Equation of pair of straight lines Questions in English

(A) Given lines are ${x^2} - 9{y^2} = 0$ and $x = 4$.
We have ${x^2} - 9{y^2} = 0$,which can be factored as $(x - 3y)(x + 3y) = 0$.
So the equations of the lines are:
$x - 3y = 0$ ..... $(i)$
$x + 3y = 0$ ..... $(ii)$
$x = 4$ ..... $(iii)$
By solving these equations,we find the vertices of the triangle:
Intersection of $(i)$ and $(ii)$ is $A(0, 0)$.
Intersection of $(i)$ and $(iii)$ is $C(4, 4/3)$.
Intersection of $(ii)$ and $(iii)$ is $B(4, -4/3)$.
Now,we calculate the lengths of the sides:
$AB = \sqrt{(4 - 0)^2 + (-4/3 - 0)^2} = \sqrt{16 + 16/9} = \sqrt{160/9} = \frac{4\sqrt{10}}{3}$
$AC = \sqrt{(4 - 0)^2 + (4/3 - 0)^2} = \sqrt{16 + 16/9} = \sqrt{160/9} = \frac{4\sqrt{10}}{3}$
$BC = \sqrt{(4 - 4)^2 + (4/3 - (-4/3))^2} = \sqrt{0 + (8/3)^2} = 8/3$
Since $AB = AC$,the triangle $ABC$ is an isosceles triangle.

(D) Expanding the given equation:
$(x^2 + 2xy + y^2) - (x^2 + y^2) = 0$
$2xy = 0$
This implies $x = 0$ or $y = 0$.
$x = 0$ represents the $y$-axis and $y = 0$ represents the $x$-axis.
Since the $x$-axis and $y$-axis are perpendicular to each other,the equation represents two mutually perpendicular lines.
Therefore,the correct option is $D$.

(A) Let the slopes of the lines be $m_1$ and $m_2$. Given $m_1 = m_2^2$.
From the equation $ax^2 + 2hxy + by^2 = 0$,we have $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$.
Substituting $m_1 = m_2^2$ in the sum and product:
$m_2^2 + m_2 = -\frac{2h}{b}$ $(i)$
$m_2^2 \cdot m_2 = m_2^3 = \frac{a}{b} \Rightarrow m_2 = (\frac{a}{b})^{1/3}$ $(ii)$
Substituting $m_2$ from $(ii)$ into $(i)$:
$(\frac{a}{b})^{2/3} + (\frac{a}{b})^{1/3} = -\frac{2h}{b}$
Cubing both sides:
$(\frac{a}{b})^2 + \frac{a}{b} + 3(\frac{a}{b})^{2/3}(\frac{a}{b})^{1/3} [(\frac{a}{b})^{2/3} + (\frac{a}{b})^{1/3}] = -\frac{8h^3}{b^3}$
Using $(i)$,we substitute the term in the bracket:
$\frac{a^2}{b^2} + \frac{a}{b} + 3(\frac{a}{b})(-\frac{2h}{b}) = -\frac{8h^3}{b^3}$
$\frac{a^2}{b^2} + \frac{a}{b} - \frac{6ah}{b^2} = -\frac{8h^3}{b^3}$
Multiplying by $b^3$:
$a^2b + ab^2 - 6abh = -8h^3$
$a^2b + ab^2 - 6abh + 8h^3 = 0$.

(D) The given equation is $x^2 + xy + y^2 = 0$.
Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we have $a = 1$,$2h = 1$,and $b = 1$.
The nature of the lines is determined by the discriminant $h^2 - ab$.
Here,$h = 1/2$,so $h^2 - ab = (1/2)^2 - (1)(1) = 1/4 - 1 = -3/4$.
Since $h^2 - ab < 0$,the lines represented by the equation are imaginary.

(B) The general equation of a second-degree curve is $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$.
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 $hxy + gx + fy + c = 0$ with the general form,we have $a = 0, b = 0, h' = h/2, g = g/2, f = f/2, c = c$.
Substituting these values into the condition:
$0(0)(c) + 2(f/2)(g/2)(h/2) - 0(f/2)^2 - 0(g/2)^2 - c(h/2)^2 = 0$.
This simplifies to $\frac{fgh}{4} - \frac{ch^2}{4} = 0$.
Multiplying by $4$,we get $fgh - ch^2 = 0$,which implies $h(fg - ch) = 0$.
Since $h \neq 0$ for the $xy$ term to exist,we must have $fg = ch$.

(C) The given equation is $ax^2 + 2hxy + by^2 = 0$.
To find the equation of the pair of lines perpendicular to these,we interchange the coefficients of $x^2$ and $y^2$ and change the sign of the $xy$ term.
Thus,the new equation becomes $bx^2 - 2hxy + ay^2 = 0$,which is equivalent to $ay^2 - 2hxy + bx^2 = 0$.

