Straight Lines and Pair of Straight Lines

$\left.\begin{array}{c}4 x-3 y=12 \alpha \\ 4 x+3 y=\frac{12}{\alpha}\end{array}\right\} \rightarrow 16 x^2-9 y^2=144$

$\begin{aligned} & \frac{x^2}{9}-\frac{y^2}{16}=1 \rightarrow \text { Curve :S } \\ & T: y=m x \pm \sqrt{9 m^2-16} \\ & m=\frac{4 \sqrt{2}}{3} \\ & y=\frac{4 \sqrt{2} x}{3} \pm \sqrt{32-16} \\ & 3 y=4 \sqrt{2} x \pm 12 \\ & \text { as } q>0 \\ & 3 y=4 \sqrt{2} x+12\end{aligned}$

$ \mathrm{p}=-\frac{3}{\sqrt{2}} \quad \& \quad \mathrm{q}=4 $

$ \mathrm{pq}=-6 \sqrt{2} $

Let P is the point of intersection of line $a x+b y +c=0$ and $b x+a y-c=0$

$\mathrm{P}=\left(\frac{-c}{a+b}, \frac{-c}{a+b}\right)$

Given, the distance between $(1,1)$ and $P$ is less than $2 \sqrt{2}$

$\begin{aligned} & \therefore \sqrt{\left(1+\frac{c}{a+b}\right)^2+\left(1+\frac{c}{a+b}\right)^2}<2 \sqrt{2} \\ & \Rightarrow \sqrt{2}\left|\frac{a+b+c}{a+b}\right|<2 \sqrt{2} \\ & \Rightarrow\left|\frac{a+b+c}{a+b}\right|<2 \\ & \therefore a>b>c>0 \\ & \therefore \frac{a+b+c}{a+b}<2 \\ & \Rightarrow a+b+c<2 a+2 b \\ & \Rightarrow a+b-c>0 \end{aligned}$

Hence, option (A) correct

$\therefore$ Given $a > b > c > 0$

$\therefore a-b$ is positive and c is also positive

$\Rightarrow a-b+c>0$

Hence, option (C) is also true.

Hints :

Given, $a > b > c > 0$

So, $a-b, a-c, b$ and $c$ all are positive

$\therefore \quad a-b+c>0, a-c+b>0$

We have $\left| {{{m + \sqrt 3 } \over {1 - \sqrt 3 m}}} \right| = \sqrt 3 $.

$ \Rightarrow m + \sqrt 3 = \pm (\sqrt 3 - 3m)$

$ \Rightarrow 4m = 0 \Rightarrow m = 0$

or $2m = 2\sqrt 3 \Rightarrow m = \sqrt 3 $

Therefore, the equation is

$y + 2 = \sqrt 3 (x - 3)$

$ \Rightarrow \sqrt 3 x - y - (2 + 3\sqrt 3 ) = 0$

$\Delta = \left| {\matrix{ {\cos (\beta - \alpha )} & {\sin \beta } & 1 \cr { - \sin (\beta - \alpha )} & { - \cos \beta } & 1 \cr {\cos (\beta - \alpha - \theta )} & {\sin (\beta - \theta )} & 1 \cr } } \right|$

${R_3} \to R(\cos \theta {R_1} + \sin \theta {R_2})$

$ = \left| {\matrix{ {\cos (\beta - \alpha )} & {\sin \beta } & 1 \cr { - \sin (\beta - \alpha )} & { - \cos \beta } & 1 \cr 0 & 0 & {1 - (\cos \theta + \sin \theta )} \cr } } \right|$

$ = - [1 - (\cos \theta + \sin \theta )]\cos (2\beta - \alpha ) \ne 0$

$ = 2\beta < {\pi \over 2}$ and $\alpha > 0$

$ \Rightarrow (2\beta - \alpha ) < {\pi \over 2} \Rightarrow \cos (2\beta - \alpha ) \ne 0$

Also, since $0 < \theta < {\pi \over 2} \Rightarrow \cos \theta + \sin \theta > 1$

$\Delta \ne 0$

$\Rightarrow$ P, Q and R are not collinear.

