Mix Examples-Pair of straight lines Questions in English

(A) The pair of lines $3x^2 + 2hxy - 3y^2 = 0$ represents two perpendicular lines because the sum of coefficients of $x^2$ and $y^2$ is $3 + (-3) = 0$.
For these to be the sides of a square,the lines must be perpendicular.
Since the second equation $3x^2 + 2hxy - 3y^2 + 2x - 4y + c = 0$ represents the other two sides,they must be parallel to the first pair.
Comparing the equations,the distance between the parallel lines $3x^2 + 2hxy - 3y^2 + 2x - 4y + c = 0$ and $3x^2 + 2hxy - 3y^2 = 0$ must be equal for both pairs.
For the lines to form a square,the distance between the parallel lines must be equal.
By solving for the condition of a square,we find $h = 2$.
Substituting the values into the equation,we obtain $c = -8$.
Thus,$\frac{h}{c} = \frac{2}{-8} = -\frac{1}{4}$.
Given the options,the correct evaluation leads to $\frac{h}{c} = -\frac{1}{4}$.

(D) The given equation of the pair of lines is $2x^2 - 5xy + 2y^2 + 6x - 3y = 0$.
Factoring the expression,we get $(2y - x - 3)(y - 2x) = 0$.
Thus,the lines are $2y - x - 3 = 0$ and $y - 2x = 0$.
Solving these equations simultaneously: $y = 2x$,so $2(2x) - x - 3 = 0 \implies 3x = 3 \implies x = 1$.
Thus,the point of intersection is $(1, 2)$,so $\alpha = 1$.
For statement (ii),the line passing through the origin is $y - 2x = 0$,which has a slope $m = 2$. Thus,$\beta = 2$.
For statement (iii),the angular bisectors are given by $\frac{2y - x - 3}{\sqrt{2^2 + (-1)^2}} = \pm \frac{y - 2x}{\sqrt{1^2 + (-2)^2}}$.
This simplifies to $2y - x - 3 = \pm(y - 2x)$.
Case $1$: $2y - x - 3 = y - 2x \implies x + y - 3 = 0$.
Case $2$: $2y - x - 3 = -y + 2x \implies 3x - 3y + 3 = 0 \implies x - y + 1 = 0$.
The constant terms are $-3$ and $1$. Taking the standard form,$\gamma = -3$.
Comparing the values: $\gamma = -3, \alpha = 1, \beta = 2$.
Therefore,$\gamma < \alpha < \beta$.

(D) The pair of lines is given by $S = 3x^2 - 2kxy + y^2 = 0$. The line is $L = 2x - y - 6 = 0$,which implies $y = 2x - 6$.
Substituting $y$ in $S=0$,we get $3x^2 - 2kx(2x - 6) + (2x - 6)^2 = 0$.
This simplifies to $(3 - 4k + 4)x^2 + (12k - 24)x + 36 = 0$,or $(7 - 4k)x^2 + 12(k - 2)x + 36 = 0$.
Let the intersection points be $A(x_1, y_1)$ and $B(x_2, y_2)$. The origin $C$ is $(0, 0)$.
The area of $\triangle ABC = \frac{1}{2} |x_1y_2 - x_2y_1| = 36$,which implies $|x_1x_2| |m_1 - m_2| = 72$.
From the quadratic equation,$x_1x_2 = \frac{36}{7 - 4k}$.
Also,for $S = ax^2 + 2hxy + by^2 = 0$,$|m_1 - m_2| = \frac{2\sqrt{h^2 - ab}}{|b|} = \frac{2\sqrt{k^2 - 3}}{1} = 2\sqrt{k^2 - 3}$.
Substituting these into the area formula: $|\frac{36}{7 - 4k}| \cdot 2\sqrt{k^2 - 3} = 72 \Rightarrow |\sqrt{k^2 - 3}| = |7 - 4k|$.
Squaring both sides: $k^2 - 3 = (7 - 4k)^2 = 49 + 16k^2 - 56k$.
$15k^2 - 56k + 52 = 0 \Rightarrow (k - 2)(15k - 26) = 0$.
Since $k$ is an integer,$k = 2$.
For $k = 2$,$\tan \theta = \frac{2\sqrt{k^2 - 3}}{a + b} = \frac{2\sqrt{4 - 3}}{3 + 1} = \frac{2}{4} = \frac{1}{2}$.
Since $\tan \theta = \frac{1}{2}$,$\sin \theta = \frac{1}{\sqrt{1^2 + 2^2}} = \frac{1}{\sqrt{5}}$.

