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

(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 $(3, 1)$.

(D) Given the equation: $y^3-4x^2y=0$
Factorizing the equation: $y(y^2-4x^2)=0$
Using the identity $a^2-b^2=(a-b)(a+b)$,we get: $y(y-2x)(y+2x)=0$
This gives the three lines: $L_1: y=0$,$L_2: y=2x$,and $L_3: y=-2x$.
All three lines pass through the origin $(0,0)$.
Since all three lines intersect at a single common point $(0,0)$,they are concurrent lines.

(D) The equation of the angle bisectors for the pair of lines $a x^2+2 h x y+b y^2=0$ is given by $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$.
For the first pair $x^2-2 p x y-y^2=0$,we have $a=1, b=-1, h=-p$.
The angle bisectors are $\frac{x^2-y^2}{1-(-1)}=\frac{x y}{-p}$,which simplifies to $\frac{x^2-y^2}{2}=\frac{x y}{-p}$,or $x^2-y^2+\frac{2 x y}{p}=0$.
Given that this pair of bisectors is the same as the second pair $x^2-2 q x y-y^2=0$,we compare the coefficients.
Comparing $x^2-y^2+\frac{2}{p} x y=0$ with $x^2-2 q x y-y^2=0$,we identify that the coefficient of $x y$ must match:
$\frac{2}{p} = -2 q$.
Therefore,$p q = -1$.

(C) 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 of lines $x^2-2pxy-y^2=0$,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 $\frac{x^2-y^2}{2} = \frac{xy}{-p}$,or $x^2-y^2 = -\frac{2}{p}xy$,which is $x^2 + \frac{2}{p}xy - y^2 = 0$.
It is given that this pair of bisectors is the same as the pair of lines $x^2-2qxy-y^2=0$.
Comparing the coefficients of $xy$,we get $\frac{2}{p} = -2q$.
This implies $pq = -1$,or $pq+1=0$.
Thus,option $(C)$ is correct.

(B) The given pair of straight lines is $x^2-y^2+2y-1=0$.
This can be written as $x^2-(y-1)^2=0$,which factors as $(x-y+1)(x+y-1)=0$.
Thus,the two lines are $L_1: x-y+1=0$ and $L_2: x+y-1=0$.
The angular bisectors of these lines are given by $\frac{x-y+1}{\sqrt{1^2+(-1)^2}} = \pm \frac{x+y-1}{\sqrt{1^2+1^2}}$,which simplifies to $x-y+1 = \pm(x+y-1)$.
Case $1$: $x-y+1 = x+y-1$ $\Rightarrow 2y=2$ $\Rightarrow y=1$.
Case $2$: $x-y+1 = -(x+y-1)$ $\Rightarrow x-y+1 = -x-y+1$ $\Rightarrow 2x=0$ $\Rightarrow x=0$.
The triangle is formed by the line $x+y=3$ and the lines $x=0$ and $y=1$.
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 vertices are $(0,1), (0,3),$ and $(2,1)$.
The base of the triangle along the $y$-axis is $|3-1|=2$ units.
The height of the triangle from the vertex $(2,1)$ to the $y$-axis is $|2-0|=2$ units.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 2 = 2$ sq. units.

(B) Given,the pair of lines is $x^2-y^2+2y-1=0$.
$x^2 = y^2-2y+1$
$\Rightarrow x^2 = (y-1)^2$
$\Rightarrow x^2 - (y-1)^2 = 0$
$\Rightarrow (x+y-1)(x-y+1) = 0$
So,the lines are $l_1: x+y-1=0$ and $l_2: x-y+1=0$.
The angle bisectors of these lines are found by $\frac{x+y-1}{\sqrt{1^2+1^2}} = \pm \frac{x-y+1}{\sqrt{1^2+(-1)^2}}$.
$\Rightarrow x+y-1 = \pm(x-y+1)$.
Case $1$: $x+y-1 = x-y+1$ $\Rightarrow 2y = 2$ $\Rightarrow y=1$.
Case $2$: $x+y-1 = -(x-y+1)$ $\Rightarrow x+y-1 = -x+y-1$ $\Rightarrow 2x = 0$ $\Rightarrow x=0$.
The angle bisectors are the lines $x=0$ ($Y$-axis) and $y=1$.
The triangle is formed by the lines $x+y=4$,$x=0$,and $y=1$.
The vertices of the triangle are:
$1$. Intersection of $x=0$ and $y=1$ is $(0, 1)$.
$2$. Intersection of $x=0$ and $x+y=4$ is $(0, 4)$.
$3$. Intersection of $y=1$ and $x+y=4$ is $(3, 1)$.
The base of the triangle along $x=0$ is $|4-1| = 3$.
The height of the triangle from $x=0$ to $x=3$ is $3$.
Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 3 = 4.5$ sq. units.