(A) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if the determinant of the matrix $\begin{bmatrix} a & h & g \\ h & b & f \\ g & f & c \end{bmatrix}$ is zero.
Comparing the given equation $3x^2 + 2hxy - 3y^2 - 40x + 30y - 75 = 0$ with the general form,we get:
$a = 3, b = -3, c = -75, h = h, g = -20, f = 15$.
The condition for a pair of straight lines is $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Substituting the values:
$(3)(-3)(-75) + 2(15)(-20)(h) - 3(15)^2 - (-3)(-20)^2 - (-75)(h)^2 = 0$
$675 - 600h - 675 + 1200 + 75h^2 = 0$
$75h^2 - 600h + 1200 = 0$
Dividing by $75$:
$h^2 - 8h + 16 = 0$
$(h - 4)^2 = 0$
Thus,$h = 4, 4$.

(A) The lines passing through the origin and parallel to the given lines $y = m_1x + c_1$ and $y = m_2x + c_2$ are $y = m_1x$ and $y = m_2x$.
These can be written as $(y - m_1x) = 0$ and $(y - m_2x) = 0$.
The joint equation of these two lines is given by the product $(y - m_1x)(y - m_2x) = 0$.
Expanding this,we get $y^2 - m_2xy - m_1xy + m_1m_2x^2 = 0$.
Rearranging the terms,we get $m_1m_2x^2 - (m_1 + m_2)xy + y^2 = 0$.

(A) Given equation: $ab(x^2 - y^2) + (a^2 - b^2)xy = 0$
Expanding the terms: $abx^2 - aby^2 + a^2xy - b^2xy = 0$
Rearranging the terms: $abx^2 + a^2xy - b^2xy - aby^2 = 0$
Factoring by grouping: $ax(bx + ay) - by(bx + ay) = 0$
Factoring out the common term: $(bx + ay)(ax - by) = 0$
Therefore,the lines are $bx + ay = 0$ and $ax - by = 0$.

(C) Let $X = x - 5$ and $Y = y - 6$. The equation becomes $X^2 + XY - 2Y^2 = 0$.
Factoring the quadratic expression: $X^2 + 2XY - XY - 2Y^2 = 0$
$X(X + 2Y) - Y(X + 2Y) = 0$
$(X - Y)(X + 2Y) = 0$
Substituting back $X = x - 5$ and $Y = y - 6$:
$((x - 5) - (y - 6))((x - 5) + 2(y - 6)) = 0$
$(x - y + 1)(x + 2y - 17) = 0$
This represents two straight lines. Since both factors vanish at $(5, 6)$,the lines pass through the point $(5, 6)$.

(D) second degree homogeneous equation in $x$ and $y$ is of the form $ax^2 + 2hxy + by^2 = 0$.
This equation always represents a pair of straight lines passing through the origin.
Since the provided option $(a)$ is '$A$ pair of straight lines' and option $(d)$ is '$A$ pair of straight lines passing through the origin',option $(d)$ is the most precise description.

(A) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if the determinant $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Comparing the given equation $6x^2 + 11xy - 10y^2 + x + 31y + k = 0$ with the general form:
$a = 6, b = -10, c = k, h = \frac{11}{2}, g = \frac{1}{2}, f = \frac{31}{2}$.
Substituting these values into the condition $\Delta = 0$:
$6(-10)(k) + 2(\frac{31}{2})(\frac{1}{2})(\frac{11}{2}) - 6(\frac{31}{2})^2 - (-10)(\frac{1}{2})^2 - k(\frac{11}{2})^2 = 0$
$-60k + \frac{341}{4} - 6(\frac{961}{4}) + \frac{10}{4} - \frac{121k}{4} = 0$
Multiply the entire equation by $4$:
$-240k + 341 - 5766 + 10 - 121k = 0$
$-361k - 5415 = 0$
$361k = -5415$
$k = -\frac{5415}{361} = -15$.

(B) Let the slopes of the lines be $m_1$ and $m_2$. For the equation $ax^2 + 2hxy + by^2 = 0$,we have $m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$.
Given $4ab = 3h^2$,we can write $ab = \frac{3h^2}{4}$.
Now,$(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1m_2 = \frac{4h^2}{b^2} - 4\left(\frac{a}{b}\right) = \frac{4h^2 - 4ab}{b^2}$.
Substituting $4ab = 3h^2$,we get $(m_1 - m_2)^2 = \frac{4h^2 - 3h^2}{b^2} = \frac{h^2}{b^2}$,so $m_1 - m_2 = \pm \frac{h}{b}$.
Solving $m_1 + m_2 = -\frac{2h}{b}$ and $m_1 - m_2 = \frac{h}{b}$,we get $2m_1 = -\frac{h}{b} \implies m_1 = -\frac{h}{2b}$ and $2m_2 = -\frac{3h}{b} \implies m_2 = -\frac{3h}{2b}$.
The ratio $m_1 : m_2 = \left(-\frac{h}{2b}\right) : \left(-\frac{3h}{2b}\right) = 1:3$.

