Straight Lines and Pair of Straight Lines

Given, lines $x+2 y-5=0$ and $3 x-y+1=0$

$ \begin{aligned} &\text { lie in the same region of lines }\\ &\begin{aligned} & \begin{array}{l} x+2 y-5=0 \text { and } 3 x-y+1=0 \\ \left(k^2+2(k+1)-5\right)<0 \end{array} \\ & \text { and } \quad 3 k^2-(2 k+1)+1>0 \\ & \quad \begin{array}{r} \Rightarrow\,\, k^2+2 k-3<0 \\ \text { and } \quad 3 k^2-k>0 \end{array} \\ & \text { and } \quad \begin{aligned} &(k+3)(k-1)<0 \\ & k(3 k-1)>0 \\ & k \in(-3,1) \end{aligned} \\ & \text { and } \quad k \in(-\infty, 0) \cup\left(\frac{1}{3}, \infty\right) \\ & \therefore \quad k \in(-3,0) \cup\left(\frac{1}{3}, 1\right) \end{aligned} \end{aligned} $

Equation of line are $x+y-1=0$ and $x-y+1=0$

Solving equation we get, $(0,1)$

$ \therefore R(0,1) $

Perpendicular distance from $P$ to line $x+y-1=0$ and $x-y+1=0$ are $\frac{1}{\sqrt{2}}$ and $\frac{1}{\sqrt{2}}$ respectively

$ \therefore \quad P S=P Q $

and $\triangle P S R$ and $\triangle P Q R$ are congruent triangle

∴ Area of quadrilateral $P Q R S=P Q \times P S$

$ =\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{1}{2} $

Equation of curves are,

$ x^2+y^2+g x+c=0 $

and $x^2+y^2+2 f y-c=0$

Subtract Eq. (i) from Eq. (ii), we get

$ 2 f y-g x=2 c \Rightarrow \frac{2 f y-g x}{2 c}=1 $

Homogeneousing the Eq. (i), we get

$ \begin{aligned} & x^2+y^2+g x \frac{(2 f y-g x)}{2 c}+\frac{C(2 f y-g x)^2}{4 c^2}=0 \\ & \Rightarrow 4 c\left(x^2+y^2\right)+4 g f x y-2 g^2 x^2+4 f^2 y^2 +g^2 x^2-4 g f x y=0 \\ & \Rightarrow \quad\left(4 c-2 g^2+g^2\right) x^2+\left(4 c+4 f^2\right) y^2=0 \end{aligned} $

∴ A pair of straight formed right angle, then

$ \begin{array}{r} 4 c-g^2+4 f^2+4 c=0 \\ g^2-4 f^2=8 c \end{array} $

We have, pair of straight line is

$ \begin{array}{r} x^2-y^2-x+3 y-2=0 \\ (x-y+1)(x+y-2)=0 \\ x-y+1=0 \text { and } x+y-2=0 \end{array} $

Solving these equation, we get

$ x=\frac{1}{2} \text { and } y=\frac{3}{2} $

∴ Intersection point $P\left(\frac{1}{2}, \frac{3}{2}\right)$

$ \alpha=O P=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{3}{2}\right)^2}=\frac{\sqrt{10}}{2} $

Perpendicular distance from origin to line $x-y+1=0$

and $x+y-2=0$ are $\frac{1}{\sqrt{2}}$ and $\frac{2}{\sqrt{2}}$ respectively,

$ \begin{aligned} &\begin{aligned} \beta & =\frac{1}{\sqrt{2}} \times \frac{2}{\sqrt{2}}=1 \\ \Rightarrow \quad \alpha \beta & =\frac{\sqrt{10}}{2} \times 1=\sqrt{\frac{5}{2}} \end{aligned}\end{aligned} $

Given equation of line $x-y=0$ Distance from $P(\alpha, \beta)$ to line $x-y=0$ is

$ b=\left|\frac{\alpha-\beta}{\sqrt{2}}\right| \Rightarrow(\alpha-\beta)^2=2 b^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Distance between $P(\alpha, \beta)$ and $A(2,1)$ is

$ a^2=(\alpha-2)^2+(\beta-1)^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Distance from origin to $A(2,1)$ is $C^2=5$

Also $\quad a=b c$

$ \therefore \quad a^2=5 b^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) $

