 - 108 + 9 - k = 0$.
$-13k - 117 = 0$.
$-13k = 117$.
$k = -9$.

(C) Let the slopes of the lines $ax^{2}+2hxy+by^{2}=0$ be $m_{1}$ and $m_{2}$.
According to the problem,$m_{1} = nm_{2}$.
We know that $m_{1}+m_{2} = -\frac{2h}{b}$ and $m_{1}m_{2} = \frac{a}{b}$.
Substituting $m_{1} = nm_{2}$ into the sum and product equations:
$m_{2}(n+1) = -\frac{2h}{b} \implies m_{2} = -\frac{2h}{b(n+1)}$.
$nm_{2}^{2} = \frac{a}{b}$.
Substituting the value of $m_{2}$ into the product equation:
$n \left( -\frac{2h}{b(n+1)} \right)^{2} = \frac{a}{b}$.
$n \left( \frac{4h^{2}}{b^{2}(n+1)^{2}} \right) = \frac{a}{b}$.
$4nh^{2} = ab(n+1)^{2}$.

(B) Let the point be $P(h, k)$.
$P_{1}$ is the length of the perpendicular from $P$ to the $x$-axis $(y=0)$,so $P_{1} = |k|$.
$P_{2}$ is the length of the perpendicular from $P$ to the line $x-y=0$,so $P_{2} = \frac{|h-k|}{\sqrt{1^{2}+(-1)^{2}}} = \frac{|h-k|}{\sqrt{2}}$.
Given that $P_{1} = 2P_{2}$,we have:
$|k| = 2 \cdot \frac{|h-k|}{\sqrt{2}}$
$|k| = \sqrt{2} |h-k|$
Squaring both sides:
$k^{2} = 2(h-k)^{2}$
$k^{2} = 2(h^{2} + k^{2} - 2hk)$
$k^{2} = 2h^{2} + 2k^{2} - 4hk$
$2h^{2} - 4hk + k^{2} = 0$
Replacing $(h, k)$ with $(x, y)$,the locus is:
$2x^{2} - 4xy + y^{2} = 0$

(B) The given pair of lines is $23x^2 - 48xy + 3y^2 = 0$.
The area of the triangle formed by the pair of lines $ax^2 + 2hxy + by^2 = 0$ and the line $lx + my + n = 0$ is given by the formula:
$\text{Area} = \frac{n^2 \sqrt{h^2 - ab}}{|am^2 - 2hlm + bl^2|}$
Here,$a = 23$,$h = -24$,$b = 3$,$l = 2$,$m = 3$,and $n = 5$.
First,calculate $\sqrt{h^2 - ab} = \sqrt{(-24)^2 - (23)(3)} = \sqrt{576 - 69} = \sqrt{507} = 13 \sqrt{3}$.
Next,calculate the denominator $|am^2 - 2hlm + bl^2| = |23(3)^2 - 2(-24)(2)(3) + 3(2)^2| = |23(9) + 144(2) + 3(4)| = |207 + 288 + 12| = |507| = 507$.
Now,substitute the values into the area formula:
$\text{Area} = \frac{5^2 \times 13 \sqrt{3}}{507} = \frac{25 \times 13 \sqrt{3}}{507} = \frac{25 \times 13 \sqrt{3}}{39 \times 13} = \frac{25}{39} \sqrt{3} = \frac{25}{13 \sqrt{3}} \text{ sq. units}$.

(A) The difference of the slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{2 \sqrt{h^2 - ab}}{|b|}$.
For the given equation $y^2 - 2xy \sec^2 \alpha + (3 + \tan^2 \alpha)(\tan^2 \alpha - 1) x^2 = 0$,we have $a = (3 + \tan^2 \alpha)(\tan^2 \alpha - 1)$,$2h = -2 \sec^2 \alpha$ (so $h = -\sec^2 \alpha$),and $b = 1$.
The difference of slopes is $\frac{2 \sqrt{(-\sec^2 \alpha)^2 - (3 + \tan^2 \alpha)(\tan^2 \alpha - 1)(1)}}{|1|}$.
$= 2 \sqrt{\sec^4 \alpha - (3\tan^2 \alpha - 3 + \tan^4 \alpha - \tan^2 \alpha)}$.
$= 2 \sqrt{\sec^4 \alpha - (\tan^4 \alpha + 2\tan^2 \alpha - 3)}$.
Since $\sec^2 \alpha = 1 + \tan^2 \alpha$,then $\sec^4 \alpha = (1 + \tan^2 \alpha)^2 = 1 + 2\tan^2 \alpha + \tan^4 \alpha$.
Substituting this,the expression becomes $2 \sqrt{1 + 2\tan^2 \alpha + \tan^4 \alpha - \tan^4 \alpha - 2\tan^2 \alpha + 3} = 2 \sqrt{4} = 4$.
Thus,both $A$ and $R$ are true and $R$ is the correct explanation of $A$.

