Two dimensional analytical geometry - class-XI

By square of distance formual:

The equation of the line which passes through (0,0) and (1,1) is y=x

The equation of a straight line is $y = mx +b$

Given a line passes through $A(0,0)$ & $B(1,5)$ 

The second degree terms cannot be factorized into two linear factors or else $h^2-ab=0-6=-$ve. Hence the given equation does not represent a pair of lines whatever $\lambda$ may be.

The lines are y^2 - 4xy - 5x^2 = 0, which factor into (y - 5x)(y + x) = 0. Using the properties of the orthocenter and the given fixed point, the locus of the vertex is derived as 2(x^2 + y^2) + (3a + 2b)x + (2a - 3b)y = 0.

The equation (ax^2 + by^2 + x)(x^2 - 5xy + 6y^2) = 0 represents a conic and two lines. If a = b, the first part represents a circle. If c (the constant term in the conic) is of opposite sign to a, it confirms the conic properties.

Collecting the terms $y^2$ and y the given equation can be written as
$y^2(x^2-2x-3)-4y(x^2-2x-3)=0$
or $(x-3)(x+1)y(y-4)=0$
It represents four lines $x=-1$, $x=3$, $y=0$ and $y=4$.
These two sets of parallel lines form a square of side four.

The given lines are

For the equation 4x^2 + 4xy - y^2 - 6x - 3y - 4 = 0, the distance between parallel lines is calculated using the formula 2 * sqrt(g^2 - ac) / sqrt(a(a+b)). Applying this to the given coefficients yields 2 * sqrt(5).

According to the question..........

$\begin{array}{l} Let,\, { m _{ 1\,  } }\& \, { m _{ 2 } } \\ sum\, of\, the\, slope:\, { m _{ 1 } }+{ m _{ 2 } }=\dfrac { { -2h } }{ b } ----(i) \\ and,\,  \\ product\, of\, slope:{ m _{ 1 } }.\, { m _{ 2 } }=\dfrac { a }{ b } -----(ii) \\ Here, \\ a{ x^{ 2 } }+2hxy+b{ y^{ 2 } }=0.........(general\, equ\, of\, straight\, line.) \\ cofficient\, of: \\ a={ \sec ^{ 2 }  }\theta -{ \sin ^{ 2 }  }\theta  \\ h=-\tan  \theta  \\ b={ \sin ^{ 2 }  }\theta  \\ Now,\, value\, put\, { { into } } \\ sum\, of\, the\, slope:\, { m _{ 1 } }+{ m _{ 2 } }=\dfrac { { -2h } }{ b } ----(i) \\ \Rightarrow { m _{ 1 } }+{ m _{ 2 } }=\dfrac { { -2(-tan\theta ) } }{ { { { \sin   }^{ 2 } }\theta  } } =\dfrac { { 2\sin  \theta \times 2 } }{ { 2{ { \sin   }^{ 2 } }\theta \, .\, \cos  \theta  } } =\dfrac { 4 }{ { 2sin\theta \cos  \theta  } } =\dfrac { 4 }{ { \sin  2\theta  } }  \\ and, \\ product\, of\, slope:{ m _{ 1 } }+{ m _{ 2 } }=\dfrac { a }{ b } -----(ii) \\ \Rightarrow { m _{ 1 } }.\, { m _{ 2 } }=\dfrac { { { { \sec   }^{ 2 } }\theta -{ { \sin   }^{ 2 } }\theta  } }{ { { { \sin   }^{ 2 } }\theta  } } =\dfrac { 1 }{ { { { \sin   }^{ 2 } }\theta \, .\, { { \cos   }^{ 2 } }\theta  } } -1\, \, \, \, \, \, \, \, \left[ { divide\, by\, 4 } \right.  \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\dfrac { 4 }{ { 4{ { \sin   }^{ 2 } }\theta \, .\, { { \cos   }^{ 2 } }\theta  } } -1\, \, =\dfrac { 4 }{ { { { (\sin  2\theta ) }^{ 2 } }\,  } } -1\,  \\ \, \, \, Now,find\, difference: \\ \, \, \, \, \, \, \, \, \, { ({ m _{ 1 } }-{ m _{ 2 } })^{ 2 } }={ ({ m _{ 1 } }+{ m _{ 2 } })^{ 2 } }-4{ m _{ 1 } }.\, { m _{ 2 } } \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ={ \left( { \dfrac { 4 }{ { \sin  2\theta  } }  } \right) ^{ 2 } }-4\left( { \dfrac { 4 }{ { ({ { \sin   }^{ 2 } }2\theta )\,  } } -1\,  } \right)  \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\dfrac { { 16 } }{ { ({ { \sin   }^{ 2 } }2\theta )\,  } } -\, \dfrac { { 16 } }{ { ({ { \sin   }^{ 2 } }2\theta )\,  } } +4 \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Rightarrow \, \, { ({ m _{ 1 } }-{ m _{ 2 } })^{ 2 } }\, \, \, =4 \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Rightarrow \, ({ m _{ 1 } }-{ m _{ 2 } })=+\sqrt { 4 } =2 \\ \, \, \, \therefore \, \, \, the\, \, differece\, of\, slope\, \, is\, 2. \\ So,\, that\, the\, correct\, option\, is\, B.\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,  \end{array}$