(C) The given equation is $a(b - c)x^2 - xy(a(b - c) + c(a - b)) + c(a - b)y^2 = 0$.
This is a quadratic equation in $x$ and $y$ of the form $Ax^2 + Bxy + Cy^2 = 0$.
We can factorize this as $(a(b - c)x - c(a - b)y)(x - y) = 0$.
Expanding this: $a(b - c)x^2 - a(b - c)xy - c(a - b)xy + c(a - b)y^2 = 0$.
$a(b - c)x^2 - xy(a(b - c) + c(a - b)) + c(a - b)y^2 = 0$.
Thus,the lines are $a(b - c)x - c(a - b)y = 0$ and $x - y = 0$.

(A) Given the homogeneous equation of the second degree: $ax^2 + 2hxy + by^2 = 0$.
Dividing by $x^2$,we get $a + 2h(\frac{y}{x}) + b(\frac{y}{x})^2 = 0$.
Let $m = \frac{y}{x}$,then $bm^2 + 2hm + a = 0$.
This is a quadratic equation in $m$ with roots $m_1$ and $m_2$.
By the properties of roots of a quadratic equation,the sum of roots $m_1 + m_2 = \frac{-(\text{coefficient of } m)}{\text{coefficient of } m^2} = \frac{-2h}{b}$.
The product of roots $m_1m_2 = \frac{\text{constant term}}{\text{coefficient of } m^2} = \frac{a}{b}$.
Thus,the correct relation is $m_1 + m_2 = \frac{-2h}{b}$ and $m_1m_2 = \frac{a}{b}$.

(A) The given equation is $4x^2 + 12xy + 9y^2 = 0$.
Comparing this with the general homogeneous equation of the second degree $ax^2 + 2hxy + by^2 = 0$,we get $a = 4$,$2h = 12$ (so $h = 6$),and $b = 9$.
The nature of the lines is determined by the value of $h^2 - ab$.
Calculating $h^2 - ab = (6)^2 - (4)(9) = 36 - 36 = 0$.
Since $h^2 - ab = 0$,the lines represented by the equation are real and coincident.

(A) The general equation of a pair of straight lines is given by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
Comparing this with the given equation $2x^2 - 10xy + 12y^2 + 5x - 16y - 3 = 0$,we have $a = 2$,$2h = -10$ (so $h = -5$),and $b = 12$.
The equation of the pair of lines passing through the origin and perpendicular to the lines represented by the homogeneous part of the given equation is $bx^2 - 2hxy + ay^2 = 0$.
Substituting the values,we get $12x^2 - (-10)xy + 2y^2 = 0$,which simplifies to $12x^2 + 10xy + 2y^2 = 0$.
Dividing the entire equation by $2$,we obtain $6x^2 + 5xy + y^2 = 0$.

(C) general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if the determinant $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
For option $(c)$,the equation is $4x^2 - 4xy + y^2 - 4 = 0$.
Here,$a = 4, h = -2, b = 1, g = 0, f = 0, c = -4$.
Calculating the determinant: $\Delta = (4)(1)(-4) + 2(0)(-2)(0) - 4(0)^2 - 1(0)^2 - (-4)(-2)^2 = -16 + 0 - 0 - 0 + 16 = 0$.
Since $\Delta = 0$,the equation $4x^2 - 4xy + y^2 = 4$ represents a pair of straight lines (specifically,parallel lines $(2x - y)^2 = 4$ or $2x - y = \pm 2$).

(A) The general equation of a pair of straight lines is given by $Ax^2 + 2Hxy + By^2 = 0$.
Comparing the given equation ${a^2}{x^2} - a(b + c)xy + bc{y^2} = 0$ with the standard form,we have $A = a^2$,$2H = -a(b + c)$,and $B = bc$.
The lines are coincident if $H^2 - AB = 0$.
Substituting the values,we get $\left\{ -\frac{a(b + c)}{2} \right\}^2 - (a^2)(bc) = 0$.
$\frac{a^2(b + c)^2}{4} - a^2bc = 0$.
$a^2(b^2 + 2bc + c^2) - 4a^2bc = 0$.
$a^2(b^2 - 2bc + c^2) = 0$.
$a^2(b - c)^2 = 0$.
Thus,$a = 0$ or $b = c$.

