Straight Lines and Pair of Straight Lines

We set up a coordinate system so that the two parallel lines are given by

$ y = 0 \quad \text{and} \quad y = 5, $

since their distance is 5 units. Choose point

$ P = (0,1) $

so that the distance from $P$ to the line $y=0$ is 1 unit (and its distance to the line $y=5$ is 4 units).

Let point

$ Q = (a,0) $

be on the line $y = 0$, and let point

$ R = (b,5) $

be on the line $y = 5$. Since triangle $PQR$ is equilateral with side length $s$, we require:

$ PQ = PR = QR = s. $

A convenient method is to “rotate” $Q$ about $P$ by an angle of $60^\circ$ to obtain $R$. In complex-number (or vector) terms, if we translate so that $P$ is at the origin, then the rotation is given by

$ e^{i60^\circ} = \cos 60^\circ + i \sin 60^\circ = \frac{1}{2} + i \frac{\sqrt{3}}{2}. $

Thus, writing $Q$ in vector form relative to $P$, we have

$ Q - P = (a, -1). $

Rotating this by $60^\circ$ gives

$ R - P = \left(a\cos60^\circ - (-1)\sin60^\circ,\; a\sin60^\circ + (-1)\cos60^\circ\right). $

Substituting the values $\cos60^\circ = \frac{1}{2}$ and $\sin60^\circ = \frac{\sqrt{3}}{2}$, we obtain

$ \begin{aligned} R - P &= \left(\frac{a}{2} + \frac{\sqrt{3}}{2},\; \frac{a\sqrt{3}}{2} - \frac{1}{2}\right), \\ \text{so} \quad R &= \left( \frac{a+\sqrt{3}}{2},\; 1 + \frac{a\sqrt{3}}{2} - \frac{1}{2} \right) = \left( \frac{a+\sqrt{3}}{2},\; \frac{a\sqrt{3}+1}{2} \right). \end{aligned} $

Since $R$ lies on $y = 5$, its $y$-coordinate must equal 5:

$ \frac{a\sqrt{3}+1}{2} = 5. $

Solve for $a$:

$ \begin{aligned} a\sqrt{3} + 1 &= 10, \\ a\sqrt{3} &= 9, \\ a &= \frac{9}{\sqrt{3}} = 3\sqrt{3}. \end{aligned} $

Now, the side length $s$ (which is the distance $PQ$) is given by

$ \begin{aligned} s^2 &= PQ^2 = \left(3\sqrt{3} - 0\right)^2 + \left(0 - 1\right)^2 \\ &= (3\sqrt{3})^2 + 1^2 \\ &= 27 + 1 \\ &= 28. \end{aligned} $

Thus, the square of side $QR$ is

$ (QR)^2 = s^2 = 28. $

Equation of incident ray : $a x+b y+1=0$

Using mirror image,

$\frac{m-7}{2}=\frac{n-2}{1}=\frac{-2(14+2-6)}{5}$

$\begin{array}{l|l} \frac{m-7}{2}=-4 & n-2=-4 \\ m=-8+7 & n=-2 \\ m=-1 & \end{array}$

${ }^*$ Note: It can be observed from diagram $A, P, B$' are collinear.

Equation of Incident Ray,

Using two-point form,

$\begin{aligned} & (y-10)=\frac{10+2}{3+1}(x-3) \\ & (y-10)=\frac{12}{4}(x-3) \\ & y-10=+3(x-3) \\ & y-10=+3 x-9 \\ & 3 x-y+1=0 \end{aligned}$

On comparing,

$\begin{aligned} & a=3 \\ & b=-1 \end{aligned}$

Let circumcentre, orthocentre and centroid of a triangle $P Q R$ are $C_1, H_1$ and $G_1$ respectively

$\Rightarrow G_1 \equiv(0,0)$ orthocentre of $\triangle A B C$ is $(0,0)$

$\begin{aligned} & m_{A H_2}=+\frac{b}{a} \Rightarrow a+b=0 \\ & \text { eq }{ }^{\text {n }} \text { of lines } H_2 C \text { is } y=\frac{3}{2} x \\ & \Rightarrow \text { point } C \equiv\left(\frac{1}{4}, \frac{3}{8}\right) \text { lies on } a x+b y-1=0 \\ & \Rightarrow \frac{a}{4}+\frac{3}{8} b-1=0 \Rightarrow \frac{a}{4}-\frac{3}{8} a-1=0 \\ & \Rightarrow a=-8, b=8 \\ & |a-b|=16 \end{aligned}$