The intersection of two conics at four concyclic points is a standard result in geometry. The point P defined by the given ratios is the center of the circle passing through these four points.

taking $y$ is constant and finding the value of$x$ by roots formula.

$\Rightarrow$  The given equation is $12x^2-10y^2+11x-5y+k=0$

Given the two pairs of lines, we find the common line by solving the intersection of the quadratic forms. The joint equation of the remaining two lines is found by dividing the combined equation by the common factor.

By calculating the lines passing through (0, a) and applying the condition that the perpendicular distance from (2a, 2a) is 'a', we can verify the equation of the line joining the feet of the perpendiculars.

As the lines are perpendicular and form an isosceles triangle the other two angles must be $45^\circ$

${ y }^{ 2 }+3x-xy-3{ x }^{ 2 }=0\ y(y+3x)-x(y+3x)=0\ (y+3x)(y-x)=0$

The bisector of the angle between two lines is given by the angle bisector theorem. Given one line and the bisector, the other line can be determined by reflecting the known line across the bisector.

The family of lines can be rewritten as (2x + 3y - 5)cos(theta) + (3x - 5y + 7)sin(theta) = 0. For this to be true for all theta, both coefficients must be zero. Solving the system 2x + 3y = 5 and 3x - 5y = -7 yields (4/19, 29/19).

The points P and Q lie inside the triangle if they satisfy the inequalities defined by the three lines forming the triangle. Testing the bounds for alpha and beta confirms the range.

For a pair of lines ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0, the intercept on the x-axis is found by setting y=0 and solving the resulting quadratic in x. The distance between the roots is the intercept length.

Given equation of pair of straight lines is 

The condition for the general second-degree equation to represent a pair of lines is abc + 2fgh - af^2 - bg^2 - ch^2 = 0. Rearranging this gives 2fgh = af^2 + bg^2 + ch^2 - abc. The provided option A is a standard simplification for specific cases.

Given equation of pair of straight lines be $x^2+\lambda x{y}+2{y^2}+3{x}-5{y}+2=0$

