Straight Lines and Pair of Straight Lines

Slope of $PQ = {{3 - 4} \over {k - 1}} = {{ - 1} \over {k - 1}}$

$\therefore$ Slope of perpendicular bisector of

$PQ = \left( {k - 1} \right)$

Also mid point of

$PQ\left( {{{k + 1} \over 2},{7 \over 2}} \right).$

Equation of perpendicular bisector is

$y - {7 \over 2} = \left( {k - 1} \right)\left( {x - {{k + 1} \over 2}} \right)$

$ \Rightarrow 2y - 7 = 2\left( {k - 1} \right)x - \left( {{k^2} - 1} \right)$

$ \Rightarrow 2\left( {k - 1} \right)x - 2y + \left( {8 - {k^2}} \right) = 0$

$\therefore$ $y$-intercept $ = {{8 - {k^2}} \over { - 2}} = - 4$

$ \Rightarrow $ $8 - {k^2} = - 8$ or ${k^2} = 16 \Rightarrow k = \pm 4$

Given : The vertices of a right angled triangle $A\left( {1,k} \right),$

$B\left( {1,1} \right)$ and $C\left( {2,1} \right)$ and area of $\Delta ABC = 1$ square unit



We know that, area of night angled triangle

$ = {1 \over 2} \times BC \times AB = 1 = {1 \over 2}\left( 1 \right)\left| {\left( {k - 1} \right)} \right|$

$ \Rightarrow \pm \left( {k - 1} \right) = 2 \Rightarrow k = - 1,3$

Equation of bisectors of lines, $xy=0$ are $y = \pm x$



$\therefore$ Put $y = \pm \,x$ in the given equation

$m{y^2} + \left( {1 - {m^2}} \right)xy - m{x^2} = 0$

$\therefore$ $m{x^2} + \left( {1 - {m^2}} \right){x^2} - m{x^2} = 0$

$ \Rightarrow 1 - {m^2} = 0 \Rightarrow m = \pm 1$

Given : The coordinates of points $P,Q,R$ are $(-1,0),$

$\left( {0,0} \right),\,\left( {3,3\sqrt 3 } \right)$ respectively



Slope of $QR$ $ = {{{y_2} - {y_1}} \over {{x_2} - {x_1}}} = {{3\sqrt 3 } \over 3}$

$ \Rightarrow \tan \theta = \sqrt 3 \Rightarrow \theta = {\pi \over 3}$

$ \Rightarrow \angle RQX = {\pi \over 3}$

$\therefore$ $\angle RQP = \pi - {\pi \over 3} = {{2\pi } \over 3}$

Let $QM$ bisects the $\angle PQR,$

$\therefore$ Slope of the line $QM=tan$ ${{2\pi } \over 3} = - \sqrt 3 $

$\therefore$ Equation of line $QM$ is $\left( {y - 0} \right) = - \sqrt 3 \left( {x - 0} \right)$

$ \Rightarrow y = - \sqrt 3 \,x \Rightarrow \sqrt 3 x + y = 0$

Clearly for point $P,$



${a^2} - 3a < 0$ and ${a^2} - {a \over 2} > 0 \Rightarrow {1 \over 2} < a < 3$

As is the mid point of $PQ,$ therefore

${{a + 0} \over 2} = 3,{{0 + b} \over 2} = 4 \Rightarrow a = 6,b = 8$

$\therefore$ Equation of line is ${x \over 6} + {y \over 8} = 1$

or $4x + 3y = 24$

Vertex of triangle is $\left( {1,\,1} \right)$ and midpoint of sides through -

this vertex is $\left( { - 1,\,2} \right)$ and $\left( {3,2} \right)$



$ \Rightarrow $ vertex $B$ and $C$ come out to be $\left( { - 3,3} \right)$ and $\left( {5,3} \right)$

$\therefore$ centroid is ${{1 - 3 + 5} \over 3},{{1 + 3 + 5} \over 3} \Rightarrow \left( {1,{7 \over 3}} \right)$

$a,b,c$ are in $H.P. \Rightarrow {1 \over a}.{1 \over b},{1 \over c}$ are in $A.P.$

$ \Rightarrow {2 \over b} = {1 \over a} + {1 \over c}$

$ \Rightarrow {1 \over a} - {2 \over b} + {1 \over c} = 0$

$\therefore$ ${x \over a} + {y \over a} + {1 \over c} = 0$ passes through $\left( {1, - 2} \right)$

The line passing through the intersection of lines

$ax + 2by = 3b = 0$

and $bx - 2ay - 3a = 0$ is

$ax + 2by + 3b + \lambda \left( {bx - 2ay - 3a} \right) = 0$

$ \Rightarrow \left( {a + b\lambda } \right)x + \left( {2b - 2a\lambda } \right)y + 3b - 3\lambda a = 0$

As this line is parallel to $x$-axis.

