Bisectors of the angle between the lines, Point of intersection of the lines Questions in English

(C) The general quadratic equation is $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
The condition for the equation to represent a pair of straight lines is $\Delta = 0$,where $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
For the lines to be parallel,the condition is $h^2 = ab$.
For the lines to be coincident,the distance between the parallel lines must be zero.
The distance $d$ between parallel lines is given by $d = 2\sqrt{\frac{g^2 - ac}{a(a+b)}}$ or $d = 2\sqrt{\frac{f^2 - bc}{b(a+b)}}$.
Setting $d = 0$ implies $g^2 = ac$ and $f^2 = bc$.
Thus,the combined condition for coincident lines is $\Delta = 0, h^2 = ab, g^2 = ac$,and $f^2 = bc$.
Therefore,the correct option is $C$.

(C) The equation of the pair of lines is ${x^2} - 2hxy - 2{y^2} = 0$.
Comparing this with $ax^2 + 2h'xy + by^2 = 0$,we have $a = 1$,$2h' = -2h$,and $b = -2$.
The equation of the bisectors of the angles between the lines is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h'}$.
Substituting the values,we get $\frac{x^2 - y^2}{1 - (-2)} = \frac{xy}{-h} \Rightarrow \frac{x^2 - y^2}{3} = \frac{xy}{-h}$.
This simplifies to $-h(x^2 - y^2) = 3xy$,or $hx^2 + 3xy - hy^2 = 0$.
Since $x - 2y = 0$ is a bisector,it must satisfy this equation.
Substituting $x = 2y$,we get $h(2y)^2 + 3(2y)y - hy^2 = 0$.
$4hy^2 + 6y^2 - hy^2 = 0 \Rightarrow 3hy^2 + 6y^2 = 0$.
$3y^2(h + 2) = 0$.
For this to hold for all $y$,$h + 2 = 0$,so $h = -2$.

(A) The given equation is $4x^2 - 16xy - 7y^2 = 0$.
Comparing this with $ax^2 + 2hxy + by^2 = 0$,we get $a = 4$,$2h = -16$ (so $h = -8$),and $b = -7$.
The equation of the bisectors of the angle between the lines is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.
Substituting the values,we get $\frac{x^2 - y^2}{4 - (-7)} = \frac{xy}{-8}$.
This simplifies to $\frac{x^2 - y^2}{11} = \frac{xy}{-8}$.
Cross-multiplying gives $-8(x^2 - y^2) = 11xy$.
Rearranging the terms,we get $8x^2 + 11xy - 8y^2 = 0$.

(A) The given equation is $(y - mx)^2 = (x + my)^2$.
Expanding both sides,we get $y^2 + m^2x^2 - 2mxy = x^2 + m^2y^2 + 2mxy$.
Rearranging the terms,we get $(m^2 - 1)x^2 - 4mxy + (1 - m^2)y^2 = 0$.
This is of the form $Ax^2 + 2Hxy + By^2 = 0$,where $A = m^2 - 1$,$H = -2m$,and $B = 1 - m^2$.
The equation of the bisectors of the angles between the lines is given by $\frac{x^2 - y^2}{A - B} = \frac{xy}{H}$.
Substituting the values,we get $\frac{x^2 - y^2}{(m^2 - 1) - (1 - m^2)} = \frac{xy}{-2m}$.
$\frac{x^2 - y^2}{2(m^2 - 1)} = \frac{xy}{-2m}$.
$-m(x^2 - y^2) = (m^2 - 1)xy$.
$-mx^2 + my^2 = (m^2 - 1)xy$.
$mx^2 + (m^2 - 1)xy - my^2 = 0$.

(C) The given equation is ${x^2} - {y^2} = 0$.
Comparing this with the general equation of a pair of straight lines ${ax^2} + {2hxy} + {by^2} = 0$,we have $a = 1$,$h = 0$,and $b = -1$.
The equation of the bisectors of the angles between the lines is given by $\frac{{x^2 - y^2}}{{a - b}} = \frac{{xy}}{h}$.
Substituting the values,we get $\frac{{x^2 - y^2}}{{1 - (-1)}} = \frac{{xy}}{0}$.
This implies $xy = 0$.

(C) The equation of the pair of lines is $ax^2 - 2hxy + by^2 = 0$. The equation of the bisectors of the angles between these lines is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{-h}$.
This can be rewritten as $h(x^2 - y^2) + (a - b)xy = 0$.
Since $y = mx$ is one of the bisectors,it must satisfy the equation of the bisectors.
Substituting $y = mx$ into the equation $h(x^2 - y^2) + (a - b)xy = 0$,we get:
$h(x^2 - (mx)^2) + (a - b)x(mx) = 0$
$h(x^2 - m^2x^2) + (a - b)mx^2 = 0$
Dividing by $x^2$ (assuming $x \neq 0$):
$h(1 - m^2) + m(a - b) = 0$.

(B) The coordinate axes are given by the equations $x = 0$ and $y = 0$.
The bisectors of the angles between the coordinate axes are the lines $y = x$ and $y = -x$.
These can be rewritten as $y - x = 0$ and $y + x = 0$.
The combined equation is obtained by multiplying these two equations:
$(y - x)(y + x) = 0$
$y^2 - x^2 = 0$
Multiplying by $-1$,we get $x^2 - y^2 = 0$.