$ax^{2} + 2hxy + by^{2} = 0$

$ x=x'\cos  \theta -y'\sin  \theta =x'\left( { \dfrac { { \sqrt { 3 }  } }{ 2 }  } \right) -\frac { { y' } }{ 2 }  \ y=x'\sin  \theta +y'\cos  \theta =x'\left( { \dfrac { 1 }{ 2 }  } \right) +y'\left( { \dfrac { { \sqrt { 3 }  } }{ 2 }  } \right)  \ { x^{ 2 } }+2\sqrt { 3 } xy-{ y^{ 2 } }=2{ a^{ 2 } } \ \dfrac { { { { \left[ { \sqrt { 3 } x'-2y' } \right]  }^{ 2 } } } }{ 4 } -\dfrac { { { { \left[ { x'-\sqrt { 3 } y' } \right]  }^{ 2 } } } }{ 4 } +2\sqrt { 3 } \dfrac { { \left[ { \sqrt { 3 } x'-y' } \right]  } }{ 2 } \dfrac { { { { \left[ { x'-\sqrt { 3 } y' } \right]  }^{ 2 } } } }{ 2 } =2{ a^{ 2 } } \ \dfrac { { 2x{ '^{ 2 } }-2y{ '^{ 2 } } } }{ 4 } -\sqrt { 3 } x'y'+\dfrac { { \sqrt { 3 }  } }{ 2 } \left[ { \sqrt { 3 } x'-y' } \right] \left[ { x'+\sqrt { 3 } y } \right] =2{ a^{ 2 } } \ -\sqrt { 3 } x'y'+\dfrac { { \sqrt { 3 }  } }{ 2 } \left[ { \sqrt { 3 } x{ '^{ 2 } }-\sqrt { 3 } y{ '^{ 2 } }+2x'y' } \right] =2{ a^{ 2 } } \ 2x{ '^{ 2 } }-2y{ '^{ 2 } }=2{ a^{ 2 } } \ x{ '^{ 2 } }-2y{ '^{ 2 } }=2{ a^{ 2 } } $

$x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})$
$(x^{2}-xy+y^{2})$ does not factorize so this equation is not combined equation of line.

For the general second-degree equation to represent lines intersecting on the y-axis, the intersection point (0, y0) must satisfy the equation. Setting x=0 gives by^2 + 2fy + c = 0. The condition for the lines to intersect on the y-axis is derived from the general condition for intersection and the specific coordinate constraints.

Given equation is 

The equation (xy + 12 - 4x - 4y)^2 = (2x - 2y)^2 factors into (xy + 12 - 4x - 4y - 2x + 2y)(xy + 12 - 4x - 4y + 2x - 2y) = 0. This simplifies to (x-2)(y-6) = 0 and (x-6)(y-2) = 0, representing a rectangle with vertices (2,2), (6,2), (6,6), and (2,6). The center of this rectangle is (4,4), and any line passing through the center bisects the area.

For a general equation ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 to represent perpendicular lines, the sum of the coefficients of x^2 and y^2 must be zero. Thus, a + b = 0. Given a = 2, we have 2 + b = 0, so b = -2.

As $\displaystyle s=ax^{2}+2hxy +by^{2}+2gx +2fy+c=0 $ represent a pair of line $\displaystyle \therefore \begin{vmatrix}a &h  &g \h  &b  &f \g  &f  &c \end{vmatrix}=0$
or $\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0....(1)$ Now say point ofintersection onY axis be $\displaystyle (0,y _{1} $ and  point of intersection of pair of line be obtained by solving the equations $\displaystyle \frac{\partial s}{\partial x}=0=\frac{\partial s}{\partial y}$ $\displaystyle \therefore \frac{\partial s}{\partial x}=0\Rightarrow ax+by+g=0$ $\displaystyle \begin{matrix}\Rightarrow  \ \Rightarrow  \end{matrix} \left{\begin{matrix}hy _{1}+g=0 \by _{1}+f=0 \end{matrix}\right.>  ()$ and $\displaystyle \frac{\partial s}{\partial y}=0\Rightarrow bx+by+f=0$  On compairing the equation given in () we get $\displaystyle bg=fh $ and  $\displaystyle bg^{2}=fgh ....(2) $ Again $\displaystyle ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ meet at y-axis $\displaystyle \therefore x=0$ $\displaystyle \Rightarrow by^{2}+2fy+c=0$ whose roots must be equal $\displaystyle \Rightarrow by^{2}+2fy+c=0$ whose roots must be equal $\displaystyle \therefore f^{2}=bc af^{2}=abc ......(3)$ Now using (2) and (3) in equation (I) we have $\displaystyle abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$ $\displaystyle \Rightarrow (abc-af^{2})+(fgh-bg^{2})+fgh-ch^{2}=0$ $\displaystyle \Rightarrow 0+0+fgh-ch^{2}=0 \therefore ch^{2}=fgh .....(4) $ Now adding (2) and (4) $\displaystyle 2fgh=ch^{2}+bg^{2}$

The line is at distance 4 from the origin. Using the normal form x cos(theta) + y sin(theta) = 4, and the condition that the normal makes 60 degrees with x+y=0 (which has a normal vector (1,1) at 45 degrees), the angle of the normal is 45 +/- 60 degrees. Calculating the intercepts and checking the positive intercept condition leads to the correct equation.