(C) The product of the perpendiculars from the origin $(0,0)$ to the pair of lines represented by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is given by $p = \left| \frac{c}{\sqrt{(a-b)^2 + 4h^2}} \right|$.
For $xy+x+y+1=0$:
$a=0, b=0, h=1/2, c=1$.
$p_1 = \left| \frac{1}{\sqrt{(0-0)^2 + 4(1/2)^2}} \right| = \left| \frac{1}{\sqrt{1}} \right| = 1$.
For $x^2-y^2+2x+1=0$:
$a=1, b=-1, h=0, c=1$.
$p_2 = \left| \frac{1}{\sqrt{(1-(-1))^2 + 4(0)^2}} \right| = \left| \frac{1}{\sqrt{4}} \right| = 1/2$.
For $2x^2+3xy-2y^2+2x+1=0$:
$a=2, b=-2, h=3/2, c=1$.
$p_3 = \left| \frac{1}{\sqrt{(2-(-2))^2 + 4(3/2)^2}} \right| = \left| \frac{1}{\sqrt{16 + 9}} \right| = \frac{1}{\sqrt{25}} = 1/5$.
Comparing the values: $1/5 < 1/2 < 1$,which means $p_3 < p_2 < p_1$.

(C) The given equation of the pair of lines is $15 x^2 + 31 x y + 14 y^2 = 0$.
Factoring the quadratic expression: $15 x^2 + 10 x y + 21 x y + 14 y^2 = 0$ $\Rightarrow 5 x(3 x + 2 y) + 7 y(3 x + 2 y) = 0$ $\Rightarrow (3 x + 2 y)(5 x + 7 y) = 0$.
Thus,the two lines are $L_1: 3 x + 2 y = 0$ and $L_2: 5 x + 7 y = 0$.
The perpendicular distance from $(2,3)$ to $3 x + 2 y = 0$ is $d_1 = \frac{|3(2) + 2(3)|}{\sqrt{3^2 + 2^2}} = \frac{12}{\sqrt{13}}$.
The perpendicular distance from $(2,3)$ to $5 x + 7 y = 0$ is $d_2 = \frac{|5(2) + 7(3)|}{\sqrt{5^2 + 7^2}} = \frac{31}{\sqrt{74}}$.
Since $p_1 > p_2$,we have $p_1 = \frac{31}{\sqrt{74}}$ and $p_2 = \frac{12}{\sqrt{13}}$.
Now,calculating $p_1^2 + \frac{1}{74} - p_2^2 + \frac{1}{13} = \frac{31^2}{74} + \frac{1}{74} - \frac{12^2}{13} + \frac{1}{13} = \frac{961+1}{74} - \frac{144-1}{13} = \frac{962}{74} - \frac{143}{13} = 13 - 11 = 2$.

(B) The given equation is $8x^2+8xy+2y^2+26x+13y+15=0$ ...$(i)$
We can rewrite the equation as $2(4x^2+4xy+y^2)+13(2x+y)+15=0$.
Let $2x+y=t$. Then the equation becomes $2t^2+13t+15=0$.
Factoring the quadratic: $2t^2+10t+3t+15=0 \Rightarrow 2t(t+5)+3(t+5)=0$.
So,$(2t+3)(t+5)=0$.
Substituting $t=2x+y$ back,we get $(2(2x+y)+3)(2x+y+5)=0$.
This simplifies to $(4x+2y+3)(2x+y+5)=0$.
Thus,the two lines are $4x+2y+3=0$ and $2x+y+5=0$.
Dividing the first line by $2$,we get $2x+y+1.5=0$.
The distance $d$ between parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is given by $d = \frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$.
Here,$A=2, B=1, C_1=5, C_2=1.5$.
$d = \frac{|5-1.5|}{\sqrt{2^2+1^2}} = \frac{3.5}{\sqrt{5}} = \frac{7}{2\sqrt{5}}$.