(B) The given equation is $3 x^2-5 x y+4 y^2=0$.
Comparing this with the general form $a x^2+2 h x y+b y^2=0$,we get $a=3$,$2 h=-5$,and $b=4$.
The equation of the bisectors of the angle between the pair of lines is given by $\frac{x^2-y^2}{a-b} = \frac{x y}{h}$.
Substituting the values,we get $\frac{x^2-y^2}{3-4} = \frac{x y}{-5/2}$.
This simplifies to $\frac{x^2-y^2}{-1} = \frac{2 x y}{-5}$.
Multiplying both sides by $-5$,we get $5(x^2-y^2) = 2 x y$.

(A) The given equation is $x^2-2 m x y-y^2=0$.
Comparing this with the general form $a x^2+2 h x y+b y^2=0$,we get $a=1, h=-m, b=-1$.
The equation of the angle bisectors is given by $\frac{x^2-y^2}{a-b} = \frac{x y}{h}$.
Substituting the values,we get $\frac{x^2-y^2}{1-(-1)} = \frac{x y}{-m}$.
This simplifies to $\frac{x^2-y^2}{2} = \frac{x y}{-m}$,which implies $-m(x^2-y^2) = 2 x y$.
Rearranging the terms,we get $m x^2 + 2 x y - m y^2 = 0$.
Dividing by $m$ (assuming $m \neq 0$),we get $x^2 + \frac{2}{m} x y - y^2 = 0$.
Comparing this with the given bisector equation $x^2-2 n x y-y^2=0$,we have $-2n = \frac{2}{m}$.
Therefore,$-n = \frac{1}{m}$,which gives $mn = -1$ or $mn+1=0$.

(B) The equation of the pair of bisectors of the angle between the pair of straight lines $a x^2+2 h x y+b y^2=0$ is given by $\frac{x^2-y^2}{a-b}=\frac{x y}{h}$.
For the pair $x^2-2 p x y-y^2=0$,we have $a=1, b=-1, h=-p$.
The equation of the bisectors is $\frac{x^2-y^2}{1-(-1)}=\frac{x y}{-p}$.
This simplifies to $\frac{x^2-y^2}{2}=\frac{x y}{-p}$,which implies $-p(x^2-y^2)=2xy$,or $p x^2+2 x y-p y^2=0$.
Dividing by $p$,we get $x^2+\frac{2}{p} x y-y^2=0$.
We are given that this pair is $x^2-2 q x y-y^2=0$.
Comparing the coefficients of $xy$,we have $-2 q = \frac{2}{p}$.
Therefore,$-p q = 1$,which gives $p q = -1$.

(D) The general equation of a pair of straight lines passing through the origin is given by $ax^2 + 2hxy + by^2 = 0$.
The equation of the pair of angle bisectors for these lines is given by the formula:
$\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$.
This formula is derived by considering the lines $y = m_1x$ and $y = m_2x$ and finding the bisectors of the angles between them.
Thus,the correct option is $D$.

(A) Given the equation of the pair of straight lines: $x^2+4xy-y^2=0$.
Comparing this with the general form $ax^2+2hxy+by^2=0$,we get $a=1$,$b=-1$,and $2h=4$,which implies $h=2$.
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: $\frac{x^2-y^2}{1-(-1)} = \frac{xy}{2} \Rightarrow \frac{x^2-y^2}{2} = \frac{xy}{2}$.
This simplifies to $x^2-xy-y^2=0$.
Since $y=mx$ is one of the bisectors,we substitute $y=mx$ into the equation: $x^2-x(mx)-(mx)^2=0$.
Dividing by $x^2$ (assuming $x \neq 0$),we get $1-m-m^2=0$,or $m^2+m-1=0$.
Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $m = \frac{-1 \pm \sqrt{1^2-4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}$.
Therefore,$2m = -1 \pm \sqrt{5}$.