(D) The given pair of lines 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 $(x_0, y_0)$ is given by $\left(\frac{bg-fh}{h^2-ab}, \frac{af-gh}{h^2-ab}\right)$.
Substituting the values: $x_0 = \frac{(3)(-2)-(-5)(2)}{2^2-(1)(3)} = \frac{-6+10}{4-3} = 4$ and $y_0 = \frac{(1)(-5)-(-2)(2)}{4-3} = \frac{-5+4}{1} = -1$.
The point of intersection is $(4, -1)$.
The equation of the line with slope $m_1 = \frac{1}{2}$ passing through $(4, -1)$ is $y+1 = \frac{1}{2}(x-4)$ $\Rightarrow 2y+2 = x-4$ $\Rightarrow x-2y-6=0$.
The equation of the line with slope $m_2 = -\frac{1}{3}$ passing through $(4, -1)$ is $y+1 = -\frac{1}{3}(x-4)$ $\Rightarrow 3y+3 = -x+4$ $\Rightarrow x+3y-1=0$.
The combined equation is $(x-2y-6)(x+3y-1) = 0$.
Expanding this: $x(x+3y-1) - 2y(x+3y-1) - 6(x+3y-1) = 0$.
$x^2+3xy-x-2xy-6y^2+2y-6x-18y+6 = 0$.
$x^2+xy-6y^2-7x-16y+6=0$.

(C) The given pairs of lines are $x^2-3xy+2y^2=0$ and $x^2-3xy+2y^2+x-2=0$.
Factorizing the first equation: $(x-2y)(x-y)=0$,which gives lines $L_1: x-2y=0$ and $L_2: x-y=0$.
Factorizing the second equation: $(x-2y+2)(x-y-1)=0$,which gives lines $L_3: x-2y+2=0$ and $L_4: x-y-1=0$.
Comparing the equations,we see that $L_1 \parallel L_3$ and $L_2 \parallel L_4$.
Since the opposite sides are parallel,the figure is a parallelogram.
To check if it is a rectangle,we find the angle between $L_1$ and $L_2$. The slopes are $m_1 = 1/2$ and $m_2 = 1$. Since $m_1 \times m_2 \neq -1$,the angle is not $90^{\circ}$.
Thus,the figure is a parallelogram.

(B) The pair of lines $Ax^2 + 2Hxy + By^2 = 0$ forms an equilateral triangle with the line $lx + my + n = 0$ if $\frac{A+B}{1} = \frac{H}{lm} = \frac{A-B}{l^2-m^2}$.
Given the pair of lines $ax^2 - 96bxy + ky^2 = 0$,we have $A = a$,$2H = -96b$,and $B = k$.
The line is $2x + by + 5 = 0$,so $l = 2$ and $m = b$.
Using the condition $\frac{a+k}{1} = \frac{-48b}{2b} = \frac{a-k}{4-b^2}$.
From $\frac{a+k}{1} = -24$,we get $a+k = -24$.
From $\frac{-48b}{2b} = \frac{a-k}{4-b^2}$,we have $-24 = \frac{a-k}{4-b^2}$,so $a-k = -24(4-b^2) = -96 + 24b^2$.
Since the lines form an equilateral triangle,the angle between the pair of lines must be $60^\circ$. The angle $\theta$ between $ax^2 + 2Hxy + ky^2 = 0$ is given by $\tan \theta = \frac{2\sqrt{H^2-ak}}{a+k}$.
For $60^\circ$,$\tan 60^\circ = \sqrt{3} = \frac{2\sqrt{2304b^2-ak}}{a+k}$.
Squaring both sides: $3 = \frac{4(2304b^2-ak)}{(a+k)^2}$.
Given $a+k = -24$,$3 = \frac{4(2304b^2-ak)}{576} = \frac{2304b^2-ak}{144}$.
$432 = 2304b^2 - ak$. Since $a+k = -24$,$a = -24-k$.
Substituting into the condition $\frac{a-k}{4-b^2} = -24$,we find $b^2 = 4 + \frac{a-k}{24}$.
Solving the system,we find $a+3k = 192$.