The equation $\displaystyle ax^{2}+2hxy+by^{2}=0$

Find points A and B by intersecting the line with L1 and L2. Use the section formula to find point P on the line 2x+7y-1=0 such that AP:PB=2:1. The required line is parallel to L1 and L2 (3x+4y+c=0) and passes through P.

$(2x+3y+p)(3x+4y+q)=0$

$6x^{2}+8xy+2qx+9xy+12y^{2}+3qy+3px+4py+pq=0$

$6x^{2}+17xy+12y^{2}+x(2q+3p)+y(3q+4p)+pq=0$
$\rightarrow 6x^{2}+17xy+12y^{2}+22x+31y+20=0$

Hence comparison gives us
$pq=20$
$3p+2q=22$
$4p+3q=31$
$(4p+3q)-(3p+2q)=31-22$
$p+q=9$.
Therefore
$p^{2}+q^{2}=(p+q)^{2}-2pq$
$=81-2(20)$
$=81-40$
$=41$

Since the given equation represents a pair of parallel lines, we have
$\displaystyle h^{2}=9 \times 4\Rightarrow h= \pm 6$
and $\displaystyle \begin{vmatrix}
9 & h & 3\ 
 h& 4 & f\ 
 3& f & -3
\end{vmatrix}=0$
$\displaystyle \Rightarrow 9\left ( -12-f^{2} \right )-h\left ( -3h-3f \right )+3\left ( hf-12 \right )=0$
$\displaystyle \Rightarrow 3h^{2}+6hf-9f^{2}-144=0$
$\displaystyle \Rightarrow 108 \pm 36f-9f^{2}-144=0 \ \ \ \left ( \because h= \pm 6 \right )$
$\displaystyle \Rightarrow 9f^{2} \mp 36f+36=0 \ \ \ \   (if \   h= \pm 6)$
$\displaystyle \Rightarrow f=2 \ \ \ \ if \ \ \ \ \ ( h=6)$
and $\displaystyle \Rightarrow f=-2 \ \ \ \ \ if \ \ \ \ \ \ (h=-6)$

$x^{ 2 }+xy-2y^{ 2 }-4x+7y-5=0\ \Rightarrow \left( x-y+1 \right) \left( x+2y-5 \right) =0$
Intersection point is $\left( 1,2 \right) $
Equation of line passing through $\left( 1,2 \right) $ and perpendicular to $x-y+1=0$ is
$x+y-3=0$
And equation of line passing through $\left( 1,2 \right) $ and perpendicular to $x+2y-5=0$ is
$2x-y=0$
Hence their joint equation is
$\left( x+y-3 \right) \left( 2x-y \right) =0$

As ${ h }^{ 2 }=4=1\times 4=ab$ and $\displaystyle b{ g }^{ 2 }=4\times \left( \frac { 1 }{ 4 }  \right) =1\times \left( 1 \right) =a{ f }^{ 2 }$
Therefore the lines in ${ x }^{ 2 }-4xy+4{ y }^{ 2 }+x-2y-6=0$ are parallel
and the distance between the two lines
$\displaystyle =\frac { 2\sqrt { { g }^{ 2 }-ac }  }{ \sqrt { a\left( a+b \right)  }  } =\frac { 2\sqrt { \dfrac { 1 }{ 4 } -1\times \left( -6 \right)  }  }{ \sqrt { 1\left( 1+4 \right)  }  } =\frac { \sqrt { 25 }  }{ \sqrt { 5 }  } =\sqrt { 5 } $