(A) Let the given equations be:
$x^2-16pxy-y^2=0 \quad (i)$
$x^2-16qxy-y^2=0 \quad (ii)$
The equation of the angle bisectors of a pair of lines $ax^2+2hxy+by^2=0$ is given by $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$.
For equation $(i)$,$a=1, b=-1, h=-8p$. The bisectors are $\frac{x^2-y^2}{1-(-1)} = \frac{xy}{-8p} \implies \frac{x^2-y^2}{2} = \frac{xy}{-8p} \implies -8px^2+8py^2=2xy \implies -8px^2-2xy+8py^2=0 \quad (iii)$.
For equation $(ii)$,$a=1, b=-1, h=-8q$. The bisectors are $\frac{x^2-y^2}{1-(-1)} = \frac{xy}{-8q} \implies \frac{x^2-y^2}{2} = \frac{xy}{-8q} \implies -8qx^2-2xy+8qy^2=0 \quad (iv)$.
Since each pair bisects the angle between the other,equation $(i)$ must be the same as equation $(iv)$ and equation $(ii)$ must be the same as equation $(iii)$.
Comparing $(i)$ and $(iv)$: $\frac{1}{-8q} = \frac{-16p}{-2} = \frac{-1}{8q}$. This leads to $1 = -64pq$,so $pq = \frac{-1}{64}$.

(C) The given equation is $ax^2+2hxy-2ay^2+3x+15y-9=0$.
Since the lines intersect at $(1,1)$,the point $(1,1)$ must satisfy the equation:
$a(1)^2+2h(1)(1)-2a(1)^2+3(1)+15(1)-9=0$
$a+2h-2a+3+15-9=0$
$-a+2h+9=0 \implies a-2h=9$ (Equation $1$).
For a general second-degree equation $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$ to represent a pair of lines,the condition is $ABC+2FGH-AF^2-BG^2-CH^2=0$.
Here,$A=a, H=h, B=-2a, G=3/2, F=15/2, C=-9$.
Substituting these values:
$a(-2a)(-9) + 2(15/2)(h)(3/2) - a(15/2)^2 - (-2a)(3/2)^2 - (-9)(h)^2 = 0$
$18a^2 + \frac{45h}{2} - \frac{225a}{4} + \frac{18a}{4} + 9h^2 = 0$
$18a^2 + 9h^2 + \frac{45h}{2} - \frac{207a}{4} = 0$
Using $a=2h+9$ from Equation $1$,substituting into the condition and solving for $h$ yields $h=-3$ and $a=3$.
Thus,$ah = 3 \times (-3) = -9$.
Wait,re-evaluating the intersection point condition: the partial derivatives with respect to $x$ and $y$ must be zero at $(1,1)$.
$f_x = 2ax+2hy+3 = 0 \implies 2a+2h+3=0$
$f_y = 2hx-4ay+15 = 0 \implies 2h-4a+15=0$
Solving these: $2a+2h=-3$ and $-4a+2h=-15$.
Subtracting: $6a=12 \implies a=2$.
Then $2(2)+2h=-3 \implies 2h=-7 \implies h=-3.5$.
$ah = 2 \times (-3.5) = -7$.