(B) The given equations are:
$(p^2-q^2) x^2+(q^2-r^2) xy+(r^2-p^2) y^2=0$ ...$(i)$
$(l-m) x^2+(m-n) xy+(n-l) y^2=0$ ...(ii)
For any equation of the form $Ax^2+Bxy+Cy^2=0$,if $A+B+C=0$,then $(x-y)$ is a factor of the equation.
For equation $(i)$: $(p^2-q^2) + (q^2-r^2) + (r^2-p^2) = 0$. Thus,$(x-y)$ is a factor.
For equation (ii): $(l-m) + (m-n) + (n-l) = 0$. Thus,$(x-y)$ is a factor.
Since both equations represent a pair of lines passing through the origin and share the factor $(x-y)$,the common line is $x-y=0$.

(D) The general equation of a pair of lines passing through the origin is $ax^2 + 2hxy + by^2 = 0$.
For the lines to be coincident,the condition is $h^2 - ab = 0$.
Comparing $4x^2 + hxy + y^2 = 0$ with $ax^2 + 2h'xy + by^2 = 0$,we have $a = 4$,$2h' = h$,and $b = 1$.
The condition for coincidence is $(h')^2 - ab = 0$.
Substituting $h' = \frac{h}{2}$,we get $(\frac{h}{2})^2 - (4)(1) = 0$.
$\frac{h^2}{4} = 4$.
$h^2 = 16$.
$h = \pm 4$.
Since the options provided are limited,the correct value is $4$ or $-4$.

(A) The slope of the line $x+y+1=0$ is $m = -1$. Let the slope of the required lines be $m'$.
Since the angle between the lines is $45^{\circ}$,we use the formula $\tan \theta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$.
$\tan 45^{\circ} = |\frac{m' - (-1)}{1 + m'(-1)}| \Rightarrow 1 = |\frac{m'+1}{1-m'}|$.
This gives two cases: $\frac{m'+1}{1-m'} = 1$ or $\frac{m'+1}{1-m'} = -1$.
Case $1$: $m'+1 = 1-m'$ $\Rightarrow 2m' = 0$ $\Rightarrow m' = 0$.
The equation of the line is $y-4 = 0(x-3) \Rightarrow y-4 = 0$.
Case $2$: $m'+1 = -(1-m')$ $\Rightarrow m'+1 = -1+m'$ $\Rightarrow 1 = -1$,which implies the line is vertical $(m' = \infty)$.
The equation of the line is $x-3 = 0$.
The combined equation is $(x-3)(y-4) = 0 \Rightarrow xy-4x-3y+12 = 0$.

(D) The given equation is $2x^2 + hxy + 6y^2 = 0$.
Comparing this with the general form $ax^2 + 2h'xy + by^2 = 0$,we have $a = 2$,$2h' = h$,and $b = 6$.
Let the slopes of the two lines be $m_1$ and $m_2$.
Given that $m_1 = 3m_2$.
We know that the product of the slopes $m_1 m_2 = \frac{a}{b} = \frac{2}{6} = \frac{1}{3}$.
Substituting $m_1 = 3m_2$ into the product,we get $(3m_2)m_2 = \frac{1}{3}$ $\Rightarrow 3m_2^2 = \frac{1}{3}$ $\Rightarrow m_2^2 = \frac{1}{9}$ $\Rightarrow m_2 = \pm \frac{1}{3}$.
Thus,$m_1 = 3(\pm \frac{1}{3}) = \pm 1$.
The sum of the slopes is $m_1 + m_2 = -\frac{2h'}{b} = -\frac{h}{6}$.
Substituting the values of $m_1$ and $m_2$: $\pm 1 \pm \frac{1}{3} = -\frac{h}{6}$.
For the positive case: $1 + \frac{1}{3} = \frac{4}{3} = -\frac{h}{6} \Rightarrow h = -8$.
For the negative case: $-1 - \frac{1}{3} = -\frac{4}{3} = -\frac{h}{6} \Rightarrow h = 8$.
Therefore,$h = \pm 8$.

(A) The given equation is $8x^2 + axy + y^2 = 0$.
Let the slopes of the two lines be $m_1$ and $m_2$.
For a homogeneous equation $Ax^2 + Bxy + Cy^2 = 0$,the sum of slopes is $m_1 + m_2 = -B/C$ and the product of slopes is $m_1 m_2 = A/C$.
Here,$A = 8$,$B = a$,and $C = 1$.
So,$m_1 + m_2 = -a$ and $m_1 m_2 = 8$.
Given that one slope is thrice the other,let $m_1 = 3m_2$.
Substituting this into the product equation: $(3m_2) \times m_2 = 8$ $\Rightarrow 3m_2^2 = 8$ $\Rightarrow m_2^2 = 8/3$.
Substituting into the sum equation: $3m_2 + m_2 = -a$ $\Rightarrow 4m_2 = -a$ $\Rightarrow m_2 = -a/4$.
Squaring both sides: $m_2^2 = a^2/16$.
Equating the two values of $m_2^2$: $a^2/16 = 8/3$ $\Rightarrow a^2 = 128/3$ $\Rightarrow a = \pm \sqrt{128/3} = \pm 8 \sqrt{2/3}$.
Since the options provide the positive value,$a = 8 \sqrt{\frac{2}{3}}$.