From Eqs. (i), (ii) and (iii), we get

$ \begin{aligned} & (\alpha-2)^2+(\beta-1)^2=\frac{5(\alpha-\beta)^2}{2} \\ & \begin{aligned} 2\left(\alpha^2-4 \alpha+4+\beta^2-2 \beta+1\right] & \\ =5\left[\alpha^2+\beta^2-2 \alpha \beta\right] & \\ \Rightarrow 3\left(\alpha^2+\beta^2\right)-10 \alpha \beta+8 \alpha+4 \beta & -10=0 \end{aligned} \end{aligned} $

∴ Locus of $P$ is

$ 3\left(x^2+y^2\right)-10 x y+8 x+4 y-10=0 $

We have,

The point $(4,1)$ undergoes the following transformations

(i) Reflection in the line $x-y=0$

∵ Reflection of point $(4,1)$ to line $x-y=0$ is $(1,4)$

(ii) Shifting through a distance of 2 units along the positive $X$-axis then $(1+2,4) \equiv(3,4)$

(iii) Projection of $(3,4)$ in $X$-axis is $(3,0)$

∴ The new coordinates is $(3,0)$

Given line

$ 3 x+4 y=5 \text { and } $

$4 x-3 y=15$

Slope of line $3 x+4 y=5$ is

i.e.,        $ m_1=\frac{-3}{4} $

Slope of line $4 x-3 y=15$

i.e.,        $ m_2=\frac{4}{3} $

Let the slope of line $B C=m$

Given,$\,\,\,\,A B=A C$

$\because\,\,\,\,\tan \alpha=\tan \beta$

$ \begin{aligned} \frac{m-\frac{4}{3}}{1+\frac{4 m}{3}} & =\frac{-\frac{3}{4}-m}{1-\frac{3 m}{4}} \\ \frac{3 m-4}{3+4 m} & =\frac{-3-4 m}{4-3 m} \\ 12 m^2-9 m^2-16+12 m & \\ =-9-12 m-12 m-16 m^2 & \\ \Rightarrow \quad 7 m^2+48 m-7 & =0 \\ (7 m-1)(m+7) & =0 \\ m=1 / 7, m & =-7 \end{aligned} $

Equation of line passing through $(1,2)$ and slope $\frac{1}{7}$ is,

$ =y-2=\frac{1}{7}(x-1)=x-7 y+13=0 $

Equation of line passing through $(1,2)$ and slope $=-7$ is

$ \begin{aligned} y-2 & =-7(x-1) \\ 7 x+y & =9 \end{aligned} $

We know that the equation of two straight lines which passes through a point ( $x_1, y_1$ ) and made a given angle $\alpha$ with the given straight line

$ \begin{aligned} & y=m x+c \text { are } \\ & y-y_1=\frac{m \pm \tan \alpha}{1 \mp m \tan \alpha}\left(x-x_1\right) \end{aligned} $

Here, $x_1=1, y_1=1, \alpha=60^{\circ}$ and $m=-1$

(slope of line $x+y-3=0$ )

So, equation of lines are

$ y-1=\frac{-1-\tan 60^{\circ}}{1+(-1) \tan 60^{\circ}}(x-1) $

and $y-1=\frac{-1+\tan 60^{\circ}}{1-(-1) \tan 60^{\circ}}(x-1)$

$ \begin{aligned} & \Rightarrow \quad y-1=\frac{-1-\sqrt{3}}{1-\sqrt{3}}(x-1) \\ & \text { and } \quad y-1=\frac{-1+\sqrt{3}}{1+\sqrt{3}}(x-1) \\ & \Rightarrow \quad(y-1)(1-\sqrt{3})=(-1-\sqrt{3})(x-1) \\ & \text { and }(y-1)(1+\sqrt{3})=(-1+\sqrt{3})(x-1) \\ & \Rightarrow \quad(1+\sqrt{3}) x+(1-\sqrt{3}) y-1 +\sqrt{3}-1-\sqrt{3}=0 \end{aligned} $

and $\quad(1-\sqrt{3}) x+(1+\sqrt{3}) y-1-\sqrt{3}$

$ \begin{array}{lr} & -1+\sqrt{3}=0 \\ \Rightarrow & (1+\sqrt{3}) x+(1-\sqrt{3}) y-2=0 \\ \text { and } & (1+\sqrt{3}) x+(1-\sqrt{3}) y-2=0 \end{array} $