(D) The equation of the pair of angle bisectors for $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$ ... $(i)$.
The second equation is $(a + \lambda)x^2 + 2hxy + (b + \lambda)y^2 = 0$.
The equation of the pair of angle bisectors for this equation is $\frac{x^2 - y^2}{(a + \lambda) - (b + \lambda)} = \frac{xy}{h}$,which simplifies to $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$ ... $(ii)$.
Since equation $(i)$ and equation $(ii)$ are identical for any value of $\lambda$,the bisectors are coincident for any real number $\lambda$.

(A) The given equation is $\sqrt{3}x^2 - 4xy + \sqrt{3}y^2 = 0$.
Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we have $a = \sqrt{3}$,$h = -2$,and $b = \sqrt{3}$.
The equation of the bisectors of the angles between the lines is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.
Substituting the values,we get $\frac{x^2 - y^2}{\sqrt{3} - \sqrt{3}} = \frac{xy}{-2}$.
Since the denominator on the left is $0$,the equation becomes $x^2 - y^2 = 0$.

(A) The general equation of a pair of lines passing through the origin is $ax^2 + 2hxy + by^2 = 0$.
Comparing this with the given equation ${x^2} + 2xy \cot \theta + {y^2} = 0$,we get $a = 1$,$h = \cot \theta$,and $b = 1$.
The equation of the bisectors of the angles between the lines is given by the formula $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.
Substituting the values,we get $\frac{x^2 - y^2}{1 - 1} = \frac{xy}{\cot \theta}$.
Since the denominator on the left side is $0$,the equation becomes $x^2 - y^2 = 0$.

(A) The equation of the pair of angle bisectors for the pair of lines $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.
For the first pair,the equation of bisectors is $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.
For the second pair,the equation of bisectors is $\frac{x^2 - y^2}{a' - b'} = \frac{xy}{h'}$.
Since the bisectors are the same,the ratios of the coefficients must be equal:
$\frac{a - b}{h} = \frac{a' - b'}{h'}$.
Cross-multiplying gives: $(a - b)h' = (a' - b')h$.

(C) The equation of the bisectors of the angle between the lines represented by $px^2 - 2rxy + qy^2 = 0$ is given by:
$\frac{x^2 - y^2}{p - q} = \frac{xy}{-r}$ .....$(i)$
To check if $y = mx$ is a bisector,we substitute $y = mx$ into equation $(i)$:
$\frac{x^2 - (mx)^2}{p - q} = \frac{x(mx)}{-r}$
$\frac{x^2(1 - m^2)}{p - q} = \frac{mx^2}{-r}$
Dividing by $x^2$ (assuming $x \neq 0$):
$\frac{1 - m^2}{p - q} = \frac{m}{-r}$
$-r(1 - m^2) = m(p - q)$
$r(1 - m^2) + m(p - q) = 0$
Since this matches the given condition,$y = mx$ is a bisector.

(D) The equation $a(x - 1)^2 + 2h(x - 1)y + by^2 = 0$ represents a pair of straight lines intersecting at the point $(1, 0)$.
By shifting the origin to $(1, 0)$,we substitute $x = X + 1$ and $y = Y$,which transforms the equation into $aX^2 + 2hXY + bY^2 = 0$ $(i)$.
The given bisector $2x + y - 2 = 0$ becomes $2(X + 1) + Y - 2 = 0$,which simplifies to $2X + Y = 0$.
Since the angle bisectors of a pair of straight lines are always perpendicular to each other,the slope of the second bisector must be the negative reciprocal of the slope of the first bisector.
The slope of $2X + Y = 0$ is $m_1 = -2$. Thus,the slope of the second bisector $m_2$ is $1/2$.
The equation of the second bisector passing through $(1, 0)$ is $Y - 0 = \frac{1}{2}(X - 0)$,which simplifies to $X - 2Y = 0$.
Substituting back $X = x - 1$ and $Y = y$,we get $(x - 1) - 2y = 0$,or $x - 2y - 1 = 0$.

(C) The given equation is $2x^2 + 7xy + 3y^2 + 8x + 14y + 8 = 0$.
Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get:
$a = 2, b = 3, 2h = 7 \implies h = 3.5, 2g = 8 \implies g = 4, 2f = 14 \implies f = 7, c = 8$.
The point of intersection $(x, y)$ is given by the formulas:
$x = \frac{hf - bg}{ab - h^2} = \frac{(3.5)(7) - (3)(4)}{(2)(3) - (3.5)^2} = \frac{24.5 - 12}{6 - 12.25} = \frac{12.5}{-6.25} = -2$.
$y = \frac{hg - af}{ab - h^2} = \frac{(3.5)(4) - (2)(7)}{(2)(3) - (3.5)^2} = \frac{14 - 14}{6 - 12.25} = 0$.
Thus,the point of intersection is $(-2, 0)$.

(C) Let $X = x + 2$ and $Y = y - 2$.
The given equation becomes $2X^2 + 3XY - 2Y^2 = 0$.
This is a homogeneous equation of the second degree in $X$ and $Y$,which represents a pair of straight lines passing through the origin $(X, Y) = (0, 0)$.
Substituting back the values of $X$ and $Y$:
$x + 2 = 0 \implies x = -2$
$y - 2 = 0 \implies y = 2$
Thus,the point of intersection is $(-2, 2)$.