$\frac{\mathrm{c}}{\sin 30^{\circ}}=\frac{4 \sqrt{3}}{\sin 120^{\circ}}[$ By sine rule $]$

$ 2 c=8 \Rightarrow c=4 $

$\begin{gathered}\mathrm{AB}=|(\mathrm{b}+1)|=4 \\\\ \mathrm{~b}=3, \mathrm{~m}_{\mathrm{AB}}=0 \\\\ \mathrm{~m}_{\mathrm{BC}}=\frac{-1}{\sqrt{3}} \\\\ \mathrm{BC}:-\mathrm{y}=\frac{-1}{\sqrt{3}}(\mathrm{x}-3) \\\\ \sqrt{3} \mathrm{y}+\mathrm{x}=3\end{gathered}$

Point of intersection : $y=x+3, \sqrt{3} y+x=3$

$\begin{aligned} & ({\sqrt{3}+1}) y=6 \\\\ & y=\frac{6}{\sqrt{3}+1} \\\\ & x=\frac{6}{\sqrt{3}+1}-3 \\\\ & =\frac{6-3 \sqrt{3}-3}{\sqrt{3}+1} \\\\ & =3 \frac{(1-\sqrt{3})}{(1+\sqrt{3})}=\frac{-6}{(1+\sqrt{3})^2}\end{aligned}$

$\frac{\beta^4}{\alpha^2}=36$

To find the maximum number of points of intersection of pairs of lines from the given set, we need to consider how the lines are arranged based on the given conditions.

Firstly, there are 10 lines (${L}_1, {L}_3, ..., {L}_{19}$) that are parallel to each other. Since parallel lines do not intersect with each other, these 10 lines will not contribute to the number of intersection points among themselves.

Secondly, there are 10 lines (${L}_2, {L}_4, ..., {L}_{20}$) that all pass through a given point $P$. Although these lines intersect at $P$, they only contribute one unique point of intersection to the total count.

To calculate the maximum number of intersection points, we need to consider the total number of ways to pick pairs of lines from the 20 lines available without restrictions, then subtract the combinations that do not result in intersections, which includes the combinations of parallel lines among themselves and the concurrent lines through point $P$.

This calculation is represented as:

$Total = ^{20}C_2 - ^{10}C_2 - ^{10}C_2 + 1$

Here, $^{20}C_2$ calculates the total number of ways to pick any two lines out of 20, which includes intersecting and non-intersecting lines. $^{10}C_2$ is subtracted twice: once for the set of parallel lines (${L}_1, {L}_3, ..., {L}_{19}$) that don't intersect among themselves and once more for the set of concurrent lines (${L}_2, {L}_4, ..., {L}_{20}$) intersecting only at point $P$. Since all the concurrent lines intersect at the same point, we add 1 back to include this intersection point.

Carrying out this calculation gives us the total number of distinct intersection points as $101$.

$\begin{aligned} & \mathrm{P} \equiv\left(\frac{\alpha-2}{2}, \frac{\beta-1}{2}\right) \equiv\left(\frac{\gamma+1}{2}, \frac{\delta}{2}\right) \\ & \frac{\alpha-2}{2}=\frac{\gamma+1}{2} \text { and } \frac{\beta-1}{2}=\frac{\delta}{2} \\ & \Rightarrow \alpha-\gamma=3 \ldots .(1), \beta-\delta=1 \ldots \ldots (2) \end{aligned}$

Also, $(\gamma, \delta)$ lies on $3 x-2 y=6$

$3 \gamma-2 \delta=6$ ..... (3)

and $(\alpha, \beta)$ lies on $2 x-y=5$

$\Rightarrow 2 \alpha-\beta=5 \text {. }$

Solving (1), (2), (3), (4)

$\begin{aligned} & \alpha=-3, \beta=-11, \gamma=-6, \delta=-12 \\ & |\alpha+\beta+\gamma+\delta|=32 \end{aligned}$

$\begin{aligned} & 2 x-y+3=0 \\ & 6 x+3 y+1=0 \\ & \alpha x+2 y-2=0 \end{aligned}$

Will not form a $\Delta$ if $\alpha x+2 y-2=0$ is concurrent with $2 x-y+3=0$ and $6 x+3 y+1=0$ or parallel to either of them so

Case-1: Concurrent lines

$\left|\begin{array}{ccc} 2 & -1 & 3 \\ 6 & 3 & 1 \\ \alpha & 2 & -2 \end{array}\right|=0 \Rightarrow \alpha=\frac{4}{5}$