We have,

(A)

${L_1}:x + 3y - 5 = 0$

${L_2}:3x - ky - 1 = 0$

${L_3}:5x + 2y - 12 = 0$

Point of intersection of L$_1$ and L$_2$ is (2, 1)

If lines are concurrent, then (2, 1) will satisfy L$_2$

$\Rightarrow 3(2)-k(1)-1=0$

$k=5$

(A) $\to$ (iv)

(B) If L$_1$ and L$_2$ are parallel

${3 \over 1} = {{ - k} \over 3} \ne {{ - 1} \over { - 5}} \Rightarrow k - 9$

If L$_2$ and L$_3$ are parallel

${3 \over 5} = {{ - k} \over 2}$

$k = {{ - 6} \over 5}$

(B) $\to$ (i, ii)

(C) As lines form a triangle, they cannot be concurrent and no two of them are parallel.

$\Rightarrow k\ne5,-9,\frac{-6}{5}$

$\Rightarrow k=\frac{5}{6}$

(C) $\to$ (iii)

(D) If the lines do not form a triangle then

$k=5,-9,\frac{-6}{5}$

(D) $\to$ (i, ii, iv)

Let a and b be non-zero real numbers.

Therefore, the given equation $(a{x^2} + b{y^2} + c)({x^2} - 5xy + 6{y^2}) = 0$ implies either

${x^2} - 5xy + 6{y^2} = 0$

$(x - 2y)(x - 3y) = 0$

$x = 2y$ and $x = 3y$ represent two straight line passing through origin or $a{x^2} + b{y^2} + c = 0$ when c = 0 and a and b are of same signs then

$a{x^2} + b{y^2} + c = 0$

$y=0$

Which is a point specified as the origin. When a = b and c is of sign opposite to that of a $a{x^2} + b{y^2} + c = 0$ represent a circle.

Hence, the given equation,

$(a{x^2} + b{y^2} + c)({x^2} - 5xy + 6{y^2}) = 0$

May represent two straight lines and a circle.

As S is not the centre of the circumcircle, there for PS $\ne$ ST and QS $\ne$ SR

Also, (PS)(ST) = (QS)(SR)

[$\therefore$ from the properties of two intersecting chords in a circle]

$\Rightarrow$ A.M $\ge$ G.M

We get ${{{1 \over {PS}} + {1 \over {ST}}} \over 2} \ge \sqrt {{1 \over {PS}} \times {1 \over {ST}}} $

or

${1 \over {PS}} + {1 \over {ST}} \ge {2 \over {\sqrt {QS \times SR} }}$

${{QS + SR} \over 2} \ge \sqrt {QS \times SR} $ ..... (i)

${1 \over {\sqrt {QS \times SR} }} \ge {2 \over {QR}}$ .... (ii)

$ \Rightarrow {1 \over {PS}} + {1 \over {ST}} \ge {4 \over {QR}}$ From (i) and (ii)

$\because$ ar. $(\triangle \mathrm{OPR}=$ ar. $(\triangle \mathrm{PQR})=$ ar. $(\triangle \mathrm{OQR})$

R should be the centroid of $\triangle \mathrm{OPQ}$

$ \begin{aligned}R(x, y) & =\left(\frac{0+3+6}{3}, \frac{0+4+0}{3}\right) \\ & =\left(\frac{9}{3}, \frac{4}{3}\right) \\ & =\left(3, \frac{4}{3}\right) \end{aligned}$