(D) The general equation of a pair of straight lines passing through the origin is $ax^2 + 2h'xy + by^2 = 0$.
Comparing this with the given equation $2x^2 - 2hxy + 2y^2 = 0$,we have $a = 2$,$b = 2$,and $2h' = -2h$,so $h' = -h$.
For the lines to be coincident,the condition is $h'^2 - ab = 0$.
Substituting the values,we get $(-h)^2 - (2)(2) = 0$.
$h^2 - 4 = 0$.
$h^2 = 4$.
Therefore,$h = \pm 2$.

(A) Given the equation of the pair of straight lines is $ax^2 + 2hxy + by^2 = 0$.
Since $y = mx$ is one of the lines,we substitute $y = mx$ into the given equation:
$ax^2 + 2h(mx)x + b(mx)^2 = 0$
$ax^2 + 2hmx^2 + bm^2x^2 = 0$
Factoring out $x^2$ (assuming $x \neq 0$):
$x^2(a + 2hm + bm^2) = 0$
Since $x^2 \neq 0$,we have:
$bm^2 + 2hm + a = 0$.

(A) The given equation is $2x^2 - xy - 6y^2 + 7x + 21y - 15 = 0$.
Any line parallel to the pair of lines represented by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is given by $ax^2 + 2hxy + by^2 + k = 0$ for some constant $k$.
Since the required lines pass through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$ into the equation $2x^2 - xy - 6y^2 + k = 0$,which gives $k = 0$.
Therefore,the equation of the lines passing through the origin and parallel to the given lines is $2x^2 - xy - 6y^2 = 0$.

(C) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
Comparing this with $2x^2 + 4xy - py^2 + 4x + qy + 1 = 0$,we get $a = 2, h = 2, b = -p, g = 2, f = q/2, c = 1$.
For the lines to be perpendicular,the condition is $a + b = 0$.
$2 + (-p) = 0 \Rightarrow p = 2$.
For the equation to represent a pair of straight lines,the determinant condition is $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Substituting the values: $2(-p)(1) + 2(q/2)(2)(2) - 2(q/2)^2 - (-p)(2)^2 - 1(2)^2 = 0$.
With $p = 2$: $2(-2)(1) + 4q - 2(q^2/4) - (-2)(4) - 4 = 0$.
$-4 + 4q - q^2/2 + 8 - 4 = 0$.
$-q^2/2 + 4q = 0 \Rightarrow q^2 - 8q = 0$.
$q(q - 8) = 0 \Rightarrow q = 0$ or $q = 8$.
Thus,$p = 2$ and $q = 0$ or $8$.

(D) The general second-degree equation $Ax^2 + 2Bxy + Cy^2 + Dx + Ey + F = 0$ represents a pair of straight lines if and only if the determinant of the matrix $\begin{bmatrix} A & B & D/2 \\ B & C & E/2 \\ D/2 & E/2 & F \end{bmatrix}$ is equal to $0$.
This condition is given by $ACF + 2B(D/2)(E/2) - A(E/2)^2 - C(D/2)^2 - F(B^2) = 0$.
This condition does not uniquely determine the sign of $B^2 - AC$. For a pair of straight lines,$B^2 - AC$ can be positive (intersecting lines),zero (parallel lines),or negative (imaginary lines). Therefore,none of the given options are universally correct.

(B) The given equation is $xy + a^2 = a(x + y).$
Rearranging the terms,we get $xy - ax - ay + a^2 = 0.$
This can be factored as $x(y - a) - a(y - a) = 0.$
Thus,$(x - a)(y - a) = 0.$
This represents two straight lines: $x = a$ and $y = a.$
Since the equation represents two intersecting straight lines,the correct option is $B$.

(C) Let the slopes of the two lines be $m_1$ and $m_2$. Given $m_2 = \lambda m_1$.
From the equation $ax^2 + 2hxy + by^2 = 0$,we have $m_1 + m_2 = -\frac{2h}{b}$ and $m_1 m_2 = \frac{a}{b}$.
Substituting $m_2 = \lambda m_1$ in the sum: $m_1(1 + \lambda) = -\frac{2h}{b} \Rightarrow m_1 = -\frac{2h}{b(1 + \lambda)}$.
Substituting $m_2 = \lambda m_1$ in the product: $\lambda m_1^2 = \frac{a}{b} \Rightarrow m_1^2 = \frac{a}{b\lambda}$.
Equating the squares: $\left(-\frac{2h}{b(1 + \lambda)}\right)^2 = \frac{a}{b\lambda}$.
$\frac{4h^2}{b^2(1 + \lambda)^2} = \frac{a}{b\lambda}$.
$4\lambda h^2 = ab(1 + \lambda)^2$.