Case-2 : Parallel lines

$\begin{aligned} & -\frac{\alpha}{2}=\frac{-6}{3} \text { or }-\frac{\alpha}{2}=2 \\ & \Rightarrow \alpha=4 \text { or } \alpha=-4 \\ & P=16+16+\frac{16}{25} \\ & {[P]=\left[32+\frac{16}{25}\right]=32} \end{aligned}$

$ \begin{aligned} & L_1: 3 y-2 x=3 \\\\ & L_2: x-y+1=0 \\\\ & L_3: \alpha x+\beta y+17=0 \end{aligned} $

Point of intersection of $L_1 $ and $ L_2$ is $(0,1)$, should lie on $L_3 \Rightarrow \beta=-17$

Any point, say $\left(\frac{-3}{2}, 0\right)$ on $L_1$ should be equidistant from the lines $L_2 $ and $ L_3$

$ \begin{aligned} & \Rightarrow\left|\frac{\frac{-3}{2}-0+1}{\sqrt{1^2+1^2}}\right|=\left|\frac{\frac{-3 \alpha}{2}+0+17}{\sqrt{\alpha^2+(-17)^2}}\right| \\\\ & \Rightarrow (\alpha-7)(\alpha-17)=0 \end{aligned} $

For $\alpha=17, L_2 $ and $ L_3$ coincides $\Rightarrow \alpha=7$

$ \begin{aligned} \alpha^2+\beta^2-\alpha-\beta & =(7)^2+(-17)^2-7+17 \\\\ & =348 \end{aligned} $

We have, $A B C D$ is a parallelogram

Let equation of $A B$ be $2 x-3 y=-23 \ldots$ (i)

and equation of $B C$ be $5 x+4 y=23\ldots$ (ii)

Equation of $A C$ is $3 x+7 y=23 \ldots$ (iii)

Solving Eqs. (i) and (ii), we get

$x=-1$, and $y=7$

$\therefore$ Co-ordinate of $B$ is $(-1,7)$

On solving Eqs. (ii) and (iii), we get

$ x=3, y=2 $

$\therefore$ Co-ordinate of $C$ is $(3,2)$

On solving Eqs. (i) and (iii), we get $x=-4$ and $y=5$

$\therefore$ Co-ordinate of $A$ is $(-4,5)$.

Let $E$ be the intersection point of diagonal co-ordinate of

$E$ is $\left(\frac{-4+3}{2}, \frac{5+2}{2}\right)$ or $\left(-\frac{1}{2}, \frac{7}{2}\right)$

$\because E$ is mid-point of $A C$

$ \begin{aligned} & \text { Equation of } B D \text { is } y-7=\left(\frac{7-\frac{7}{2}}{-1+\frac{1}{2}}\right)(x+1) \\\\ & \Rightarrow 7 x+y=0 \end{aligned} $

Distance of $A$ from diagonal $B D=\frac{|7 \times(-4)+5|}{\sqrt{7^2+1^2}}$

$ \therefore d=\frac{23}{\sqrt{50}} $

Hence, $50 d^2=(23)^2=529$

$ \because \frac{21 a}{1-2 p}+1+\frac{3 p+21 a}{p+1}=6 $

$ \begin{aligned} & \therefore 4 p^{2}-21 a p+8 p+42 a-5=0\quad...(1) \end{aligned} $

And $\frac{-42 a}{1-2 p}-2+\frac{21 a-3}{p+1}=3 a$

$ \therefore 4 p^{2}-81 a p+6 a p^{2}-24 a+8 p-5=0 \quad...(2) $

From equation (1) - equation (2) we get;

$ 60 a p+66 a-6 a p^{2}=0 $

$ \begin{aligned} \because a \neq 0 \Rightarrow p^{2}-10 p-11=0 \\\\ p=-1 \text { or } 11 \Rightarrow p=11 . \end{aligned} $

When $p=11$ then $a=3$

Coordinate of $B=(-3,6)$

And coordinate of $C=(8,5)$

$\therefore B C^{2}=122$

Slope of $AH = {{a + 2} \over 1}$

Slope of $BC = - {1 \over p}$

$\therefore$ $p = a + 2$ ...... (i)

Coordinate of $C = \left( {{{18p - 30} \over {p + 1}},\,{{15p - 33} \over {p + 1}}} \right)$

Slope of $HC = {{{{15P - 33} \over {p + 1}} - a} \over {{{18p - 30} \over {p + 1}} - 2}} = {{15p - 33 - (p - 2)(p + 1)} \over {18p - 30 - 2p - 2}}$