$lx+my+x=0$
$\displaystyle y=\frac{-x-lx}{m}$ ---1
$\therefore $$ax^{2}+2hxy+by^{2}=1$
Put $\displaystyle y=\frac{-x-lx}{m}$ in the above equation
$\displaystyle ax^{2}+2hx(\frac{-h-lx}{m})+b(\frac{h+lx}{m})^{2}=1$
$\displaystyle ax^{2}-\frac{2xhx}{m}-\frac{2hlx^{2}}{m}+\frac{bx^{2}}{m^{2}}+\frac{bl^{2}x^{2}}{m^{2}}+\frac{2bxlx}{m}=1$
$\displaystyle (a-\frac{2hl}{m}+\frac{bl^{2}}{m^{2}})x^{2}+(\frac{2bxl}{m}-\frac{2xh}{m})x+\frac{bx^{2}}{m^{2}}-1=0$
$\displaystyle \therefore $$x _{2}+x _{2}=\displaystyle \dfrac{\dfrac{2xh-2bxl}{m}}{a-\dfrac{2hl}{m}+\dfrac{bl^{2}}{m^{2}}}$

$x^{2}-3x-4=0$
$(x-4)(x+1)=0$

Let the line be x cos(alpha) + y sin(alpha) = p. The sum of perpendiculars from (x1, y1) and (x2, y2) is |x1 cos(alpha) + y1 sin(alpha) - p| + |x2 cos(alpha) + y2 sin(alpha) - p| = 2k. This is a standard locus problem where the line touches a circle.

Given

Given pairs of lines

The slope of the line, whose end points is $\left( {2,5} \right)$ and $\left( { - 3,6} \right)$ is,

$m = \frac{{6 - 5}}{{ - 3 - 2}}$

$ =  - \frac{1}{5}$

The two non-vertical lines are perpendicular to each other if they have the slopes as negative reciprocals of each other.

So, the slope of the line that is perpendicular to the line joining the points $\left( {2,5} \right)$ and $\left( { - 3,6} \right)$ is,

$ =  - \frac{1}{m}$

$ = 5$

The equation of line passing through the point $\left( { - 3,5} \right)$ with slope 5 is,

$\left( {y - 5} \right) = 5\left( {x - \left( { - 3} \right)} \right)$

$y - 5 = 5x + 15$

$5x - y + 20 = 0$

Therefore, the required equation of line is $5x - y + 20 = 0$.

The equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents the general equation of pair of lines which are parallel to each other, then the distance between them is given by,

For $x < 0$, the equation is
$(y - 20) (y+2x + 20) =0$
Hence, the area is $20 \times 40 = 800$ sq. units.
For $x\geq0$ the equation simplifies to  $y^2-400+2xy+40x=(y-20)(y+20)+2x(y+20)=(y+20)(y+2x-20)=0$
Thus, obtained lines make a quadrilateral $ABCD,$
Area of quadrilateral $ABCD =2\times$ Area of triangle $BCD $$=2\times\dfrac{1}{2}\times$ base $\times $height$ = 20\times 40=800$ sq.units

The above given equation
$9x^2-24xy+16y^2-12x+16y-12=0$ can be factorized as
$(3x-4y+2)(3x-4y-6)=0$
Since both the lines are parallel the distance between them will be
$d=\left|\dfrac{C _{2}-C _{1}}{\sqrt{3^2+4^2}}\right|$
$=\left|\dfrac{2-(-6)}{5}\right|$
$=\dfrac{8}{5}$

$x^2y^2-9y^2-6x^2y+54y=0$
$y^2(x^2-9)-6y(x^2-9)=0$
$(y^2-6y)(x^2-9)=0$
$y(y-6)(x+3)(x-3)=0$
Therefore the lines are
$y=0$
$y=6$
$x=3$
$x=-3$
The above set of lines represent a square of side $6$ units.

Let $\displaystyle y={ m } _{ 1 }x$ and $\displaystyle y={ m } _{ 2 }x$ be the two lines given by $\displaystyle{ x }^{ 2 }+2hxy+b{ y }^{ 2 }=0$ so that

The lines will be
$y-1=\tan A(x-1)$
and $y-1=\cot A(x-1)$
Therefore their joint equation will be
$(y-1-\cot A(x-1))(y-1-\tan A(x-1))=0$
$(y-1)^{2}-(\cot A+ \tan A)(x-1)(y-1)+(x-1)^{2}=0$
$y^2-2y+1-(\cot A+\tan A)(xy-x-y+1)+(x^2-2x+1)=0$
$x^2+y^2-(\cot A+\tan A)(xy)+((\cot A+\tan A)-2)(x+y-1)=0$
Comparing coefficients we get
$\cot A+\tan A=a+2$
$\dfrac {1}{\sin A \cos A}=a+2$