(A) The equations of the given lines are:
$ax^2 + 2hxy + by^2 = 0$ ..... $(i)$
$a'x^2 + 2h'xy + b'y^2 = 0$ ..... $(ii)$
Let the common line be $y = mx$. Substituting this into both equations:
$a + 2hm + bm^2 = 0$ ..... $(iii)$
$a' + 2h'm + b'm^2 = 0$ ..... $(iv)$
Using the method of cross-multiplication to eliminate $m$:
$\frac{m^2}{2ha' - 2h'a} = \frac{-m}{ab' - a'b} = \frac{1}{2bh' - 2b'h}$
From the first and third terms: $m^2 = \frac{2(ha' - h'a)}{2(bh' - b'h)} = \frac{ha' - h'a}{bh' - b'h}$
From the second and third terms: $m = -\frac{ab' - a'b}{2(bh' - b'h)}$
Squaring the expression for $m$: $m^2 = \frac{(ab' - a'b)^2}{4(bh' - b'h)^2}$
Equating the two expressions for $m^2$:
$\frac{ha' - h'a}{bh' - b'h} = \frac{(ab' - a'b)^2}{4(bh' - b'h)^2}$
$(ab' - a'b)^2 = 4(ha' - h'a)(bh' - b'h) = 4(ah' - a'h)(hb' - h'b)$

(A) The general second-degree equation is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
Comparing this with $2y^2 - xy - x^2 + 6x - 8 = 0$,we have $a = -1, h = -1/2, b = 2, g = 3, f = 0, c = -8$.
The condition for the equation to represent a pair of straight lines is $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Substituting the values: $\Delta = (-1)(2)(-8) + 2(0)(3)(-1/2) - (-1)(0)^2 - (2)(3)^2 - (-8)(-1/2)^2$.
$\Delta = 16 + 0 - 0 - 18 - (-8)(1/4) = 16 - 18 + 2 = 0$.
Since $\Delta = 0$,the equation represents a pair of straight lines.

(D) Given equation: ${x^2} - 5xy + 6{y^2} = 0$
Splitting the middle term:
${x^2} - 2xy - 3xy + 6{y^2} = 0$
Factoring by grouping:
$x(x - 2y) - 3y(x - 2y) = 0$
$(x - 2y)(x - 3y) = 0$
Thus,the lines are $x - 2y = 0$ and $x - 3y = 0$.
Rewriting these as $y = \frac{1}{2}x$ and $y = \frac{1}{3}x$,or $2y - x = 0$ and $3y - x = 0$.
Comparing with the given options,none of the options match the derived equations.

(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 $ax^2 + by^2 + cx + cy = 0$ with the general equation,we have:
$A = a, B = b, C = 0, H = 0, G = c/2, F = c/2$.
Substituting these values into the condition:
$a(b)(0) + 2(c/2)(c/2)(0) - a(c/2)^2 - b(c/2)^2 - 0(0)^2 = 0$.
This simplifies to:
$-a(c^2/4) - b(c^2/4) = 0$.
Multiplying by $-4$,we get:
$ac^2 + bc^2 = 0$.
$c^2(a + b) = 0$.
Therefore,$c(a + b) = 0$.

(A) Given the equation $ax^2 + (a + b)xy + by^2 + x + y = 0$.
We can factorize this expression by grouping terms:
$ax^2 + axy + by^2 + bxy + x + y = 0$
$ax(x + y) + by(y + x) + 1(x + y) = 0$
$(ax + by + 1)(x + y) = 0$.
Thus,the lines represented by the equation are $ax + by + 1 = 0$ and $x + y = 0$.

(D) The given equation is $x^2 - 7xy + 12y^2 = 0$.
Factorizing the quadratic expression: $x^2 - 3xy - 4xy + 12y^2 = 0$.
$x(x - 3y) - 4y(x - 3y) = 0$.
$(x - 3y)(x - 4y) = 0$.
This represents two straight lines $x - 3y = 0$ and $x - 4y = 0$ passing through the origin.
Since the slopes are $m_1 = 1/3$ and $m_2 = 1/4$,and $m_1 \times m_2 \neq -1$,the lines are non-perpendicular and intersect at the origin.
Therefore,the correct option is $D$.

(A) The given equation is ${y^2} - ({x^2} - 2x + 1) = 0$.
This can be rewritten as ${y^2} - {(x - 1)^2} = 0$.
Using the identity ${a^2} - {b^2} = (a - b)(a + b)$,we get $(y - (x - 1))(y + (x - 1)) = 0$.
This simplifies to $(y - x + 1)(y + x - 1) = 0$.
Since the equation can be factored into two linear equations,it represents a pair of straight lines.