$\therefore$ $a + b\lambda = 0 \Rightarrow \lambda = - a/b$

$ \Rightarrow ax + 2by + 3b - {a \over b}\left( {bx - 2ay - 3a} \right) = 0$

$ \Rightarrow ax + 2by + 3b - ax + {{2{a^2}} \over b}y + {{3{a^2}} \over b} = 0$

$y\left( {2b + {{2{a^2}} \over b}} \right) + 3b + {{3{a^2}} \over b} = 0$

$y\left( {{{2{b^2} + 2{a^2}} \over b}} \right) = - \left( {{{3{b^2} + 3{a^2}} \over b}} \right)$

$y = {{ - 3\left( {{a^2} + {b^2}} \right)} \over {2\left( {{b^2} + {a^2}} \right)}} = {{ - 3} \over 2}$

So it is $3/2$ units below $x$-axis.

Let the lines be $y = {m_1}x$ and $y = {m_2}x$ then

${m_1} + {m_2} = - {{2c} \over 7}$ and ${m_1}{m_2} = - {1 \over 7}$

Given ${m_1} + {m_2} = 4m{}_1{m_2}$

$ \Rightarrow {{2c} \over 7} = - {4 \over 7} \Rightarrow c = 2$

$3x+4y=0$ is one of the lines of the pair

$6{x^2} - xy + 4c{y^2} = 0,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$

Put $y = - {3 \over 4}x,$

we get $6{x^2} + {3 \over 4}{x^2} + 4c{\left( { - {3 \over 4}x} \right)^2} = 0$

$ \Rightarrow 6 + {3 \over 4} + {{9c} \over 4} = 0 \Rightarrow c = - 3$

Let the required line be ${x \over a} + {y \over b} = 1.......\left( 1 \right)$

then $a+b=-1$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.........\left( 2 \right)$

$(1)$ passes through $(4,3), $ $ \Rightarrow {4 \over a} + {3 \over b} = 1$

$ \Rightarrow 4b + 3a = ab\,\,...............\left( 3 \right)$

Eliminating $b$ from $(2)$ and $(3),$ we get

${a^2} - 4 = 0 \Rightarrow a = \pm 2 \Rightarrow b = - 3$, $1$

$\therefore$ Equation of straight lines are

${x \over 2} + {y \over { - 3}} = 1$

or ${x \over { - 2}} + {y \over 1} = 1$

Let the vertex $C$ be $(h,k),$ then the

centroid of $\Delta ABC$ is $\left( {{{2 + (- 2) + h} \over 3},{{ - 3 + 1 + k} \over 3}} \right)$

or $\left( {{h \over 3},{{ - 2 + k} \over 3}} \right).$ It lies on $2x+3y=1$

$ \Rightarrow {{2h} \over 3} - 2 + k = 1$

$ \Rightarrow 2h + 3k = 9$

$ \therefore $ Locus of $C$ is $2x+3y=9$

$x = {{a\cos t + b\sin t + 1} \over 3}$

$ \Rightarrow a\cos t + b\sin t = 3x - 1$

$y = {{a\sin t - b\cos t} \over 3}$

$ \Rightarrow a\sin t - b\cos t = 3y$

Since, the points $\left(a_1, b_1\right)$ and $\left(a_2, b_2\right)$ satisfy the equation. So, that

$ \begin{aligned} & a_1\left(a_1-a_2\right)+b_1\left(b_1-b_2\right)+c=0 ~~........(1) \\\\ & \text { and } a_2\left(a_1-a_2\right)+b_2\left(b_1-b_2\right)+c=0 ~~.........(2) \\\\ & \text { On adding Eqs. (i) and (ii), we get } \\\\ & \left(a_1+a_2\right)\left(a_1-a_2\right)+\left(b_1+b_2\right)\left(b_1-b_2\right)+2 c=0 \\\\ & \Rightarrow 2 c=-\left(a_1^2-a_2^2+b_1^2-b_2^2\right) \\\\ & \Rightarrow c=\frac{1}{2}\left(a_2^2+b_2^2-a_1^2-b_1^2\right) \end{aligned} $

Equation of bisectors of second pair of straight lines is,

$q{x^2} + 2xy - q{y^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

It must be identical to the first pair

${x^2} - 2\,pxy - {y^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(... 2 \right)$

from $(1)$ and $(2)$ ${q \over 1} = {2 \over { - 2p}} = {{ - q} \over { - 1}}$

$ \Rightarrow pq = - 1.$

Taking co-ordinates as

$\left( {{x \over r},{y \over r}} \right);\left( {x,y} \right)\,\,\& \,\,\left( {xr,yr} \right)$

Then slope of line joining

$\left( {{x \over r},{y \over r}} \right),\left( {x,y} \right) = {{y\left( {1 - {1 \over r}} \right)} \over {x\left( {1 - {1 \over r}} \right)}} = {y \over x}$

and slope of line joining $(x,y)$ and $(xr, yr)$

$ = {{y\left( {r - 1} \right)} \over {x\left( {r - 1} \right)}} = {y \over x}$

$\therefore$ ${m_1} = {m_2}$

$ \Rightarrow $ Points lie on the straight line.