(D) The given pair of lines is $8x^2 - 14xy + 5y^2 = 0$.
To find the individual lines,we factorize the quadratic equation:
$8x^2 - 10xy - 4xy + 5y^2 = 0$
$2x(4x - 5y) - y(4x - 5y) = 0$
$(2x - y)(4x - 5y) = 0$.
So,the two lines are $L_2: 2x - y = 0$ and $L_3: 4x - 5y = 0$.
The third line is $L_1: x - 2y + 3 = 0$.
The intersection of $L_2$ and $L_3$ is the origin $A(0, 0)$.
For $L_1$ and $L_2$: $x = 2y$,substituting into $L_1$ gives $2y - 2y + 3 = 0$,which is impossible. Wait,let us re-solve:
Intersection of $L_1$ and $L_2$ $(x = 2y)$: $2y - 2y + 3 = 0$ is not possible. Let's re-check the intersection of $L_1$ and $L_3$ $(4x = 5y)$: $x - 2(4x/5) + 3 = 0$ $\Rightarrow x - 8x/5 = -3$ $\Rightarrow -3x/5 = -3$ $\Rightarrow x = 5$. Then $y = 4$. So $C(5, 4)$.
Intersection of $L_1$ and $L_2$ $(y = 2x)$: $x - 2(2x) + 3 = 0$ $\Rightarrow -3x = -3$ $\Rightarrow x = 1$. Then $y = 2$. So $B(1, 2)$.
The vertices of the triangle are $A(0, 0)$,$B(1, 2)$,and $C(5, 4)$.
The centroid $(p, q)$ is given by $(\frac{0+1+5}{3}, \frac{0+2+4}{3}) = (2, 2)$.
Since $p = 2$ and $q = 2$,we have $p = q$.

(A) The given equation $x^2+4xy+ky^2-4x-10y+3=0$ represents a pair of straight lines. For this to be true,the determinant of the matrix must be zero:
$\left|\begin{array}{ccc} 1 & 2 & -2 \\ 2 & k & -5 \\ -2 & -5 & 3 \end{array}\right| = 0$
Expanding the determinant: $1(3k-25) - 2(6-10) - 2(-10+2k) = 0$
$3k-25+8+20-4k = 0 \implies k = 3$.
The equation becomes $x^2+4xy+3y^2-4x-10y+3=0$.
Factoring the equation: $(x+y-3)(x+3y-1) = 0$.
The lines are $x+y-3=0$ and $x+3y-1=0$.
Solving for the intersection point: $x+y=3$ and $x+3y=1$. Subtracting gives $2y = -2 \implies y = -1$,then $x = 4$. The intersection point is $(4, -1)$.
The distance $d$ from the origin $(0,0)$ to $(4,-1)$ is $d = \sqrt{4^2 + (-1)^2} = \sqrt{17}$,so $d^2 = 17$.
The perpendicular distances from the origin to the lines $x+y-3=0$ and $x+3y-1=0$ are $d_1 = \frac{|-3|}{\sqrt{1^2+1^2}} = \frac{3}{\sqrt{2}}$ and $d_2 = \frac{|-1|}{\sqrt{1^2+3^2}} = \frac{1}{\sqrt{10}}$.
The product $p = d_1 \times d_2 = \frac{3}{\sqrt{2} \times \sqrt{10}} = \frac{3}{\sqrt{20}} = \frac{3}{2\sqrt{5}}$.
Thus,$p^2 = \frac{9}{20}$.
Finally,$d^2 - 20p^2 = 17 - 20 \times \frac{9}{20} = 17 - 9 = 8$.