(B) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Comparing the given equation $\lambda x^2 - 5xy + 2y^2 + 5x - 7y + 3 = 0$ with the general form,we have:
$a = \lambda, b = 2, c = 3, h = -5/2, g = 5/2, f = -7/2$.
Substituting these values into the condition:
$\lambda(2)(3) + 2(-7/2)(5/2)(-5/2) - \lambda(-7/2)^2 - 2(5/2)^2 - 3(-5/2)^2 = 0$
$6\lambda + 175/4 - 49\lambda/4 - 25/2 - 75/4 = 0$
Multiplying by $4$ to clear the denominators:
$24\lambda + 175 - 49\lambda - 50 - 75 = 0$
$-25\lambda + 50 = 0$
$25\lambda = 50 \Rightarrow \lambda = 2$.

(C) The given equation is $x^2 + 2xy \cot \theta - y^2 = 0$.
This is a homogeneous equation of degree $2$ of the form $ax^2 + 2hxy + by^2 = 0$,where $a = 1$,$h = \cot \theta$,and $b = -1$.
The lines represented by this equation are given by $y = \frac{h \pm \sqrt{h^2 - ab}}{b} x$.
Substituting the values: $y = \frac{\cot \theta \pm \sqrt{\cot^2 \theta - (1)(-1)}}{-1} x$.
$y = -(\cot \theta \pm \sqrt{\csc^2 \theta}) x$.
$y = -(\cot \theta \pm \csc \theta) x$.
For the positive sign: $y = -(\frac{\cos \theta + 1}{\sin \theta}) x \Rightarrow x \sin \theta + y(\cos \theta + 1) = 0$.
For the negative sign: $y = -(\frac{\cos \theta - 1}{\sin \theta}) x \Rightarrow x \sin \theta + y(\cos \theta - 1) = 0$.
Thus,one of the lines is $x \sin \theta + y(\cos \theta + 1) = 0$.

(A) Given equation: $pq(x^2 - y^2) + (p^2 - q^2)xy = 0$
Expanding the terms: $pqx^2 - pqy^2 + p^2xy - q^2xy = 0$
Rearranging the terms: $pqx^2 + p^2xy - q^2xy - pqy^2 = 0$
Grouping the terms: $px(qx + py) - qy(qx + py) = 0$
Factoring out the common term: $(qx + py)(px - qy) = 0$
Thus,the two lines are $qx + py = 0$ and $px - qy = 0$.
Comparing with the given options,the correct equation is $px - qy = 0$.

(B) The given equation of the pair of straight lines is $3x^2 - 8xy + 5y^2 = 0$.
Factoring this,we get $3x^2 - 3xy - 5xy + 5y^2 = 0$,which simplifies to $3x(x - y) - 5y(x - y) = 0$,or $(3x - 5y)(x - y) = 0$.
Thus,the individual lines are $L_1: 3x - 5y = 0$ and $L_2: x - y = 0$.
$A$ line perpendicular to $ax + by + c = 0$ is of the form $bx - ay + k = 0$.
For $L_1: 3x - 5y = 0$,the perpendicular line is $5x + 3y + k_1 = 0$. Since it passes through $(1, 2)$,$5(1) + 3(2) + k_1 = 0 \implies 5 + 6 + k_1 = 0 \implies k_1 = -11$. So,$5x + 3y - 11 = 0$.
For $L_2: x - y = 0$,the perpendicular line is $x + y + k_2 = 0$. Since it passes through $(1, 2)$,$1 + 2 + k_2 = 0 \implies k_2 = -3$. So,$x + y - 3 = 0$.
The combined equation is $(5x + 3y - 11)(x + y - 3) = 0$.

(A) The general quadratic equation $f(x, y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
If $h^2 = ab$,the angle $\theta$ between the lines is given by $\tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b}$.
Since $h^2 = ab$,we have $\tan \theta = 0$,which implies $\theta = 0$.
When $\Delta = 0$ and $h^2 = ab$,the lines are parallel.
If $g^2 = ac$ and $f^2 = bc$ also hold,the lines are coincident; however,in the general case where only $\Delta = 0$ and $h^2 = ab$ are given,the lines are parallel.

(C) The general equation of a pair of straight lines passing through the origin is given by $ax^2 + 2hxy + by^2 = 0$.
For this equation to represent a pair of coincident lines,the condition $h^2 - ab = 0$ must be satisfied.
Comparing the given equation $x^2 + 4xy + ky^2 = 0$ with the general form,we have $a = 1$,$2h = 4$ (so $h = 2$),and $b = k$.
Substituting these values into the condition $h^2 - ab = 0$:
$(2)^2 - (1)(k) = 0$
$4 - k = 0$
$k = 4$.