(A) The point of intersection $(x_0, y_0)$ of the lines represented by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is given by the solution of the partial derivatives $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$.
These are $2ax + 2hy + 2g = 0 \implies ax + hy + g = 0$ and $2hx + 2by + 2f = 0 \implies hx + by + f = 0$.
Solving these equations using Cramer's rule,we get $x_0 = \frac{hf - bg}{ab - h^2}$ and $y_0 = \frac{gf - ah}{ab - h^2}$.
The square of the distance from the origin $(0, 0)$ is $x_0^2 + y_0^2 = \frac{(hf - bg)^2 + (gf - ah)^2}{(ab - h^2)^2}$.
Expanding the numerator: $h^2f^2 + b^2g^2 - 2hfbg + g^2f^2 + a^2h^2 - 2gfah = f^2(g^2 + h^2) + h^2(a^2 + b^2) - 2hfg(a + b)$.
Using the condition for the pair of lines $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$,the expression simplifies to $\frac{c(a + b) - f^2 - g^2}{ab - h^2}$.

(A) The equation of the pair of bisectors of the angle between the lines $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.
For the given equation $x^2 - 2pxy - y^2 = 0$,we have $a = 1, h = -p, b = -1$.
Substituting these values,the equation of the bisectors is $\frac{x^2 - y^2}{1 - (-1)} = \frac{xy}{-p}$.
$\frac{x^2 - y^2}{2} = \frac{xy}{-p}$ $\Rightarrow -p(x^2 - y^2) = 2xy$ $\Rightarrow -px^2 + py^2 = 2xy$ $\Rightarrow px^2 + 2xy - py^2 = 0$.
Comparing this with the given bisector equation $x^2 - 2qxy - y^2 = 0$,we have the ratio of coefficients:
$\frac{p}{1} = \frac{2}{-2q} = \frac{-p}{-1}$.
From $\frac{p}{1} = \frac{2}{-2q}$,we get $p = -\frac{1}{q}$,which implies $pq = -1$ or $pq + 1 = 0$.

(C) The given equation is $x^2 - y^2 + 2y = 1$,which can be written as $x^2 - (y^2 - 2y + 1) = 0$,or $x^2 - (y - 1)^2 = 0$.
This represents the pair of lines $(x - y + 1)(x + y - 1) = 0$.
The angle bisectors of these lines are given by the lines $x = 0$ and $y = 1$.
The triangle is formed by the lines $x = 0$,$y = 1$,and $x + y = 3$.
The vertices of the triangle are the intersection points of these lines:
$1$. Intersection of $x = 0$ and $y = 1$ is $(0, 1)$.
$2$. Intersection of $x = 0$ and $x + y = 3$ is $(0, 3)$.
$3$. Intersection of $y = 1$ and $x + y = 3$ is $(2, 1)$.
The base of the triangle along $y = 1$ is $|2 - 0| = 2$.
The height of the triangle along $x = 0$ is $|3 - 1| = 2$.
Therefore,the area of the triangle is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 2 = 2$.

(D) Let $\varphi(x, y) = 2x^2 - 7xy - 4y^2 - x + 22y - 10 = 0$.
To find the point of intersection,we take the partial derivatives with respect to $x$ and $y$ and set them to zero.
$\frac{\partial \varphi}{\partial x} = 4x - 7y - 1 = 0$ (Equation $1$)
$\frac{\partial \varphi}{\partial y} = -7x - 8y + 22 = 0$ (Equation $2$)
From Equation $1$,$4x = 7y + 1 \implies x = \frac{7y + 1}{4}$.
Substituting this into Equation $2$:
$-7(\frac{7y + 1}{4}) - 8y + 22 = 0$
$-49y - 7 - 32y + 88 = 0$
$-81y + 81 = 0 \implies y = 1$.
Substituting $y = 1$ into $4x - 7(1) - 1 = 0$:
$4x - 8 = 0 \implies x = 2$.
Thus,the point of intersection is $(2, 1)$.

(A) The bisectors of the angles between the lines in the new position are the same as the bisectors of the angles between the lines in the original position.
The equation of the pair of bisectors for $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.
Here,$a = 1$,$h = -p$,and $b = -1$.
Substituting these values,we get:
$\frac{x^2 - y^2}{1 - (-1)} = \frac{xy}{-p}$
$\frac{x^2 - y^2}{2} = \frac{xy}{-p}$
$-p(x^2 - y^2) = 2xy$
$-px^2 + py^2 = 2xy$
$px^2 + 2xy - py^2 = 0$

(A) The given equation is $2x^{2} - 7xy + 3y^{2} = 0$.
Comparing this with $ax^{2} + 2hxy + by^{2} = 0$,we have $a = 2$,$2h = -7 \implies h = -7/2$,and $b = 3$.
The equation of the bisectors of the angles between the lines is given by the formula:
$\frac{x^{2} - y^{2}}{a - b} = \frac{xy}{h}$
Substituting the values:
$\frac{x^{2} - y^{2}}{2 - 3} = \frac{xy}{-7/2}$
$\frac{x^{2} - y^{2}}{-1} = \frac{2xy}{-7}$
$7(x^{2} - y^{2}) = 2xy$
$7x^{2} - 7y^{2} = 2xy$
$7x^{2} - 2xy - 7y^{2} = 0$.