(C) The given equation is $2x^2 + 2hxy + 2y^2 - x + y - 1 = 0$.
Comparing this with the general form $ax^2 + 2h'xy + by^2 + 2gx + 2fy + c = 0$,we have $a = 2$,$h' = h$,$b = 2$,$g = -1/2$,$f = 1/2$,and $c = -1$.
The angle $\theta$ between the lines is given by $\tan \theta = |\frac{2\sqrt{h^2 - ab}}{a + b}|$.
Given $\tan \theta = 3/4$,we have $3/4 = |\frac{2\sqrt{h^2 - 4}}{2 + 2}| = |\frac{2\sqrt{h^2 - 4}}{4}| = \frac{\sqrt{h^2 - 4}}{2}$.
Squaring both sides,$9/16 = (h^2 - 4)/4$,which gives $9/4 = h^2 - 4$,so $h^2 = 25/4$,implying $h = 5/2$ (since $h > 0$).
The point of intersection $(x, y)$ is found by solving $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$.
$\frac{\partial f}{\partial x} = 4x + 2hy - 1 = 0 \implies 4x + 5y - 1 = 0$.
$\frac{\partial f}{\partial y} = 2hx + 4y + 1 = 0 \implies 5x + 4y + 1 = 0$.
Solving these equations: $16x + 20y - 4 = 0$ and $25x + 20y + 5 = 0$.
Subtracting gives $9x + 9 = 0$,so $x = -1$.
Substituting $x = -1$ into $4x + 5y - 1 = 0$ gives $-4 + 5y - 1 = 0$,so $5y = 5$,$y = 1$.
The point of intersection is $(-1, 1)$.

(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 $abc+2fgh-af^2-bg^2-ch^2=0$.
For the given equation $3x^2+7xy+2y^2+2gx+2fy+2=0$,we have $a=3, h=\frac{7}{2}, b=2, g=g, f=f, c=2$.
Substituting these values into the condition:
$3(2)(2) + 2f g(\frac{7}{2}) - 3f^2 - 2g^2 - 2(\frac{7}{2})^2 = 0$
$12 + 7fg - 3f^2 - 2g^2 - \frac{49}{2} = 0$
$24 + 14fg - 6f^2 - 4g^2 - 49 = 0$
$6f^2 + 4g^2 - 14fg = -25$ ---$(i)$
To find the point of intersection $(x, y)$,we partially differentiate the equation with respect to $x$ and $y$:
$\frac{\partial}{\partial x}(3x^2+7xy+2y^2+2gx+2fy+2) = 6x+7y+2g = 0$ ---(ii)
$\frac{\partial}{\partial y}(3x^2+7xy+2y^2+2gx+2fy+2) = 7x+4y+2f = 0$ ---(iii)
Solving (ii) and (iii) for $x$ and $y$ using Cramer's rule:
$x = \frac{-(2g)(4) - (-2f)(7)}{(6)(4) - (7)(7)} = \frac{-8g+14f}{-25} = \frac{8g-14f}{25}$
$y = \frac{(6)(-2f) - (7)(-2g)}{-25} = \frac{-12f+14g}{-25} = \frac{12f-14g}{25}$
The square of the distance from the origin is $x^2+y^2 = \frac{2}{5}$.
$\left(\frac{8g-14f}{25}\right)^2 + \left(\frac{12f-14g}{25}\right)^2 = \frac{2}{5}$
$(64g^2 + 196f^2 - 224fg + 144f^2 + 196g^2 - 336fg) = \frac{2}{5} \times 625 = 250$
$260g^2 + 340f^2 - 560fg = 250$
$26g^2 + 34f^2 - 56fg = 25$ ---(iv)
From $(i)$,$6f^2 + 4g^2 - 14fg = -25$,so $24f^2 + 16g^2 - 56fg = -100$ ---$(v)$
Subtracting $(v)$ from (iv):
$(26g^2 - 16g^2) + (34f^2 - 24f^2) = 25 - (-100)$
$10g^2 + 10f^2 = 125$
$f^2 + g^2 = \frac{125}{10} = \frac{25}{2}$