Intersection of $\mathrm{L}_{1}$ and $\mathrm{L}_{3}$ is $\mathrm{P}=(-2,-2)$

Intersection of $\mathrm{L}_{2}$ and $\mathrm{L}_{3}$ is $\mathrm{Q}=(1,-2)$

Now, Intersection of $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ is $\mathrm{O}(0,0)$ equation of angular bisector of $\triangle \mathrm{OPQ}$ will be $(\sqrt{5}+2 \sqrt{2}) x=(\sqrt{5}-\sqrt{2}) y$

In $\triangle \mathrm{OPQ}$, angle of bisector of $\mathrm{O}$ divides $\mathrm{PQ}$ in the ratio of OP : OQ which is $2 \sqrt{2}: \sqrt{5}$ but it does not divide triangle into two similar triangle. Statement 1 is true, statements 2 is false.

The given system can be written as

(A) AX = 0

Where $A = \left( {\matrix{ a & b & c \cr b & c & a \cr c & a & b \cr } } \right),X = \left( {\matrix{ x \cr y \cr z \cr } } \right)$

$|A| = \left| {\matrix{ a & b & c \cr b & c & a \cr c & a & b \cr } } \right| = (a + b + c)\left| {\matrix{ 1 & b & c \cr 1 & c & a \cr 1 & a & b \cr } } \right|$

$ = (a + b + c)\left| {\matrix{ 1 & b & c \cr 0 & {c - b} & {a - c} \cr 0 & {a - b} & {b - c} \cr } } \right|$

R$_2$ $\to$ R$_2$ $-$ R$_1$, R$_3$ $\to$ R$_3$ $-$ R$_1$

$ = (a + b + c)(bc + ac + ab - {a^2} - {b^2} - {c^2})$

$ = {1 \over 2}(a + b + c)[{(b - c)^2} + {(c - a)^2} + {(a - b)^2}]$

If $a + b + c \ne 0$ and ${a^2} + {b^2} + {c^2} = bc + ca + ab$ then $a = b = c \ne 0$

Thus, three equation represent identical planes $x + y + z = 0$

(B) If $a + b + c = 0$ we get two equations as

$ax + by = (a + b)z$

$bx + cy = (b + c)z$

Eliminating y, we have $(ac - {b^2})x = (ac - {b^2})z$

$x = z$

putting $x = z$ in first equation, we get

$ax + by + cx = 0$

$by = - (a + c)x = bx$

$y = x$

hence, $x = y = z$

(C) If $a + b + c \ne 0$

${a^2} + {b^2} + {c^2} \ne ab + bc + ca$ then $|A| \ne 0$

So, only solution is

$x = y = z = 0$

(D) If $a + b + c = 0,{a^2} + {b^2} + {c^2} = ab + bc + ca$

${(b - c)^2} + {(c - a)^2} + {(a - b)^2} = 0$

$a = b = c$

$a = b = c = 0$

Thus, the system represents the whole three dimensional space.

Locus of point is $2 x= \pm(y-1)$.

Here the triangle formed by a line parallel to $X$-axis passing through $\mathrm{P}(h, k)$ and the straight line $y=x$ and $y=2-x$ could be shown below.

Here, area of $\Delta$ABC = $4h^2$

and also $\triangle$ABC is right angled at A.

$\frac{1}{2} \mathrm{AB} \cdot \mathrm{AC}=4 h^{2}$

Where, $\quad \mathrm{AB}=\sqrt{2(k-1)^{2}}$

$=\sqrt{2}[k-1]$

and

$\begin{aligned} \mathrm{AC}= & \sqrt{(2-k-1)^{2}+(k-1)^{2}} \\ = & \sqrt{2(k-1)^{2}}=\sqrt{2}[k-1] \\& \because(1-k)^{2}=(k-1)^{2}\end{aligned}$

Now, $\frac{1}{2} \times \sqrt{2}|k-1| \cdot \sqrt{2}|k-1|=4 h^{2}$

$\Rightarrow \quad(k-1)^{2}=4 h^{2}$

$\Rightarrow \quad 2 h= \pm(k-1)$

The locus of a point is

$2 x= \pm(y-1)$