(C) Let the given equation be $f(x, y) = 2x^2 + 3xy + y^2 - 7x - 5y + 6 = 0$.
To find the point of intersection $(h, k)$,we take partial derivatives with respect to $x$ and $y$ and set them to zero.
$\frac{\partial f}{\partial x} = 4x + 3y - 7 = 0$ (Equation $1$)
$\frac{\partial f}{\partial y} = 3x + 2y - 5 = 0$ (Equation $2$)
Multiply Equation $1$ by $2$ and Equation $2$ by $3$:
$8x + 6y - 14 = 0$
$9x + 6y - 15 = 0$
Subtracting the first from the second:
$(9x - 8x) + (6y - 6y) - (15 - 14) = 0 \implies x - 1 = 0 \implies x = 1$.
Substitute $x = 1$ into Equation $1$:
$4(1) + 3y - 7 = 0 \implies 3y - 3 = 0 \implies y = 1$.
Thus,the point of intersection is $(1, 1)$.

(A) The given pair of lines is $my^2 + (1 - m^2)xy - mx^2 = 0$.
This can be rewritten as $-mx^2 + (1 - m^2)xy + my^2 = 0$,or $mx^2 - (1 - m^2)xy - my^2 = 0$.
The lines $xy = 0$ represent the coordinate axes,$x = 0$ and $y = 0$.
The angle bisectors of the lines $x = 0$ and $y = 0$ are given by $x^2 - y^2 = 0$,which implies $x = y$ and $x = -y$.
If one of the lines $my^2 + (1 - m^2)xy - mx^2 = 0$ is a bisector,it must be either $x - y = 0$ or $x + y = 0$.
For $x - y = 0$,the equation is $x^2 - xy - xy + y^2 = 0$ (not matching) or by substituting $y = x$ into the original equation:
$m(x^2) + (1 - m^2)x^2 - mx^2 = 0 \Rightarrow (1 - m^2)x^2 = 0$,which implies $m^2 = 1$,so $m = \pm 1$.
However,if $m = 1$,the equation becomes $y^2 - x^2 = 0$,i.e.,$(y - x)(y + x) = 0$,which are the bisectors of $xy = 0$.
If $m = -1$,the equation becomes $-y^2 + 2xy + x^2 = 0$,which are not the bisectors.
Thus,$m = \pm 1$. Given the options,$m = 1$ is the correct choice.

(D) The given equation of the pair of straight lines is $xy - x - y + 1 = 0$.
Factoring the expression,we get $(x - 1)(y - 1) = 0$.
This implies the two lines are $x - 1 = 0$ and $y - 1 = 0$.
The intersection point of these two lines is $(1, 1)$.
Since the three lines are concurrent,the point $(1, 1)$ must satisfy the equation of the third line $ax + 2y - 3 = 0$.
Substituting $x = 1$ and $y = 1$ into the equation: $a(1) + 2(1) - 3 = 0$.
$a + 2 - 3 = 0$.
$a - 1 = 0$,which gives $a = 1$.

(D) The locus of points equidistant from two lines is the pair of angle bisectors of the lines represented by the equation $ax^2 + 2hxy + by^2 = 0$.
Comparing the given equation $x^2 \cos^2 \theta - xy \sin^2 \theta - y^2 \sin^2 \theta = 0$ with $ax^2 + 2hxy + by^2 = 0$,we have $a = \cos^2 \theta$,$2h = -\sin^2 \theta$,and $b = -\sin^2 \theta$.
The equation of the angle bisectors is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.
Substituting the values,we get $\frac{x^2 - y^2}{\cos^2 \theta - (-\sin^2 \theta)} = \frac{xy}{-\frac{1}{2} \sin^2 \theta}$.
$\frac{x^2 - y^2}{\cos^2 \theta + \sin^2 \theta} = \frac{2xy}{-\sin^2 \theta}$.
Since $\cos^2 \theta + \sin^2 \theta = 1$,we have $x^2 - y^2 = -\frac{2xy}{\sin^2 \theta}$.
$x^2 - y^2 = -2xy \csc^2 \theta$.
Therefore,$x^2 - y^2 + 2xy \csc^2 \theta = 0$.

(B) The equation of the pair of lines is $ax^2 + 2hxy + by^2 = 0$. The equation of the bisectors of the angles between these lines is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.
For the first pair $x^2 - 2mxy - y^2 = 0$,we have $a = 1, h = -m, b = -1$. The bisectors are $\frac{x^2 - y^2}{1 - (-1)} = \frac{xy}{-m}$ $\Rightarrow \frac{x^2 - y^2}{2} = \frac{xy}{-m}$ $\Rightarrow -m(x^2 - y^2) = 2xy$ $\Rightarrow mx^2 + 2xy - my^2 = 0$.
For the second pair $x^2 - 2nxy - y^2 = 0$,we have $a = 1, h = -n, b = -1$. The bisectors are $\frac{x^2 - y^2}{1 - (-1)} = \frac{xy}{-n} \Rightarrow nx^2 + 2xy - ny^2 = 0$.
Since each pair bisects the angle between the other,the bisectors of the first pair must be the second pair,and vice versa. Comparing $mx^2 + 2xy - my^2 = 0$ with $x^2 - 2nxy - y^2 = 0$,we get $\frac{m}{1} = \frac{2}{-2n} = \frac{-m}{-1}$.
From $\frac{m}{1} = \frac{2}{-2n}$,we get $m = -\frac{1}{n}$,which implies $mn = -1$.