(C) The general equation of a pair of straight lines $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of lines if the determinant $\Delta = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0$.
Comparing with $2x^2 + kxy - 6y^2 + 3x + y + 1 = 0$,we have $a=2, h=k/2, b=-6, g=3/2, f=1/2, c=1$.
The condition $\Delta = 0$ gives $\begin{vmatrix} 2 & k/2 & 3/2 \\ k/2 & -6 & 1/2 \\ 3/2 & 1/2 & 1 \end{vmatrix} = 0$.
Multiplying rows by $2$,we get $\frac{1}{8} \begin{vmatrix} 4 & k & 3 \\ k & -12 & 1 \\ 3 & 1 & 2 \end{vmatrix} = 0$.
Expanding the determinant: $4(-24 - 1) - k(2k - 3) + 3(k + 36) = 0$.
$-100 - 2k^2 + 3k + 3k + 108 = 0$ $\Rightarrow -2k^2 + 6k + 8 = 0$ $\Rightarrow k^2 - 3k - 4 = 0$.
$(k - 4)(k + 1) = 0$. Since $k > 0$,we have $k = 4$.
The equation becomes $2x^2 + 4xy - 6y^2 + 3x + y + 1 = 0$.
Factoring the expression: $(2x - 2y + 1)(x + 3y + 1) = 0$.
Solving the system $2x - 2y + 1 = 0$ and $x + 3y + 1 = 0$:
From the second equation,$x = -3y - 1$. Substituting into the first: $2(-3y - 1) - 2y + 1 = 0$ $\Rightarrow -6y - 2 - 2y + 1 = 0$ $\Rightarrow -8y = 1$ $\Rightarrow y = -1/8$.
Then $x = -3(-1/8) - 1 = 3/8 - 8/8 = -5/8$.
The point of intersection is $\left(-\frac{5}{8}, -\frac{1}{8}\right)$.

(C) The given pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$. Comparing this with $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=1, b=-35, g=-2, f=22, c=-12$.
The point of intersection $(x, y)$ of the pair of lines is given by the formula $\left(\frac{hf-bg}{ab-h^2}, \frac{gh-af}{ab-h^2}\right)$.
Substituting the values: $x = \frac{(1)(22)-(-35)(-2)}{(1)(-35)-(1)^2} = \frac{22-70}{-36} = \frac{-48}{-36} = \frac{4}{3}$.
$y = \frac{(-2)(1)-(1)(22)}{(1)(-35)-(1)^2} = \frac{-2-22}{-36} = \frac{-24}{-36} = \frac{2}{3}$.
The point of intersection is $(\frac{4}{3}, \frac{2}{3})$.
Since the lines are concurrent,this point must satisfy the equation $5x+ky-8=0$.
$5(\frac{4}{3}) + k(\frac{2}{3}) - 8 = 0$.
$\frac{20}{3} + \frac{2k}{3} = 8$.
$20 + 2k = 24$.
$2k = 4 \Rightarrow k = 2$.

(D) The point of intersection of the pair of lines $f(x, y) = x^2+xy+2y^2-3x+2y+4=0$ can be found by solving the partial derivatives of $f(x, y)$ with respect to $x$ and $y$ equal to zero.
Taking partial derivative with respect to $x$:
$\frac{\partial f}{\partial x} = 2x + y - 3 = 0 \quad \dots (i)$
Taking partial derivative with respect to $y$:
$\frac{\partial f}{\partial y} = x + 4y + 2 = 0 \quad \dots (ii)$
Solving the system of linear equations $(i)$ and $(ii)$:
From $(i)$,$y = 3 - 2x$.
Substituting into $(ii)$: $x + 4(3 - 2x) + 2 = 0$
$x + 12 - 8x + 2 = 0$
$-7x + 14 = 0 \implies x = 2$
Substituting $x = 2$ into $y = 3 - 2x$:
$y = 3 - 2(2) = 3 - 4 = -1$
Thus,the point of intersection is $(2, -1)$.
Hence,option $(D)$ is correct.

(D) The given equation of the pair of lines is $x^2+2xy-35y^2-4x+44y-12=0$.
Comparing this with the general form $ax^2+2hxy+by^2+2gx+2fy+c=0$,we get $a=1, h=1, b=-35, g=-2, f=22, c=-12$.
The point of intersection $(x_0, y_0)$ of the pair of lines is given by the formula:
$x_0 = \frac{hf-bg}{ab-h^2} = \frac{(1)(22)-(-35)(-2)}{(1)(-35)-(1)^2} = \frac{22-70}{-36} = \frac{-48}{-36} = \frac{4}{3}$.
$y_0 = \frac{gh-af}{ab-h^2} = \frac{(-2)(1)-(1)(22)}{(1)(-35)-(1)^2} = \frac{-2-22}{-36} = \frac{-24}{-36} = \frac{2}{3}$.
Since the lines are concurrent,the point $(\frac{4}{3}, \frac{2}{3})$ must satisfy the line $5x+\lambda y-8=0$.
$5(\frac{4}{3}) + \lambda(\frac{2}{3}) - 8 = 0$.
$\frac{20}{3} + \frac{2\lambda}{3} - 8 = 0$.
Multiply by $3$: $20 + 2\lambda - 24 = 0$.
$2\lambda - 4 = 0$.
$2\lambda = 4$.
$\lambda = 2$.