(B) The given equation is $ax^3+3bx^2y+3cxy^2+dy^3=0$.
Dividing by $x^3$ and setting $m = \frac{y}{x}$,we get the cubic equation in $m$: $dm^3+3cm^2+3bm+a=0$.
Let the roots be $m_1, m_2, m_3$.
From the properties of roots,the product of the roots is $m_1m_2m_3 = -\frac{a}{d}$.
Since two lines are perpendicular,let $m_1m_2 = -1$.
Substituting this into the product equation: $(-1)m_3 = -\frac{a}{d} \Rightarrow m_3 = \frac{a}{d}$.
Since $m_3$ is a root of the cubic equation,it must satisfy $d(\frac{a}{d})^3+3c(\frac{a}{d})^2+3b(\frac{a}{d})+a=0$.
Multiplying by $d^2$,we get $a^3+3a^2c+3abd+ad^2=0$.
Dividing by $a$ (since $a \neq 0$),we get $a^2+3ac+3bd+d^2=0$.

(C) The given equation is $5x^2 - 8xy + 3y^2 = 0$.
Dividing by $x^2$,we get $3(\frac{y}{x})^2 - 8(\frac{y}{x}) + 5 = 0$.
Let $m = \frac{y}{x}$ be the slope of the lines. Then $3m^2 - 8m + 5 = 0$.
Solving the quadratic equation $3m^2 - 8m + 5 = 0$ using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$m = \frac{8 \pm \sqrt{64 - 4(3)(5)}}{2(3)} = \frac{8 \pm \sqrt{64 - 60}}{6} = \frac{8 \pm \sqrt{4}}{6} = \frac{8 \pm 2}{6}$.
The two slopes are $m_1 = \frac{8+2}{6} = \frac{10}{6} = \frac{5}{3}$ and $m_2 = \frac{8-2}{6} = \frac{6}{6} = 1$.
Since $m_1 > m_2$,we have $m_1 = \frac{5}{3}$ and $m_2 = 1$.
Therefore,the ratio $m_1 : m_2 = \frac{5}{3} : 1 = 5 : 3$.

(C) The given equation of the pair of lines is $4x^2-5xy+y^2=0$.
Dividing by $x^2$,we get the quadratic equation in terms of $m = \frac{y}{x}$:
$m^2-5m+4=0$.
Here,$m_1$ and $m_2$ are the roots of this equation.
Thus,the sum of slopes is $m_1+m_2 = 5$ and the product of slopes is $m_1m_2 = 4$.
We know that $|m_1-m_2| = \sqrt{(m_1+m_2)^2-4m_1m_2}$.
Substituting the values,we get:
$|m_1-m_2| = \sqrt{5^2-4(4)} = \sqrt{25-16} = \sqrt{9} = 3$.

(B) Given the equation of lines: $x^2+2 h x y+2 y^2=0 \dots (i)$
Comparing with $a x^2+2 h x y+b y^2=0$,we get $a=1$ and $b=2$.
Let the slopes of the lines be $m_1$ and $m_2$.
Then,$m_1+m_2 = -\frac{2h}{b} = -\frac{2h}{2} = -h \dots (ii)$
And $m_1 m_2 = \frac{a}{b} = \frac{1}{2} \dots (iii)$
Given the ratio of slopes is $m_1:m_2 = 1:2$,so $m_2 = 2m_1$.
Substituting $m_2 = 2m_1$ into $(iii)$: $m_1(2m_1) = \frac{1}{2}$ $\Rightarrow 2m_1^2 = \frac{1}{2}$ $\Rightarrow m_1^2 = \frac{1}{4}$ $\Rightarrow m_1 = \pm \frac{1}{2}$.
If $m_1 = \frac{1}{2}$,then $m_2 = 1$. From $(ii)$,$m_1+m_2 = -h$ $\Rightarrow \frac{1}{2} + 1 = -h$ $\Rightarrow h = -\frac{3}{2}$.
If $m_1 = -\frac{1}{2}$,then $m_2 = -1$. From $(ii)$,$m_1+m_2 = -h$ $\Rightarrow -\frac{1}{2} - 1 = -h$ $\Rightarrow h = \frac{3}{2}$.
Thus,$h = \pm \frac{3}{2}$.