Co-ordinate of $A = \left( {a\,\cos \,\alpha ,\,\,a\,\sin \,\alpha } \right)$

Equation of $OB,$

$y = \tan \left( {{\pi \over 4} + \alpha } \right)x$

$CA{ \bot ^r}$ to $OB$

$\therefore$ slope of $CA=-$ $\cot \left( {{\pi \over 4} + \alpha } \right)$

Equation of $CA$

$y - a\sin \alpha = - cot\left( {{\pi \over 4} + \alpha } \right)\left( {x - a\,\cos \,\alpha } \right)$

$ \Rightarrow \left( {y - a\sin \alpha } \right)\left( {\tan \left( {{\pi \over 4} + \alpha } \right)} \right) = \left( {a\,\cos \,\alpha - x} \right)$

$ \Rightarrow \left( {y - a\sin \alpha } \right)\left( {{{\tan {\pi \over 4} + \tan \alpha } \over {1 - \tan {\pi \over 4}\tan \alpha }}} \right)\left( {a\,\cos \,\alpha - x} \right)$

$ \Rightarrow \left( {y - a\sin \alpha } \right)\left( {1 + \tan \alpha } \right) = \left( {a\cos \alpha - x} \right)\left( {1 - \tan \alpha } \right)$

$ \Rightarrow \left( {y - a\sin \alpha } \right)\left( {\cos \alpha + \sin \alpha } \right) = \left( {a\cos \alpha - x} \right)\left( {\cos \alpha - \sin \alpha } \right)$

$ \Rightarrow y\left( {\cos + \sin \alpha } \right) - a\sin \alpha \cos \alpha - a{\sin ^2}\alpha $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = a{\cos ^2}\alpha - a\cos \alpha \sin \alpha - x\left( {\cos \alpha - \sin \alpha } \right)$

$ \Rightarrow y\left( {\cos \alpha + sin\alpha } \right) + x\left( {\cos \alpha - \sin \alpha } \right) = a$

$y\left( {\sin \alpha + \cos \alpha } \right) + x\left( {\cos \alpha - \sin \alpha } \right) = a.$

Put $x=0$ in the given equation

$ \Rightarrow b{y^2} + 2fy + c = 0.$

For unique point of intersection ${f^2} - bc = 0$

$ \Rightarrow a{f^2} - abc = 0.$

Since $abc + 2fgh - a{f^2} - b{g^2} - c{h^2} = 0$

$ \Rightarrow 2fgh - b{g^2} - c{h^2} = 0$

$3a + {a^2} - 2 = 0 \Rightarrow {a^2} + 3a - 2 = 0;$

$ \Rightarrow a = {{ - 3 \pm \sqrt {9 + 8} } \over 2} = {{ - 3 \pm \sqrt {17} } \over 2}$

Equation of $AB$ is

$x\cos \alpha + y\sin \alpha = p;$

$ \Rightarrow {{x\cos \alpha } \over p} + {{y\sin \alpha } \over p} = 1;$

$ \Rightarrow {x \over {p/\cos \alpha }} + {y \over {p/\sin \alpha }} = 1$

So co-ordinates of $A$ and $B$ are

$\left( {{p \over {\cos \alpha }},0} \right)$ and $\left( {0,{p \over {\sin \alpha }}} \right);$

So coordinates of midpoint of $AB$ are

$\left( {{p \over {2\cos \,\alpha }},{p \over {2\sin \alpha }}} \right) = \left( {{x_1},{y_1}} \right)\left( {let} \right);$

${x_1} = {p \over {2\,\cos \,\alpha }}\,\,\& \,\,{y_1} = {p \over {2\sin \alpha }};$

$ \Rightarrow \cos \alpha = p/2{x_1}$ and $\sin \alpha = p/2{y_1};$

${\cos ^2}\alpha + {\sin ^2}\alpha = 1 \Rightarrow {{{p^2}} \over 4}\left( {{1 \over {{x_1}^2}} + {1 \over {{y_1}^2}}} \right) = 1$

Locus of $\left( {{x_1},{y_1}} \right)$ is ${1 \over {{x^2}}} + {1 \over {{y^2}}} = {4 \over {{p^2}}}.$

$AB = \sqrt {{{\left( {4 + 1} \right)}^2} + {{\left( {0 + 1} \right)}^2}} = \sqrt {26} ;$

$BC = \sqrt {{{\left( {3 + 1} \right)}^2} + {{\left( {5 + 1} \right)}^2}} = \sqrt {52} $

$CA = \sqrt {{{\left( {4 - 3} \right)}^2} + {{\left( {0 - 5} \right)}^2}} = \sqrt {26} ;$

In isosceles triangle side $AB=CA$

For right angled triangle, $B{C^2} = A{B^2} + A{C^2}$

So, here $BC = \sqrt {52} $ or $B{C^2} = 52$

or $\,\,\,\,\,\,\,\,\,{\left( {\sqrt {26} } \right)^2} + {\left( {\sqrt {26} } \right)^2} = 52$

So, the given triangle is right angled and also isosceles.