(A) The second pair of lines is $2x^2+xy-6y^2=0$. Factoring this,we get $2x^2+4xy-3xy-6y^2=0$,which is $(2x-3y)(x+2y)=0$. The lines are $2x-3y=0$ and $x+2y=0$.
Since the first pair $ax^2-7xy-3y^2=0$ shares exactly one line with the second,either $2x-3y=0$ or $x+2y=0$ must be a factor of $ax^2-7xy-3y^2=0$.
If $2x-3y=0$ is a factor,then $y = \frac{2}{3}x$. Substituting into $ax^2-7x(\frac{2}{3}x)-3(\frac{2}{3}x)^2=0$,we get $ax^2 - \frac{14}{3}x^2 - \frac{4}{3}x^2 = 0$,so $a - 6 = 0$,which gives $a=6$.
If $x+2y=0$ is a factor,then $x = -2y$. Substituting into $a(-2y)^2-7(-2y)y-3y^2=0$,we get $4ay^2 + 14y^2 - 3y^2 = 0$,so $4a + 11 = 0$,which gives $a = -\frac{11}{4}$ (not an integer).
Thus,$a=6$. The equation is $6x^2-7xy-3y^2=0$.
The equation of the bisectors of the angle between $Ax^2+Bxy+Cy^2=0$ is $\frac{x^2-y^2}{A-C} = \frac{xy}{B/2}$.
Here $A=6, B=-7, C=-3$. So $\frac{x^2-y^2}{6-(-3)} = \frac{xy}{-7/2} \Rightarrow \frac{x^2-y^2}{9} = \frac{-2xy}{7}$.
$7(x^2-y^2) = -18xy \Rightarrow 7x^2+18xy-7y^2=0$.

(C) Given the pair of straight lines: $3x^2+7xy+2y^2+5x+5y+2=0$.
Factorizing the equation,we get $(3x+y+2)(x+2y+1)=0$.
The lines are $L_1: 3x+y+2=0$ and $L_2: x+2y+1=0$.
Solving for the intersection point,we get $x = -3/5$ and $y = -1/5$.
The equation of the angle bisectors is given by $\frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}} = \pm \frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$.
Substituting the lines and shifting the origin to $(-3/5, -1/5)$,the equation becomes $\frac{3(x+3/5)+(y+1/5)}{\sqrt{3^2+1^2}} = \pm \frac{(x+3/5)+2(y+1/5)}{\sqrt{1^2+2^2}}$.
This simplifies to $\frac{5x+3}{\sqrt{10}} = \pm \frac{5y+1}{\sqrt{5}}$.
Squaring both sides: $\frac{(5x+3)^2}{10} = \frac{(5y+1)^2}{5} \Rightarrow (5x+3)^2 = 2(5y+1)^2$.
However,using the standard formula for angle bisectors of $ax^2+2hxy+by^2=0$,we have $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$.
Applying this to the shifted lines,we get $7(5x+3)^2-2(5x+3)(5y+1)-7(5y+1)^2=0$.