(A) The given equation of the pair of lines is $f(x, y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.
To find the point of intersection,we take partial derivatives with respect to $x$ and $y$:
$\frac{\partial f}{\partial x} = 2ax + 2hy + 2g = 0 \implies ax + hy + g = 0$
$\frac{\partial f}{\partial y} = 2hx + 2by + 2f = 0 \implies hx + by + f = 0$
Since the lines intersect on the $y$-axis,the $x$-coordinate of the point of intersection is $x = 0$.
Substituting $x = 0$ into the first partial derivative equation: $h(y) + g = 0 \implies y = -g/h$.
Now,substitute $(0, -g/h)$ into the original equation $f(x, y) = 0$:
$a(0)^2 + 2h(0)(-g/h) + b(-g/h)^2 + 2g(0) + 2f(-g/h) + c = 0$
$b(g^2/h^2) - 2fg/h + c = 0$
Multiplying by $h^2$,we get $bg^2 - 2fgh + ch^2 = 0$,which simplifies to $bg^2 + ch^2 = 2fgh$.

(A) The equation of the bisectors of the pair of straight lines $ax^2 + 2hxy + by^2 = 0$ is given by $\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.
This can be rewritten as $h(x^2 - y^2) - (a - b)xy = 0 \dots (1)$.
Since the line $y = mx$ is a bisector,it must satisfy equation $(1)$.
Substituting $y = mx$ into equation $(1)$,we get:
$h(x^2 - (mx)^2) - (a - b)x(mx) = 0$
$h(x^2 - m^2x^2) - (a - b)mx^2 = 0$
Dividing by $x^2$ (assuming $x \neq 0$):
$h(1 - m^2) - (a - b)m = 0$
$h - hm^2 - (a - b)m = 0$
$hm^2 + (a - b)m - h = 0$.
Thus,$m$ is a root of the quadratic equation $hx^2 + (a - b)x - h = 0$.

(D) The equation $ax^2 + 2hxy + by^2 = 0$ represents a pair of angle bisectors,which are always perpendicular to each other.
Therefore,the sum of the coefficients of $x^2$ and $y^2$ must be zero,i.e.,$a + b = 0$.
Since the origin $(0, 0)$ is the point of intersection of the angle bisectors,it must satisfy the equation of the lines $L_2$.
Substituting $(0, 0)$ into $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$,we get $F = 0$.
Thus,$a + b + F = 0 + 0 = 0$.
For the general second-degree equation to represent a pair of lines,the determinant $\Delta$ must be zero:
$ABC + 2hfg - af^2 - bg^2 - ch^2 = 0$.
Applying this condition to $L_2$ leads to the relation $BDE = AE^2 + CD^2$.
Option $D$ is incorrect because the angle bisectors of a pair of lines depend on the specific coefficients of those lines,not just any arbitrary $l, m, n$.

(B) The equation of the pair of angle bisectors of the homogeneous equation $ax^{2}+2hxy+by^{2}=0$ is given by the formula $\frac{x^{2}-y^{2}}{a-b} = \frac{xy}{h}$.
For the given equation $x^{2}-4xy-5y^{2}=0$,we have $a=1$,$2h=-4$ (so $h=-2$),and $b=-5$.
Substituting these values into the formula:
$\frac{x^{2}-y^{2}}{1-(-5)} = \frac{xy}{-2}$
$\frac{x^{2}-y^{2}}{6} = \frac{xy}{-2}$
Multiplying both sides by $6$:
$x^{2}-y^{2} = -3xy$
Rearranging the terms,we get:
$x^{2}+3xy-y^{2}=0$.

(D) The equation of the pair of angle bisectors for the homogeneous equation $ax^2+2hxy+by^2=0$ is given by:
$\frac{x^2-y^2}{a-b} = \frac{xy}{h}$
Comparing $2x^2+xy-3y^2=0$ with $ax^2+2hxy+by^2=0$,we get:
$a=2$,$2h=1 \implies h=1/2$,and $b=-3$.
Substituting these values into the formula:
$\frac{x^2-y^2}{2-(-3)} = \frac{xy}{1/2}$
$\frac{x^2-y^2}{5} = 2xy$
$x^2-y^2 = 10xy$
$x^2-y^2-10xy=0$

(B) The equation of the pair of angle bisectors of the homogeneous equation $ax^2+2hxy+by^2=0$ is given by $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$.
Given the equation $x^2+3xy+2y^2=0$,we have $a=1$,$2h=3$ (so $h=\frac{3}{2}$),and $b=2$.
Substituting these values into the formula:
$\frac{x^2-y^2}{1-2} = \frac{xy}{3/2}$
$\frac{x^2-y^2}{-1} = \frac{2xy}{3}$
$3(x^2-y^2) = -2xy$
$3x^2-3y^2 = -2xy$
$3x^2+2xy-3y^2=0$.