(C) The joint equation of two lines $L_1 = 0$ and $L_2 = 0$ is given by $L_1 \times L_2 = 0$.
Given lines are $x + y - 1 = 0$ and $x - y - 4 = 0$.
Therefore,the joint equation is $(x + y - 1)(x - y - 4) = 0$.

(B) The general equation of the second degree $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Comparing the given equation $x^2 - \lambda xy + 2y^2 + 3x - 5y + 2 = 0$ with the general form:
$a = 1, b = 2, c = 2, h = -\frac{\lambda}{2}, g = \frac{3}{2}, f = -\frac{5}{2}$.
Substituting these values into the condition $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$:
$(1)(2)(2) + 2(-\frac{5}{2})(\frac{3}{2})(-\frac{\lambda}{2}) - (1)(-\frac{5}{2})^2 - (2)(\frac{3}{2})^2 - (2)(-\frac{\lambda}{2})^2 = 0$.
$4 + \frac{15\lambda}{4} - \frac{25}{4} - \frac{18}{4} - \frac{\lambda^2}{2} = 0$.
Multiplying by $4$ to clear denominators:
$16 + 15\lambda - 25 - 18 - 2\lambda^2 = 0$.
$-2\lambda^2 + 15\lambda - 27 = 0$.
$2\lambda^2 - 15\lambda + 27 = 0$.
$(2\lambda - 9)(\lambda - 3) = 0$.
Thus,$\lambda = 3$ or $\lambda = \frac{9}{2}$.

(D) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if the determinant $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Comparing the given equation $2x^2 + 7xy + 3y^2 + 8x + 14y + \lambda = 0$ with the general form:
$a = 2, h = 7/2, b = 3, g = 4, f = 7, c = \lambda$.
Substituting these values into the condition $\Delta = 0$:
$(2)(3)(\lambda) + 2(7)(4)(7/2) - 2(7)^2 - 3(4)^2 - \lambda(7/2)^2 = 0$
$6\lambda + 196 - 98 - 48 - 49\lambda/4 = 0$
$6\lambda - 12.25\lambda + 50 = 0$
$-6.25\lambda = -50$
$\lambda = 8$.

(A) Let the slopes of the lines represented by $x^2 + hxy + 2y^2 = 0$ be $m_1$ and $m_2$.
Given that $m_1 = 2m_2$.
For the equation $ax^2 + hxy + by^2 = 0$,we have $m_1 + m_2 = -\frac{h}{b}$ and $m_1 m_2 = \frac{a}{b}$.
Here $a = 1$,$h = h$,and $b = 2$.
So,$m_1 + m_2 = -\frac{h}{2}$ and $m_1 m_2 = \frac{1}{2}$.
Substituting $m_1 = 2m_2$ into the sum: $3m_2 = -\frac{h}{2} \Rightarrow m_2 = -\frac{h}{6}$.
Substituting into the product: $(2m_2)(m_2) = \frac{1}{2}$ $\Rightarrow 2m_2^2 = \frac{1}{2}$ $\Rightarrow m_2^2 = \frac{1}{4}$ $\Rightarrow m_2 = \pm \frac{1}{2}$.
Now,$3(\pm \frac{1}{2}) = -\frac{h}{2}$ $\Rightarrow \pm \frac{3}{2} = -\frac{h}{2}$ $\Rightarrow h = \mp 3$.
Thus,$h = \pm 3$.

(B) The general equation of a second-degree curve is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
Comparing the given equation $L{x^2} - 10xy + 12{y^2} + 5x - 16y - 3 = 0$ with the general equation,we get:
$a = L, h = -5, b = 12, g = \frac{5}{2}, f = -8, c = -3$.
For the equation to represent a pair of straight lines,the condition is $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Substituting the values:
$L(12)(-3) + 2(-8)(\frac{5}{2})(-5) - L(-8)^2 - 12(\frac{5}{2})^2 - (-3)(-5)^2 = 0$
$-36L + 200 - 64L - 12(\frac{25}{4}) + 3(25) = 0$
$-36L + 200 - 64L - 75 + 75 = 0$
$-100L + 200 = 0$
$100L = 200$
$L = 2$.