(C) The angle bisectors of two lines remain invariant under rotation of the lines by the same angle in opposite directions.
Therefore,the bisectors of the new lines are the same as the bisectors of the original lines $x^2-2hxy-y^2=0$.
The equation of the angle bisectors for a pair of lines $ax^2+2hxy+by^2=0$ is given by $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$.
Here,$a=1$,$b=-1$,and the coefficient of $xy$ is $-2h$,so the coefficient of $xy$ in the formula is $2h' = -2h$,which means $h' = -h$.
Substituting these values into the formula:
$\frac{x^2-y^2}{1-(-1)} = \frac{xy}{-h}$
$\frac{x^2-y^2}{2} = \frac{xy}{-h}$
$-h(x^2-y^2) = 2xy$
$-hx^2+hy^2 = 2xy$
$hx^2-hy^2+2xy = 0$

(D) The given equation represents a pair of lines: $f(x, y) = 2x^2 + 5xy - 3y^2 + 2gx + 2fy + c = 0$.
For a general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,the point of intersection $(x_0, y_0)$ is found by solving the partial derivatives $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$.
$\frac{\partial f}{\partial x} = 4x + 5y + 2g = 0$.
Substituting the point of intersection $(-1, -1)$:
$4(-1) + 5(-1) + 2g = 0 \implies -4 - 5 + 2g = 0 \implies 2g = 9 \implies g = 4.5$.
$\frac{\partial f}{\partial y} = 5x - 6y + 2f = 0$.
Substituting the point of intersection $(-1, -1)$:
$5(-1) - 6(-1) + 2f = 0 \implies -5 + 6 + 2f = 0 \implies 1 + 2f = 0 \implies f = -0.5$.
Therefore,$g + f = 4.5 + (-0.5) = 4$.
Thus,the correct option is $D$.

(A) The given equation of the pair of lines is $x^2-y^2-x+3y-2=0$.
Factoring the expression,we get $(x-y+1)(x+y-2)=0$.
Thus,the lines are $L_1: x-y+1=0$ and $L_2: x+y-2=0$.
Solving for the intersection point $P$:
Adding the equations: $(x-y+1) + (x+y-2) = 2x-1=0 \implies x=\frac{1}{2}$.
Substituting $x=\frac{1}{2}$ into $x+y-2=0$,we get $y=\frac{3}{2}$.
So,$P = (\frac{1}{2}, \frac{3}{2})$.
$\alpha$ is the square of the distance from the origin $(0,0)$ to $P$:
$\alpha = (\frac{1}{2})^2 + (\frac{3}{2})^2 = \frac{1}{4} + \frac{9}{4} = \frac{10}{4} = \frac{5}{2}$.
$\beta$ is the product of the perpendicular distances from the origin to $L_1$ and $L_2$:
$d_1 = \frac{|0-0+1|}{\sqrt{1^2+(-1)^2}} = \frac{1}{\sqrt{2}}$.
$d_2 = \frac{|0+0-2|}{\sqrt{1^2+1^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.
$\beta = d_1 \times d_2 = \frac{1}{\sqrt{2}} \times \sqrt{2} = 1$.
Therefore,$\alpha \beta = \frac{5}{2} \times 1 = \frac{5}{2}$.

(A) Let $f(x, y) = 2x^2+axy+3y^2+bx+cy-3=0$. Since $(2,-1)$ is the point of intersection,the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ must vanish at $(2,-1)$.
$\frac{\partial f}{\partial x} = 4x+ay+b = 0 \implies 4(2)+a(-1)+b = 0 \implies 8-a+b = 0 \implies a-b=8$....$(i)$
$\frac{\partial f}{\partial y} = ax+6y+c = 0 \implies a(2)+6(-1)+c = 0 \implies 2a+c=6$....(ii)
Also,the point $(2,-1)$ satisfies the original equation:
$2(2)^2+a(2)(-1)+3(-1)^2+b(2)+c(-1)-3 = 0$
$8-2a+3+2b-c-3 = 0 \implies -2a+2b-c = -8 \implies 2a-2b+c = 8$....(iii)
From $(i)$,$b = a-8$. Substitute into (iii):
$2a-2(a-8)+c = 8 \implies 2a-2a+16+c = 8 \implies c = -8$.
Substitute $c=-8$ into (ii):
$2a-8 = 6 \implies 2a = 14 \implies a = 7$.
From $(i)$,$b = 7-8 = -1$.
Finally,$3a+2b+c = 3(7)+2(-1)+(-8) = 21-2-8 = 11$.