(A) The equation of the pair of angle bisectors of the pair of lines $ax^{2}+2hxy+by^{2}=0$ is given by $\frac{x^{2}-y^{2}}{a-b} = \frac{xy}{h}$.
For the pair $x^{2}-2pxy-y^{2}=0$,we have $a=1, h=-p, b=-1$.
The equation of the angle bisectors is $\frac{x^{2}-y^{2}}{1-(-1)} = \frac{xy}{-p}$,which simplifies to $\frac{x^{2}-y^{2}}{2} = \frac{xy}{-p}$.
This gives $x^{2}-y^{2} = -\frac{2}{p}xy$,or $x^{2} + \frac{2}{p}xy - y^{2} = 0$.
It is given that this pair is the same as $x^{2}-2qxy-y^{2}=0$.
Comparing the coefficients of $xy$,we get $-2q = \frac{2}{p}$.
Therefore,$pq = -1$.

(A) The given equation is $ax^2 + 2hxy + by^2 = 0$,where $a = 2$,$2h = 11$,and $b = 3$.
The joint equation of the bisectors of the angles between the lines is given by the formula:
$\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$
Substituting the values:
$\frac{x^2 - y^2}{2 - 3} = \frac{xy}{11/2}$
$\frac{x^2 - y^2}{-1} = \frac{2xy}{11}$
$11(x^2 - y^2) = -2xy$
$11x^2 - 11y^2 + 2xy = 0$
$11x^2 + 2xy - 11y^2 = 0$
Thus,the correct option is $A$.

(D) The equation of the angle bisectors of two lines represented by the general equation $ax^2+2hxy+by^2=0$ is given by $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$.
Comparing the given equation $x^2-4xy-5y^2=0$ with $ax^2+2hxy+by^2=0$,we get $a=1$,$b=-5$,and $2h=-4$,which implies $h=-2$.
Substituting these values into the formula:
$\frac{x^2-y^2}{1-(-5)} = \frac{xy}{-2}$
$\frac{x^2-y^2}{6} = \frac{xy}{-2}$
$x^2-y^2 = -3xy$
$x^2+3xy-y^2=0$.

(B) The equation of the pair of lines is $x^2-3 x y-4 y^2=0$. Comparing this with $Ax^2+2Hxy+By^2=0$,we get $A=1, H=-\frac{3}{2}, B=-4$.
The equation of the angle bisectors is given by $\frac{x^2-y^2}{A-B} = \frac{xy}{H}$.
Substituting the values,we get $\frac{x^2-y^2}{1-(-4)} = \frac{xy}{-3/2}$.
$\frac{x^2-y^2}{5} = -\frac{2xy}{3}$.
$3(x^2-y^2) = -10xy$.
$3x^2+10xy-3y^2=0$.
Comparing this with the given equation $3x^2-kxy-3y^2=0$,we have $-k=10$,which implies $k=-10$.

(A) The given equation is $3x^{2} + 2xy - y^{2} = 0$. Comparing this with the general form $ax^{2} + 2hxy + by^{2} = 0$,we get $a = 3$,$2h = 2 \implies h = 1$,and $b = -1$.
The joint equation of the pair of angle bisectors is given by the formula $\frac{x^{2} - y^{2}}{a - b} = \frac{xy}{h}$.
Substituting the values,we get $\frac{x^{2} - y^{2}}{3 - (-1)} = \frac{xy}{1}$.
$\frac{x^{2} - y^{2}}{4} = xy$.
$x^{2} - y^{2} = 4xy$.
$x^{2} - 4xy - y^{2} = 0$.

(B) The lines $x=5$ (a vertical line) and $y=3$ (a horizontal line) intersect at the point $(5, 3)$.
Since these lines are perpendicular,the angle bisectors are lines passing through $(5, 3)$ that make angles of $45^{\circ}$ and $135^{\circ}$ with the $x$-axis.
The slopes of these bisectors are $m = \tan 45^{\circ} = 1$ and $m = \tan 135^{\circ} = -1$.
The equations of the bisectors are:
$y - 3 = 1(x - 5) \Rightarrow y - x + 2 = 0$
$y - 3 = -1(x - 5) \Rightarrow y + x - 8 = 0$
The joint equation is $(y - x + 2)(y + x - 8) = 0$.
Expanding this: $y^2 + xy - 8y - xy - x^2 + 8x + 2y + 2x - 16 = 0$.
Simplifying,we get $-x^2 + y^2 + 10x - 6y - 16 = 0$,or $x^2 - y^2 - 10x + 6y + 16 = 0$.

(C) The equation of the bisectors of the angles between the lines $x^{2}-2 p x y-y^{2}=0$ is given by the formula $\frac{x^{2}-y^{2}}{a-b} = \frac{xy}{h}$.
Here,$a=1, b=-1, h=-p$.
So,$\frac{x^{2}-y^{2}}{1-(-1)} = \frac{xy}{-p}$.
$\Rightarrow \frac{x^{2}-y^{2}}{2} = \frac{-xy}{p}$.
$\Rightarrow px^{2} + 2xy - py^{2} = 0$.
Since this equation represents the same pair of lines as $x^{2}-2qxy-y^{2}=0$,we compare the coefficients:
$\frac{p}{1} = \frac{2}{-2q} = \frac{-p}{-1}$.
From $\frac{p}{1} = \frac{2}{-2q}$,we get $p = -\frac{1}{q}$,which implies $pq = -1$ or $pq+1=0$.