(C) Let the gradients of the lines be $m_1$ and $m_2$. Given $m_1 : m_2 = 1 : 3$,so let $m_1 = m$ and $m_2 = 3m$.
For the equation $ax^2 + 2hxy + by^2 = 0$,we have $m_1 + m_2 = -\frac{2h}{b}$ and $m_1 m_2 = \frac{a}{b}$.
Substituting the values,$m + 3m = 4m = -\frac{2h}{b}$,which gives $m = -\frac{h}{2b}$.
Also,$m(3m) = 3m^2 = \frac{a}{b}$.
Substituting $m = -\frac{h}{2b}$ into the second equation: $3(-\frac{h}{2b})^2 = \frac{a}{b}$.
$3(\frac{h^2}{4b^2}) = \frac{a}{b}$.
$\frac{3h^2}{4b^2} = \frac{a}{b} \Rightarrow \frac{h^2}{ab} = \frac{4b^2}{3b^2} = \frac{4}{3}$.
Alternatively,for a ratio $1 : n$,the condition is $\frac{h^2}{ab} = \frac{(n+1)^2}{4n}$.
For $n = 3$,$\frac{h^2}{ab} = \frac{(3+1)^2}{4(3)} = \frac{16}{12} = \frac{4}{3}$.

(C) The equation of the pair of lines is $ax^2 + 4xy + y^2 = 0$.
Comparing this with $Ax^2 + 2Hxy + By^2 = 0$,we have $A = a$,$2H = 4$ (so $H = 2$),and $B = 1$.
Let the slopes of the two lines be $m_1$ and $m_2$.
Then $m_1 + m_2 = -\frac{2H}{B} = -\frac{4}{1} = -4$ $(i)$.
And $m_1 m_2 = \frac{A}{B} = \frac{a}{1} = a$ $(ii)$.
Given that $m_1 = 3m_2$.
Substituting this into $(i)$,we get $3m_2 + m_2 = -4$,which implies $4m_2 = -4$,so $m_2 = -1$.
Then $m_1 = 3(-1) = -3$.
Substituting these values into $(ii)$,we get $a = m_1 m_2 = (-3)(-1) = 3$.

(B) Comparing the given equation $4x^2 + 2hxy - 7y^2 = 0$ with the standard form $ax^2 + 2hxy + by^2 = 0$,we get $a = 4$,$2h = 2h$,and $b = -7$.
Let $m_1$ and $m_2$ be the slopes of the pair of lines.
The sum of the slopes is $m_1 + m_2 = -\frac{2h}{b} = -\frac{2h}{-7} = \frac{2h}{7}$.
The product of the slopes is $m_1m_2 = \frac{a}{b} = \frac{4}{-7}$.
Given that the sum of the slopes is equal to the product of the slopes,we have $m_1 + m_2 = m_1m_2$.
Substituting the values,we get $\frac{2h}{7} = \frac{4}{-7}$.
Multiplying both sides by $7$,we get $2h = -4$.
Therefore,$h = -2$.

(C) Let the slopes of the lines be $m_1$ and $m_2$.
From the equation $ax^2 + 2hxy + by^2 = 0$,we have:
$m_1 + m_2 = -\frac{2h}{b}$ and $m_1m_2 = \frac{a}{b}$.
Given that $m_1 = 2m_2$,substitute this into the sum of slopes:
$2m_2 + m_2 = -\frac{2h}{b} \implies 3m_2 = -\frac{2h}{b} \implies m_2 = -\frac{2h}{3b}$.
Substitute $m_1 = 2m_2$ into the product of slopes:
$(2m_2)(m_2) = \frac{a}{b} \implies 2m_2^2 = \frac{a}{b}$.
Substitute $m_2 = -\frac{2h}{3b}$ into the product equation:
$2\left(-\frac{2h}{3b}\right)^2 = \frac{a}{b} \implies 2\left(\frac{4h^2}{9b^2}\right) = \frac{a}{b}$.
$\frac{8h^2}{9b^2} = \frac{a}{b} \implies \frac{8h^2}{9b} = a \implies 8h^2 = 9ab$.

(B) The general equation of a second-degree curve $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if the determinant of the matrix $\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0$.
Comparing the given equation $x^2 + kxy + y^2 - 5x - 7y + 6 = 0$ with the general form,we have $a=1, b=1, c=6, h=k/2, g=-5/2, f=-7/2$.
Setting the determinant to zero:
$\begin{vmatrix} 1 & k/2 & -5/2 \\ k/2 & 1 & -7/2 \\ -5/2 & -7/2 & 6 \end{vmatrix} = 0$
Expanding the determinant:
$1(6 - 49/4) - (k/2)(3k - 35/4) - (5/2)(-7k/4 + 5/2) = 0$
$(24-49)/4 - (3k^2/2 - 35k/8) + (35k/8 - 25/4) = 0$
$-25/4 - 3k^2/2 + 35k/8 + 35k/8 - 25/4 = 0$
$-50/4 - 3k^2/2 + 70k/8 = 0$
$-25/2 - 3k^2/2 + 35k/4 = 0$
Multiplying by $-4$: $6k^2 - 35k + 50 = 0$
$(2k - 5)(3k - 10) = 0$
Thus,$k = 5/2$ or $k = 10/3$. Given the options,$k = 10/3$ is the correct value.