$ = {{16p - {p^2} - 31} \over {16p - 32}}$

$\because$ ${{16p - {p^2} - 31} \over {16p - 32}} \times - 2 = - 1$

$\therefore$ ${p^2} - 8p + 15 = 0$

$\therefore$ $p = 3$ or $5$

But if $p = 5$ then $a = 3$ not acceptable

$\therefore$ $p = 3$

${4 \over {5 - \alpha }} = {3 \over {\alpha - 2}} \Rightarrow 4\alpha - 8 = 15 - 3\alpha $

$\alpha = {{23} \over 7}$

$A = \left( {{{23} \over 7},0} \right)\,Q = (5,4)$

$R = \left( {{{10 + {{23} \over 7}} \over 3},{8 \over 3}} \right)$

$ = \left( {{{31} \over 7},{8 \over 3}} \right)$

Bisector of angle PAQ is $X = {{23} \over 7}$

$ \Rightarrow M = \left( {{{23} \over 7},{8 \over 3}} \right)$

So, $7\alpha + 3\beta = 31$

Clearly $B$ is $\left(-\frac{3}{\sqrt{a}},+\sqrt{a}\right)$ and $C$ is $\left(-\frac{3}{\sqrt{a}},-\sqrt{a}\right)$

$ \begin{aligned} &\text { Area of } \triangle A C D=\frac{1}{2}\left|\begin{array}{ccc} \frac{3}{\sqrt{a}} & \sqrt{a} & 1 \\\\ -\frac{3}{\sqrt{a}} & -\sqrt{a} & 1 \\\\ 3 \cos \theta & a \sin \theta & 1 \end{array}\right| \\\\ &\Rightarrow \quad \Delta=\left|\begin{array}{ccc} 0 & 0 & 1 \\\\ -\frac{3}{\sqrt{a}} & -\sqrt{a} & 1 \\\\ 3 \cos \theta & a \sin \theta & 1 \end{array}\right| \\\\ &\Rightarrow \quad \Delta=|3 \sqrt{a} \sin \theta+3 \sqrt{a} \cos \theta|=3 \sqrt{a}|\sin \theta+\cos \theta| \\\\ &\Rightarrow \quad \Delta_{\max }=3 \sqrt{a} \cdot \sqrt{2}=12 \Rightarrow a=(2 \sqrt{2})^{2}=8 \end{aligned} $

Intersection point of given lines are (1, 2), (7, 5), (2, 3)


$\Delta = {1 \over 2}\left| {\matrix{ 1 & 2 & 1 \cr 7 & 5 & 1 \cr 2 & 3 & 1 \cr } } \right|$

$ = {1 \over 2}[1(5 - 3) - 2(7 - 2) + 1(21 - 10)]$

$ = {1 \over 2}[2 - 10 + 11]$

$\Delta$DEF $ = {1 \over 2}(3) = {3 \over 2}$

$\Delta$ABC = 4$\Delta$DEF $ = 4\left( {{3 \over 2}} \right) = 6$

For minimum $(P R+R Q)$

$R$ lies on $P Q^{\prime}$ (where $Q^{\prime}$ is image of $Q$ in $X$-axis)

$\Rightarrow$ Equation on $P Q^{\prime}$ is

$ 2 x+y+2=0 \Rightarrow R(-1,0) $

$ \therefore $ 50(PR² + RQ²)

= 50(20 + 5)

= 50(25)

= 1250

${\left( {\sqrt {50} } \right)^2} = {\left( {\sqrt {45} } \right)^2} + {\left( {\sqrt 5 } \right)^2}$

$\angle B = 90^\circ $

Circum-center $ = \left( {{1 \over 2},{{11} \over 2}} \right)$

Mid point of BC $ = \left( {2,{{17} \over 2}} \right)$

Line : $\left( {y - {{11} \over 2}} \right) = 2\left( {x - {1 \over 2}} \right) \Rightarrow y = 2x + {9 \over 2}$

Passing through $\left( {0,{\alpha \over 2}} \right)$

${\alpha \over 2} = {9 \over 2} \Rightarrow \alpha = 9$

x²y² = 1

$ \Rightarrow $ y² = ${1 \over {{x^2}}}$

$ \Rightarrow $ y = $ \pm {1 \over x}$

Graph of this equation,

$OA \bot OB$

$ \Rightarrow \left( {{1 \over {{p^2}}}} \right)\left( { - {1 \over {{q^2}}}} \right) = - 1$