(A) The given equation of the pair of straight lines is $xy-x+y-1=0$.
Factorizing the expression: $x(y-1)+1(y-1)=0$,which gives $(x+1)(y-1)=0$.
This represents two lines: $L_1: x+1=0$ and $L_2: y-1=0$.
The intersection point of these two lines is $(-1, 1)$.
Since the line $x+ky-3=0$ is concurrent with these lines,it must pass through the point $(-1, 1)$.
Substituting $(-1, 1)$ into the line equation: $-1 + k(1) - 3 = 0$.
$k - 4 = 0$,which implies $k = 4$.

(A) Let $\phi(x, y) = 2x^{2}-xy-15y^{2}-7x+32y-9=0$ ...$(1)$
To find the point of intersection,we take partial derivatives:
$\frac{\partial \phi}{\partial x} = 4x-y-7=0$ ...$(2)$
$\frac{\partial \phi}{\partial y} = -x-30y+32=0$ ...$(3)$
Solving equations $(2)$ and $(3)$:
From $(2)$,$y = 4x-7$.
Substitute into $(3)$: $-x - 30(4x-7) + 32 = 0$ $\Rightarrow -x - 120x + 210 + 32 = 0$ $\Rightarrow -121x + 242 = 0$ $\Rightarrow x = 2$.
Then $y = 4(2)-7 = 1$.
So,the point of intersection is $(2, 1)$.
The lines passing through $(2, 1)$ parallel to the coordinate axes are $x=2$ and $y=1$.
The joint equation is $(x-2)(y-1) = 0$.
Expanding this,we get $xy - x - 2y + 2 = 0$.

(C) The given equation is $x^2 - y^2 + x + 3y - 2 = 0$.
Comparing this with the general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a=1, h=0, b=-1, g=\frac{1}{2}, f=\frac{3}{2}, c=-2$.
The point of intersection $(x, y)$ of the pair of lines is given by the solution of the partial derivatives $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$.
$\frac{\partial}{\partial x}(x^2 - y^2 + x + 3y - 2) = 2x + 1 = 0 \Rightarrow x = -\frac{1}{2}$.
$\frac{\partial}{\partial y}(x^2 - y^2 + x + 3y - 2) = -2y + 3 = 0 \Rightarrow y = \frac{3}{2}$.
Thus,the point of intersection is $(-\frac{1}{2}, \frac{3}{2})$.

(B) The given lines are $x=5$ and $y=3$.
These lines are perpendicular to each other.
The bisectors of the angles between the lines $x=h$ and $y=k$ are given by the lines $x-h = \pm(y-k)$.
Substituting $h=5$ and $k=3$,we get $x-5 = \pm(y-3)$.
This gives two equations: $x-5 = y-3 \implies x-y-2=0$ and $x-5 = -(y-3) \implies x+y-8=0$.
The joint equation is $(x-y-2)(x+y-8) = 0$.
Expanding this: $x(x+y-8) - y(x+y-8) - 2(x+y-8) = 0$.
$x^2 + xy - 8x - xy - y^2 + 8y - 2x - 2y + 16 = 0$.
$x^2 - y^2 - 10x + 6y + 16 = 0$.

(B) The equation of a pair of lines is $Ax^2 + 2Hxy + By^2 = 0$. The angle bisectors of this pair are given by $\frac{x^2 - y^2}{A - B} = \frac{xy}{H}$.
For the first pair $x^2 - 2axy - y^2 = 0$,we have $A=1, B=-1, H=-a$. The bisectors are $\frac{x^2 - y^2}{1 - (-1)} = \frac{xy}{-a}$,which simplifies to $\frac{x^2 - y^2}{2} = \frac{xy}{-a}$,or $ax^2 + 2xy - ay^2 = 0$.
Since this must be the second pair $x^2 - 2bxy - y^2 = 0$,we compare the coefficients:
$\frac{a}{1} = \frac{2}{-2b} = \frac{-a}{-1}$.
From $\frac{a}{1} = \frac{-a}{-1}$,we get $a=a$ (always true).
From $\frac{a}{1} = \frac{2}{-2b}$,we get $a = -\frac{1}{b}$,which implies $ab = -1$.

(B) The given equation is $x^2+4xy+3y^2-4x-10y+3=0$.
Comparing with $ax^2+2hxy+by^2+2gx+2fy+c=0$,we have $a=1, h=2, b=3, g=-2, f=-5, c=3$.
The point of intersection $(x_1, y_1)$ is given by $\left(\frac{bg-fh}{h^2-ab}, \frac{af-gh}{h^2-ab}\right)$.
Substituting the values: $x_1 = \frac{3(-2)-(-5)(2)}{2^2-1(3)} = \frac{-6+10}{1} = 4$ and $y_1 = \frac{1(-5)-(-2)(2)}{2^2-1(3)} = \frac{-5+4}{1} = -1$.
So,the point of intersection is $(4, -1)$.
The equation of the line passing through $(4, -1)$ and $(2, 2)$ is given by $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.
$y - (-1) = \frac{2 - (-1)}{2 - 4}(x - 4) \Rightarrow y + 1 = \frac{3}{-2}(x - 4)$.
$-2y - 2 = 3x - 12 \Rightarrow 3x + 2y - 10 = 0$.