Given pairs of lines

$ \begin{aligned} & & 2 x^2-5 x y+2 y^2+6 x-3 y=0 \\ \Rightarrow & & (2 y-x-3)(y-2 x)=0 \end{aligned} $

∴ Pairs of line are

$ 2 y-x-3=0 \text { and } y-2 x=0 $

Solving these equation we get,

$ \begin{array}{ll} & x=1, y=2 \\ \therefore & \alpha=1 \end{array} $

Equation of line passing through origin is

$ y-2 x=0 $

∵ Slope of line $m=2$

$ \therefore \quad \beta=2 $

Equation of angular bisector of line

$ \begin{aligned} & \quad \frac{2 y-x-3}{\sqrt{5}}= \pm \frac{y-2 x}{\sqrt{5}} \\ & 2 y-x-3=y-2 x \text { or } \\ & 2 y-x-3=-y+2 x \\ & \because \quad x+y-3=0 \text { or } 3 x-3 y+3=0 \\ & \quad \quad x+y-3=0 \text { or } x-y+1=0 \\ & \quad \gamma=-3 \\ & \therefore \gamma<\alpha<\beta \text { i.e., }-3<1<2 \end{aligned} $

$ \begin{aligned} & \left(7 x^2-4 x y+8 y^2\right)^2+(4 x-8 y-32) \\ & \left(7 x^2-4 x y+8 y^2\right)=0 \end{aligned} $

$ \begin{aligned} \Rightarrow & \left(7 x^2-4 x y+8 y^2\right)\left(7 x^2-4 x y+8 y^2\right. +4 x-8 y-32) =0\end{aligned} $

Now, $7 x^2-4 x y+8 y^2+4 x-8 y-32=0$

$ \Rightarrow 7 x^2+(-4 y+4) x+8 y^2-8 y-32=0 $

Which is quadratic equation in $x$

$ \begin{aligned} & \therefore x=\frac{(4 y-4) \pm \sqrt{\begin{array}{c} \left(16 y^2+16-32 y\right)-4 \times 7 \\ \times\left(8 y^2-8 y-32\right) \end{array}}}{2 \times 7} \\ & \Rightarrow \\ & x=\frac{(2 y-2) \pm \sqrt{\begin{array}{c} 4 y^2+4-32 y \\ -56 y^2+56 y+224 \end{array}}}{7} \\ & \Rightarrow x=\frac{(2 y-2) \pm \sqrt{-52 y^2+24 y+228}}{7} \\ & \quad=\frac{(2 y-2) \pm \sqrt{-4\left(13 y^2-6 y-57\right)}}{7} \end{aligned} $

Here, $D<0$

So, lines are not possible.

Therefore, question is wrong.

We have, $R(h, k)$ is mid-point of $O M$.

$ \begin{aligned} & \therefore \quad\left(\frac{0+\alpha}{2}, \frac{0+\beta}{2}\right)=(h, k) \\ & \Rightarrow \alpha=2 h, \beta=2 k \end{aligned} $

⇒ Coordinates of $M$ are $(2 h, 2 k)$.

Now, slope of $O M=\frac{2 k-0}{2 h-0}=\frac{k}{h}$ and slope of $M Q=\frac{2 k-b}{2 h-a}$

Since, $O M \perp M Q$

$ \begin{aligned} & \therefore \quad \frac{k}{h} \times \frac{2 k-b}{2 h-a}=-1 \\ & \Rightarrow 2 k^2-b k=-2 h^2+a h \\ & \Rightarrow \quad 2 h^2+2 k^2-a h-b k=0 \end{aligned} $

∴ Locus of $R(h, k)$ is

$ 2 x^2+2 y^2-a x-b y=0 $

Substituting $x=X+\frac{3}{2}, y=Y+\frac{3}{2}$ in the equation

$ 32 x^2+8 x y+32 y^2-108 x-108 y+99=0 $

we get

$ \begin{aligned} & 32\left(X+\frac{3}{2}\right)^2+8\left(X+\frac{3}{2}\right)\left(Y+\frac{3}{2}\right) +32\left(Y+\frac{3}{2}\right)^2-108\left(X+\frac{3}{2}\right) -108\left(Y+\frac{3}{2}\right)+99=0 \\ & \Rightarrow 32\left[X^2+3 X+\frac{9}{4}\right] +8\left(\frac{2 X+3}{2}\right)\left(\frac{2 Y+3}{2}\right)+32\left[Y^2+3 Y+\frac{9}{4}\right] \\ & \Rightarrow 32 X^2+96 X+72+2(4 X Y+6 X +6 Y+9)+32 Y^2 \end{aligned} $