(A) Let the equation of the pair of lines be $f(x, y) = 2x^2 - 13xy - 7y^2 + x + 23y - 6 = 0$.
To find the point of intersection,we solve $\frac{\partial f}{\partial x} = 4x - 13y + 1 = 0$ and $\frac{\partial f}{\partial y} = -13x - 14y + 23 = 0$.
Solving these simultaneous equations,we multiply the first by $14$ and the second by $13$:
$56x - 182y + 14 = 0$
$-169x - 182y + 299 = 0$
Subtracting the two gives $225x - 285 = 0$,so $x = \frac{285}{225} = \frac{19}{15}$.
Substituting $x$ into $4x - 13y + 1 = 0$,we get $4(\frac{19}{15}) + 1 = 13y$,which simplifies to $\frac{76+15}{15} = 13y$,so $y = \frac{91}{15 \times 13} = \frac{7}{15}$.
The point of intersection is $P(\frac{19}{15}, \frac{7}{15})$.
The line $L: x+y-1=0$ divides the segment joining $O(0,0)$ and $P(\frac{19}{15}, \frac{7}{15})$ in the ratio $k:1$.
The ratio is given by $-\frac{L(O)}{L(P)} = -\frac{0+0-1}{\frac{19}{15} + \frac{7}{15} - 1} = -\frac{-1}{\frac{26}{15} - 1} = \frac{1}{\frac{11}{15}} = \frac{15}{11}$.
Thus,the ratio is $15:11$.

(C) Comparing the given equation with $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$,we get $a=2, b=2 \lambda, h=-5, g=\frac{5}{2}, f=-8, c=-3$.
Since the equation represents a pair of straight lines,the determinant condition is:
$\left|\begin{array}{ccc} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=0$ $\Rightarrow \left|\begin{array}{ccc} 2 & -5 & 5/2 \\ -5 & 2 \lambda & -8 \\ 5/2 & -8 & -3 \end{array}\right|=0$.
Expanding the determinant: $2(-6 \lambda-64)+5(15+20)+\frac{5}{2}(40-5 \lambda)=0$.
$-12 \lambda-128+175+100-12.5 \lambda=0$ $\Rightarrow -24.5 \lambda = -147$ $\Rightarrow \lambda=6$.
Now,$b=2 \lambda = 12$. The point of intersection $(x, y)$ is given by $\left(\frac{b g-f h}{h^2-a b}, \frac{a f-g h}{h^2-a b}\right)$.
Denominator $h^2-a b = (-5)^2 - (2)(12) = 25-24=1$.
$x = \frac{(12)(5/2) - (-8)(-5)}{1} = 30-40 = -10$.
$y = \frac{(2)(-8) - (5/2)(-5)}{1} = -16 + 12.5 = -3.5 = \frac{-7}{2}$.
Thus,the point of intersection is $\left(-10, \frac{-7}{2}\right)$.

(C) The point of intersection $(x_0, y_0)$ of the pair of straight lines $ax^2+2hxy+by^2+2gx+2fy+c=0$ is given by the solution of the partial derivatives $\frac{\partial}{\partial x} = 0$ and $\frac{\partial}{\partial y} = 0$.
$2ax+2hy+2g=0 \implies ax+hy+g=0$
$2hx+2by+2f=0 \implies hx+by+f=0$
Solving these equations using Cramer's rule:
$x_0 = \frac{hf-bg}{ab-h^2}$ and $y_0 = \frac{gh-af}{ab-h^2}$.
The square of the distance from the origin $(0,0)$ is $D^2 = x_0^2 + y_0^2$.
Using the condition for a pair of lines $abc+2fgh-af^2-bg^2-ch^2=0$,we can simplify the expression for $x_0^2+y_0^2$.
Substituting the values,we get $D^2 = \frac{(hf-bg)^2+(gh-af)^2}{(ab-h^2)^2} = \frac{h^2f^2+b^2g^2-2hfbg+g^2h^2+a^2f^2-2ghaf}{(ab-h^2)^2}$.
Using $abc+2fgh=af^2+bg^2+ch^2$,this simplifies to $\frac{c(a+b)-f^2-g^2}{ab-h^2}$.

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