(A) The given equation of the pair of lines is $ax^2+2hxy-ay^2+2gx+2fy+c=0$.
Let the point of intersection be $(x_0, y_0)$.
The coordinates of the point of intersection of the pair of lines $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$ are given by the formulas:
$x_0 = \frac{HF-BG}{AB-H^2}$ and $y_0 = \frac{GH-AF}{AB-H^2}$.
Here,$A=a$,$H=h$,$B=-a$,$G=g$,$F=f$,$C=c$.
Substituting these values:
$x_0 = \frac{hf-g(-a)}{a(-a)-h^2} = \frac{hf+ag}{-(a^2+h^2)}$
$y_0 = \frac{gh-af}{-(a^2+h^2)} = \frac{af-gh}{a^2+h^2}$.
The square of the distance from the origin $(0,0)$ is $x_0^2 + y_0^2$.
$x_0^2 + y_0^2 = \frac{(hf+ag)^2 + (af-gh)^2}{(a^2+h^2)^2} = \frac{h^2f^2+a^2g^2+2afgh + a^2f^2+g^2h^2-2afgh}{(a^2+h^2)^2}$
$= \frac{f^2(h^2+a^2) + g^2(a^2+h^2)}{(a^2+h^2)^2} = \frac{(f^2+g^2)(a^2+h^2)}{(a^2+h^2)^2} = \frac{f^2+g^2}{a^2+h^2}$.

(B) Factorizing the first pair of lines:
$6x^2 - xy - 12y^2 = 6x^2 - 9xy + 8xy - 12y^2 = 3x(2x - 3y) + 4y(2x - 3y) = (3x + 4y)(2x - 3y) = 0$
Factorizing the second pair of lines:
$15x^2 + 14xy - 8y^2 = 15x^2 + 20xy - 6xy - 8y^2 = 5x(3x + 4y) - 2y(3x + 4y) = (5x - 2y)(3x + 4y) = 0$
The common line is $3x + 4y = 0$.
The slope of this line is $m = -\frac{3}{4}$.
The equation of the line passing through $(-1, 1)$ with slope $m = -\frac{3}{4}$ is:
$y - 1 = -\frac{3}{4}(x + 1)$
$4y - 4 = -3x - 3$
$3x + 4y - 1 = 0$

(B) The equation of the pair of lines is given by $ax^2+2hxy+by^2=0$,where $a=3$,$2h=5$,and $b=4$.
The equation of the angle bisectors is given by the formula $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$.
Substituting the values,we get $\frac{x^2-y^2}{3-4} = \frac{xy}{5/2}$.
This simplifies to $\frac{x^2-y^2}{-1} = \frac{2xy}{5}$.
Multiplying both sides by $-1$,we get $x^2-y^2 = -\frac{2}{5}xy$.
Rearranging the terms,we get $x^2-y^2+\frac{2}{5}xy=0$.

(A) Given the equation of the pair of lines: $f(x, y) = 3x^2 - 11xy + 10y^2 - 7x + 13y + 4 = 0$.
To find the point of intersection,we solve the partial derivatives $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$.
$\frac{\partial f}{\partial x} = 6x - 11y - 7 = 0$
$\frac{\partial f}{\partial y} = -11x + 20y + 13 = 0$
Solving these simultaneous equations using the cross-multiplication method:
$\frac{x}{(-11)(13) - (20)(-7)} = \frac{-y}{(6)(13) - (-11)(-7)} = \frac{1}{(6)(20) - (-11)(-11)}$
$\frac{x}{-143 + 140} = \frac{-y}{78 - 77} = \frac{1}{120 - 121}$
$\frac{x}{-3} = \frac{-y}{1} = \frac{1}{-1}$
$x = 3$ and $y = 1$.
Thus,the point of intersection is $(3, 1)$.

(B) The given equation of the pair of straight lines is $3x^2-11xy+10y^2-7x+13y+k=0$.
Comparing this with the general equation $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get:
$a=3, h=-\frac{11}{2}, b=10, g=-\frac{7}{2}, f=\frac{13}{2}$.
The point of intersection $(x, y)$ of the pair of lines is given by the formula:
$x = \frac{hf-bg}{ab-h^2}$ and $y = \frac{gh-af}{ab-h^2}$.
First,calculate the denominator: $ab-h^2 = (3)(10) - (-\frac{11}{2})^2 = 30 - \frac{121}{4} = \frac{120-121}{4} = -\frac{1}{4}$.
Now,calculate the numerator for $x$: $hf-bg = (-\frac{11}{2})(\frac{13}{2}) - (10)(-\frac{7}{2}) = -\frac{143}{4} + 35 = \frac{-143+140}{4} = -\frac{3}{4}$.
So,$x = \frac{-3/4}{-1/4} = 3$.
Now,calculate the numerator for $y$: $gh-af = (-\frac{7}{2})(-\frac{11}{2}) - (3)(\frac{13}{2}) = \frac{77}{4} - \frac{39}{2} = \frac{77-78}{4} = -\frac{1}{4}$.
So,$y = \frac{-1/4}{-1/4} = 1$.
Therefore,the point of intersection is $(3, 1)$.