$ \begin{aligned} +96 Y+72-108 X-162-108 Y \\ -162+99=0 \\ \Rightarrow 32 X^2+32 Y^2+8 X Y-63=0 \end{aligned} $

Equation of $L_1$ :

$ \begin{aligned} & y-4=1(x-3) \\ & y-4=x-3 \\ & \quad x-y+1=0 \end{aligned} $

Let slope of $A C$ is $m$.

Since $A C=B C$

$\therefore \triangle A B C$ is isosceles

$ \begin{aligned} \therefore \quad \angle C & =\angle B \\ \left|\frac{m-2}{1+2 m}\right|=\left|\frac{2-1}{1+2 \times 1}\right| & \\ & {\left[\because \text { slope of } B C=\frac{-2}{-1}=2\right] } \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{m-2}{1+2 m}= \pm \frac{1}{3} \\ & \Rightarrow \quad \frac{m-2}{1+2 m}=\frac{1}{3} \text { or } \frac{m-2}{1+2 m}=\frac{-1}{3} \\ & \Rightarrow \quad m=7 \text { or } m=1 \end{aligned} $

But for $m=1$

$A C$ become $L_1$

$ \therefore \quad m=7 $

Equation of $A C$ is

$ \begin{gathered} y-4=7(x-3) \\ y-4=7 x-21 \\ 7 x-y-17=0 \end{gathered} $

Let slope of line is $m$

∴ Equation of line is

$ \begin{aligned} & y-2=m(x-1) \\ & y-2=m x-m \\ & m x-y+(2-m)=0 \end{aligned} $

On solving $m x-y+(2-m)$ and $x+y=4$, we get

$ (x, y)=\left(\frac{m+2}{m+1}, \frac{3 m+2}{m+1}\right) $

Now, it is given that

$ \begin{aligned} & \sqrt{\left(\frac{m+2}{m+1}-1\right)^2+\left(\frac{3 m+2}{m+1}-2\right)^2}=\frac{\sqrt{6}}{3} \\ & \Rightarrow \quad \frac{1}{(m+1)^2}+\frac{m^2}{(m+1)^2}=\frac{6}{9} \\ & \Rightarrow \quad \frac{1+m^2}{(m+1)^2}=\frac{2}{3} \\ & \Rightarrow \quad 3+3 m^2=2 m^2+4 m+2 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad m^2-4 m+1=0 \Rightarrow m=2 \pm \sqrt{3} \\ & \Rightarrow m=2+\sqrt{3}, 2-\sqrt{3} \end{aligned}\\ &\text { ∴ Angles will be } \frac{\pi}{12} \text { and } \frac{5 \pi}{12} \text {. } \end{aligned} $

It is given that lines $x+3 y-9=0$, $4 x+5 y-1=0, p x+q y+10=0$ are concurrent

$ \begin{aligned} & \therefore\left|\begin{array}{lll} 1 & 3 & -9 \\ 4 & 5 & -1 \\ p & q & 10 \end{array}\right|=0 \\ & \Rightarrow 50+q-3(40+p)-9(4 q-5 p)=0 \\ & \Rightarrow 50+q-120-3 p-36 q+45 p=0 \\ & \Rightarrow 42 p-35 q-70=0 \\ & \Rightarrow-6 p+5 q+10=0 \end{aligned} $

So, $5 x+6 y+10=0$ passes through $(q,-p)$.

The equation $(2 l-3) x^2+2 l x y-y^2=0$ will represent a pair of real and distinct lines, if

$ \begin{array}{ll} & l^2-(2 l-3)(-1)>0 \quad\left[\because h^2>a b\right] \\ \Rightarrow \quad & l^2+2 l-3>0 \\ \Rightarrow \quad & (l+3)(l-1)>0 \\ \therefore \quad & l \in \mathbf{R}-(-3,1) \end{array} $

We have,

$ \begin{array}{ll} & 2 y^2-x y-6 x^2=0 \\ \Rightarrow & 2 y^2-4 x y+3 x y-6 x^2=0 \\ \Rightarrow & 2 y(y-2 x)+3 x(y-2 x)=0 \\ \Rightarrow & (y-2 x)(2 y+3 x)=0 \end{array} $

∴ Equations of sides of triangle are

$ x+y=1, y-2 x=0,2 y+3 x=0 $

On solving the above equations, we get the vertices of triangle as $\left(\frac{1}{3}, \frac{2}{3}\right),(0,0)$ and $(-2,3)$.