$ \Rightarrow {p^2}{q^2} = 1$

$P\left( {{{p + q} \over 2},{{{1 \over p} - {1 \over q}} \over 2}} \right)$ midpoint of AB lies

On ${x^2}{y^2} = 1$

$ \Rightarrow {(p + q)^2}{\left( {{1 \over p} - {1 \over q}} \right)^2} = 16$

$ \Rightarrow {(p + q)^2}{(p - q)^2} = 16$

$ \Rightarrow {({p^2} - {q^2})^2} = 16$

$ \Rightarrow {P^2} - {1 \over {{P^2}}} = \pm 4$

$ \Rightarrow {p^4} \pm 4{p^2} - 1 = 0$

$ \Rightarrow {p^2} = {{ \pm 4 \pm \sqrt {20} } \over 2} = \pm 2 \pm \sqrt 5 $

$ \Rightarrow {p^2} = 2 + \sqrt 5 $ or $ - 2 + \sqrt 5 $

$O{B^2} = {p^2} + {1 \over {{p^2}}} = 2 + \sqrt 5 + {1 \over {2 + \sqrt 5 }}$ or $ - 2 + \sqrt 5 + {1 \over { - 2 + \sqrt 5 }} = 2\sqrt 5 $

Area $ = 4\left( {{1 \over 2}} \right)(OA)(OB) = 2{(OB)^2} = 4\sqrt 5 $

Since orthocentre and circumcentre both lies on y-axis.

$ \Rightarrow $ Centroid also lies on y-axis.

$ \Rightarrow $ $\sum {\cos \alpha = 0} $

cos$\alpha$ + cos$\beta$ + cos$\gamma$ = 0

$ \Rightarrow $ cos³ $\alpha$ + cos³ $\beta$ + cos³ $\gamma$ = 3cos$\alpha$cos$\beta$cos$\gamma$

$ \therefore $ ${{\cos 3\alpha + \cos 3\beta + \cos 3\gamma } \over {\cos \alpha \cos \beta \cos \gamma }}$

$ = {{4({{\cos }^3}\alpha + {{\cos }^3}\beta + {{\cos }^3}\gamma ) - 3(\cos \alpha + \cos \beta + \cos \gamma )} \over {\cos \alpha \cos \beta \cos \gamma }} = 12$

then, ${\left( {{{\cos 3\alpha + \cos 3\beta + \cos 3\gamma } \over {\cos \alpha \cos \beta \cos \gamma }}} \right)^2} = 144$

3x + 4y $ \le $ 100

4x + 3y $ \le $ 75

x $ \ge $ 0, y $ \ge $ 0

Feasible region is shown in the graph

Let maximum value of 6xy + y² = c

For a solution with feasible region,

6xy + y² = c and 4x + 3y = 75 must have at least one positive solution.

${y^2} + 6y\left( {{{75 - 3y} \over 4}} \right) - c = 0 $

$\Rightarrow {7 \over 2}{y^2} - {{225} \over 2}y + c = 0$

$ \Rightarrow {\left( {{{225} \over 2}} \right)^2} \ge 4.{7 \over 2}.c $

$\Rightarrow c \le {{{{225}^2}} \over {56}} \approx 904$

Apply distance between parallel line formula

$4x - 2y + \alpha = 0$

$4x - 2y + 6 = 0$

$\left| {{{\alpha - 6} \over {25}}} \right| = {1 \over {55}}$

$|\alpha - 6|\, = 2 \Rightarrow \alpha = 8,4$

sum = 12

Again

$6x - 3y + \beta = 0$

$6x - 3y + 9 = 0$

$\left| {{{\beta - 9} \over {3\sqrt 5 }}} \right| = {2 \over {\sqrt 5 }}$

$|\beta - 9|\, = 6 \Rightarrow \beta = 15,3$

sum = 18

$ \therefore $ Sum of all values of $\alpha$ and $\beta$ is = 30

P is centroid of the triangle ABC.

P = $\left( {{{1 + 6 + {3 \over 2}} \over 3},{{0 + 2 + 6} \over 3}} \right)$

= $\left( {{{17} \over 6},{8 \over 3}} \right)$

Given Q $\left( { - {7 \over 6}, - {1 \over 3}} \right)$.

$ \therefore $ PQ = $\sqrt {{{\left( {{{24} \over 6}} \right)}^2} + {{\left( {{9 \over 3}} \right)}^2}} $ = 5