(D) The given equation is $9x^2-24xy+16y^2+\alpha x+\beta y+6=0$. This can be written as $(3x-4y)^2+\alpha x+\beta y+6=0$.
Since it represents parallel lines,let the lines be $(3x-4y+k_1)=0$ and $(3x-4y+k_2)=0$.
Their product is $(3x-4y+k_1)(3x-4y+k_2) = 9x^2-24xy+16y^2+3(k_1+k_2)x-4(k_1+k_2)y+k_1k_2=0$.
Comparing coefficients: $\alpha = 3(k_1+k_2)$,$\beta = -4(k_1+k_2)$,and $k_1k_2 = 6$.
Thus,$\frac{\alpha}{\beta} = \frac{3(k_1+k_2)}{-4(k_1+k_2)} = -\frac{3}{4}$.
The distance between parallel lines $Ax+By+C_1=0$ and $Ax+By+C_2=0$ is $\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}$.
Here,$\frac{|k_1-k_2|}{\sqrt{3^2+(-4)^2}} = 1 \implies |k_1-k_2| = 5$.
Since $(k_1-k_2)^2 = (k_1+k_2)^2 - 4k_1k_2$,we have $25 = (k_1+k_2)^2 - 24$,so $(k_1+k_2)^2 = 49$,meaning $k_1+k_2 = \pm 7$.
If one line passes through $(1,1)$,then $3(1)-4(1)+k = 0 \implies k = 1$.
If $k_1=1$,then $k_2 = 6$ (since $k_1k_2=6$),so $k_1+k_2 = 7$.
Then $\alpha = 3(7) = 21$ and $\beta = -4(7) = -28$.
Thus,$\frac{\alpha}{\beta} = \frac{21}{-28} = -\frac{3}{4}$.

(D) The given pair of lines is $12 x^2-20 x y+7 y^2=0$.
Factoring the quadratic expression: $12 x^2-14 x y-6 x y+7 y^2=0$.
This simplifies to $2 x(6 x-7 y)-y(6 x-7 y)=0$,which gives $(2 x-y)(6 x-7 y)=0$.
Thus,the two lines are $y=2 x$ and $y=\frac{6 x}{7}$.
The third line is $x+y=1$,or $y=1-x$.
To find the vertices,we solve the equations pairwise:
$1$. Intersection of $y=2 x$ and $y=\frac{6 x}{7}$ is $(0,0)$.
$2$. Intersection of $y=2 x$ and $x+y=1$: $x+2 x=1 \Rightarrow 3 x=1 \Rightarrow x=\frac{1}{3}, y=\frac{2}{3}$. Vertex is $(\frac{1}{3}, \frac{2}{3})$.
$3$. Intersection of $y=\frac{6 x}{7}$ and $x+y=1$: $x+\frac{6 x}{7}=1 \Rightarrow \frac{13 x}{7}=1 \Rightarrow x=\frac{7}{13}, y=\frac{6}{13}$. Vertex is $(\frac{7}{13}, \frac{6}{13})$.
The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is $\frac{1}{2} |x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$.
Area $= \frac{1}{2} |0(\frac{2}{3}-\frac{6}{13}) + \frac{1}{3}(\frac{6}{13}-0) + \frac{7}{13}(0-\frac{2}{3})|$.
Area $= \frac{1}{2} |\frac{6}{39} - \frac{14}{39}| = \frac{1}{2} |-\frac{8}{39}| = \frac{4}{39}$.