$x^{2}-3y^{2}-2x+1=0$

The angle bisectors of ax^2 + 2hxy + by^2 = 0 are given by (x^2-y^2)/(a-b) = xy/h. For x^2-2mxy-y^2=0 and x^2-2nxy-y^2=0, the bisectors of the first are (x^2-y^2)/(1-(-1)) = xy/(-m), which simplifies to x^2-y^2 = -2xy/m, or x^2 + (2/m)xy - y^2 = 0. Comparing this to the second equation, -2n = 2/m, so mn = -1, or mn+1=0.

Equations of the given lines are $y -1 = \tan \theta (x -1) $ and $y -1 =\ cot \theta (x -1)$ 

$ax^3+3bx^2y+3cxy^2+dy^3=0$  ---(1)   represent three coincident line

Let $ y=mx$ is that line

$(y-mx)^3=0$

$y^3+3m^2xy-3y^2(mx)-m^3n^3=0$

$m^3x^3+3mny^2-3m^2xy-y^3=0$   ---(2)

Compare 1 & 2,

$\dfrac{a}{m^3}=\dfrac{b}{m}=\dfrac{c}{-m^2}=\dfrac{d}{-1}$

$\therefore \dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}$

Equations of the given lines are 
$y-1=\tan { \theta  } \left( x-1 \right) $ and $y-1=\cot { \theta  } \left( x-1 \right) $
Their combined equation is 
$\left( y-1-\tan { \theta  } \left( x-1 \right)  \right) \left( y-1-\cot { \theta  } \left( x-1 \right)  \right) =0\ \Rightarrow { x }^{ 2 }-\left( \tan { \theta  } +\cot { \theta  }  \right) xy+{ y }^{ 2 }+\left( \tan { \theta  } +\cot { \theta  } -2 \right) \left( x+y-1 \right) =0$
Comparing this with given equation we get
$\tan { \theta  } +\cot { \theta  } =a+2\ \Rightarrow \cfrac { 1 }{ \sin { \theta  } \cos { \theta  }  } =a+2\ \Rightarrow \sin { 2\theta  } =\cfrac { 2 }{ a+2 } $

Reason is true 
Assertion
$x^{ 2 }+2\sqrt { 2 } xy+2y^{ 2 }+4\sqrt { 2 } x+4y+1=0$ represents a pair of lines when $\Delta =0$
But $\Delta =\begin{vmatrix} 1 & \sqrt { 2 }  & 2\sqrt { 2 }  \ \sqrt { 2 }  & 4 & 1 \ 2\sqrt { 2 }  & 1 & 1 \end{vmatrix}=-2$

If $3x^{2}+2pxy+2y^{2}+2ax-4y+1=0.$, then.
$\Delta =abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$
$=3(2)(1)+2(-2)(a)(p)-3(-2)^{2}-2(a)^{2}-1(p)^{2}=0$
$6-4ap-12-2a^{2}-p^{2}=0$
$p^{2}+4ap+2a^{2}+6=0$
$\Rightarrow p $ is a solution of $x^{2}+4ax+2a^{2}+6=0$

The equations of $PQ$ and $PR$ are given by
$y - 1 = \dfrac {-2\mp \tan 45^{\circ}}{1\pm (-2)\tan 45^{\circ}} (x - 2)$

Slopes of the line $PQ$ and $PR$ are
$\tan \left(\theta +\dfrac\pi4\right)=\dfrac{1+\tan \theta }{1-\tan \theta }=\dfrac{1-2}{1+2}=-\dfrac{1}{3}$ and $3$
$\therefore $ Equations of $PQ$ and $PR$ are $3y + x - 5 = 0$ and $y-  3x + 5 = 0$
$\therefore $ Combined equation of $PQ$ and $PR$ is
$3x^{2}+8xy-3y^{2}-20x-10y+25=0$

We have, 
$y^3+8x^3 = 1 -6xy $