∴ Centroid of triangle

$ \begin{aligned} & =\left(\frac{\frac{1}{3}+0-2}{3}, \frac{\frac{2}{3}+0+3}{3}\right) \\ & =\left(\frac{-5}{9}, \frac{11}{9}\right) \end{aligned} $

For three lengths to form a triangle, the sum of the lengths of any two sides must be greater than the length of the third side. This is known as the triangle inequality theorem.

Let's apply this to our lengths: $5$, $5r$, and $5r^2$. We have three inequalities to check:

1. $5 + 5r > 5r^2$
2. $5 + 5r^2 > 5r$
3. $5r + 5r^2 > 5$

Solving these inequalities will give us a range of values for $r$. Any value outside of this range will mean that $r$ cannot form a triangle with the given sides.

1. 1. $1 + r > r^2$

We can rewrite this as: $r^2 - r - 1 < 0$

$\Rightarrow\left[r-\left(\frac{1-\sqrt{5}}{2}\right)\right]\left[r-\left(\frac{1+\sqrt{5}}{2}\right)\right]<0$

The roots of this quadratic equation are $\frac{1-\sqrt{5}}{2}$ and $\frac{1+\sqrt{5}}{2}$, so the inequality holds for $r$ in the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$.

1. 2. $1 + r^2 > r$

We can rewrite this as: $r^2 - r + 1 > 0$

This inequality is true for all real numbers since the left-hand side is always positive, which means the inequality holds for all real $R$.

1. 3. $r + r^2 > 1$

We can rewrite this as: $r^2 + r - 1 > 0$

Taking the intersection of the intervals from the three inequalities, we get $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$ as the valid range for $r$.

Given the solution of the inequalities $r \in \left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$, we need to determine which of the options are outside this interval.

1. Option A: $\frac{7}{4}$

The decimal representation of $\frac{7}{4}$ is 1.75, which is outside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$ or approximately (-0.618, 1.618). Hence, r cannot be equal to $\frac{7}{4}$.

1. Option B: $\frac{5}{4}$

The decimal representation of $\frac{5}{4}$ is 1.25, which is inside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$. Hence, r can be equal to $\frac{5}{4}$.

1. Option C: $\frac{3}{4}$

The decimal representation of $\frac{3}{4}$ is 0.75, which is also inside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$. Hence, r can be equal to $\frac{3}{4}$.

1. Option D: $\frac{3}{2}$

The decimal representation of $\frac{3}{2}$ is 1.5, which is also inside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$. Hence, r can be equal to $\frac{3}{2}$.

Therefore, the only value that r cannot be equal to, among the given options, is $\frac{7}{4}$. So, the correct answer is:

Option A: $\frac{7}{4}$

The number of integral points lie inside the triangle are

1. If x = 1, then y may be 1, 2, 3, ....., 39

2. If x = 2, then y may be 1, 2, 3, ....., 38

3. If x = 3, then y may be 1, 2, 3, ....., 37

$ \vdots $

39. If x = 39, then the value of y is 1.

Hence, the number of interior points are

Since the point of intersection lies on fourth quadrant and equidistant from the two axes,

i.e., let the point be (k, $-$k) and this point satisfies the two equations of the given lines.

$\therefore$ 4ak $-$ 2ak + c = 0 ......... (1)

and 5bk $-$ 2bk + d = 0 ..... (2)

From (1) we get, $k = {{ - c} \over {2a}}$

Putting the value of k in (2) we get,

$5b\left( { - {c \over {2a}}} \right) - 2b\left( { - {c \over {2a}}} \right) + d = 0$

or, $ - {{5bc} \over {2a}} + {{2bc} \over {2a}} + d = 0$ or, $ - {{3bc} \over {2a}} + d = 0$

or, $ - 3bc + 2ad = 0$ or, $3bc - 2ad = 0$