Straight Lines

$$\begin{gathered} \text{Since,} \cos x.\cos \left(\frac{\pi }{3}-x\right).\cos \left(\frac{\pi }{3}+x\right) \\ =\frac{1}{4}\cos 3x \\ f\left(x\right)=6+16\left(\frac{1}{4}\cos 3x\right)\sin 3x·\cos 6x \\ =6+4\cos 3x\sin 3x\cos 6x \\ =6+\sin 12x \\ \text{Since, }\sin \text{12x ∈ }\left[-1,1\right] \\ f\left(x\right) \text{∈ }\left[5,7\right] \\ \left(\alpha ,\beta \right)\equiv \left(5,7\right) \\ \text{ distance }=\left|\frac{15+28+12}{5}\right|=11 \end{gathered}$$

Rearrange the given equation by grouping terms with $\lambda$:

$x(3\lambda + 1) + y(7\lambda + 2) - 17\lambda - 5 = 0$

$\lambda(3x + 7y - 17) + (x + 2y - 5) = 0$

1. $x + 2y - 5 = 0 \Rightarrow x = 5 - 2y$
2. $3x + 7y - 17 = 0$

Substitute (1) into (2): $3(5 - 2y) + 7y - 17 = 0$ $15 - 6y + 7y - 17 = 0$ $y - 2 = 0 \Rightarrow y = 2$

Then, $x = 5 - 2(2) = 1$. So, the fixed point is $P(1, 2)$.

• Slope of $OP = \frac{2-0}{1-0} = 2$.
• Therefore, the slope of line $L$ is perpendicular, which is $-\frac{1}{2}$.

Equation of line $L$ passing through $(1, 2)$ with slope $-\frac{1}{2}$

: $y - 2 = -\frac{1}{2}(x - 1)$ $2y - 4 = -x + 1$ $x + 2y - 5 = 0$

Find the perpendicular distance $d$ from point $(3, 6)$ to line $x + 2y - 5 = 0$:

$d = \frac{|1(3) + 2(6) - 5|}{\sqrt{1^2 + 2^2}}$ $d = \frac{|3 + 12 - 5|}{\sqrt{5}} = \frac{10}{\sqrt{5}}$

Square both sides to find $d^2$: $d^2=\frac{100}{5}=20$

The mirror line is \(x+2 y-4=0\)
image of \(\mathrm{A}(1,3)\) is \(\frac{x-1}{1}=\frac{y-3}{2}=-2\left(\frac{1+6-4}{5}\right)\)
image of \(\mathrm{B}(3,1)\) is \(\frac{x-3}{1}=\frac{y-1}{2}=-2\left(\frac{3+2-4}{5}\right)\)
image of \((2,4)\) is \(\frac{x-2}{1}=\frac{y-4}{2}=-2\left(\frac{2+8-4}{5}\right)\)

So \(D=\left(-\frac{1}{5}, \frac{3}{5}\right), E=\left(\frac{13}{5}, \frac{1}{5}\right), F=\left(\frac{-2}{5}, \frac{-4}{5}\right)\)
Centroid \(=\left(\frac{\frac{-1}{5}+\frac{13}{5} -\frac{2}{5}}{3}, \frac{\frac{3}{5}+\frac{1}{5}-\frac{4}{5}}{3}\right)\)
\(=\left(\frac{10}{15}, 0\right)=\left(\frac{2}{3}, 0\right)\)

Let point be \(\mathrm{P}(\mathrm{x}, \mathrm{y})\)

As per the question:

$\frac{\sqrt{{\left(x-2\right)}^2+{\left(y-1\right)}^2}}{\sqrt{{\left(x-1\right)}^2+{\left(y-3\right)}^2}}=\frac{5}{4}$

$\Rightarrow \frac{{\left(x-2\right)}^2+{\left(y-1\right)}^2}{{\left(x-1\right)}^2+{\left(y-3\right)}^2}=\frac{25}{16}$

$\Rightarrow 16\left(x^2-4x+4+y^2-2y+1\right)$$=25\left(x^2-2x+1+y^2-6y+9\right)$
\(\Rightarrow 9 x^2+9 y^2+14 x-118 y+170=0\)

Comparing with \(a x^2+b y^2+c x y+d x+e y+170=0\)

\(a=9, b=9, c=0, d=14, e=-118\)

Now,
\(a^2+2 b+3 c+4 d+e\)\(=81+18+0+56-118\)
\(=155-118=37\)

Let $G$ be the centroid of triangle formed by $(1,3), (3,1)$ and $(2,4).$

$G\equiv \left(2,\frac{8}{3}\right)$
Image of $G$, with respect to $x+2y-2=0$

$\frac{\alpha -2}{1}=\frac{\beta -\frac{8}{3}}{2}=-2\frac{\left(2+\frac{16}{3}-2\right)}{1+4}=\frac{-2}{5}\left(\frac{16}{3}\right)$

$$\begin{gathered} \Rightarrow \alpha =\frac{-2}{15}, \beta =\frac{-24}{15} \\ \text{Now}, 15\left(\alpha -\beta \right)=15\left(\frac{-2}{15}+\frac{24}{15}\right)=22 \end{gathered}$$

$$\begin{gathered} Here, x = \frac{11}{2} \& y = \alpha \\ then put in x+y = 11, we get \\ \frac{11}{2}+\alpha =11\Rightarrow \alpha =5.5 \\ Now put in x+2y = 16, we get \\ \frac{11}{2}+2\alpha =16\Rightarrow \alpha =5.2 \\ Atlast put in 2x+3y = 29, we get \\ 2\times \frac{11}{2}+3\alpha =29 \Rightarrow \alpha =6 \\ Now the minimum value of \alpha is 5.2 \\ and maximum value is 6. \\ So the product of minimum and maximum = 6\times 5.2 = 31.2 \end{gathered}$$

\({ }^n C_{r-1}=28,{ }^n C_r=56 \)
\(\frac{{ }^n C_{r-1}}{{ }^n C_r}=\frac{28}{56} \)
\(\frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{n!(n-r)!}}=\frac{1}{2} \)
\(\frac{r}{(n-r+1)}=\frac{1}{2} \)
\(3 r=n+1 \ldots(i) \)
\(\frac{{ }^n C_{\mathrm{r}}}{{ }^n C_{\mathrm{r}+1}}=\frac{56}{70} \)
\(\frac{(r+1)}{(n-r)}=\frac{56}{70} \Rightarrow 9 r=4 n-5 \ldots(i i) \)
By (i) and (ii)
we get \(r=3,n=8) \)
\(A(4 \cos t, 4 \sin t) , B(2 \sin t,-2 \cos t), C\left(3 r-n, r^2-n-1\right)\)
\(A(4 \cos t, 4 \sin t), B(2 \sin t,-2 \cos t), C(1,0) \)

Let centroid of triangle \(ABC\) be \(G(x,y)\)

then \(x=\frac{4 \cos t+2 \sin t+1}{3} \Rightarrow 3 x-1=4 \cos t+2 \sin t\)

\(y=\frac{4 \sin t-2 \cos t+0}{3} \Rightarrow 3 y=4 \sin t-2 \cos t\)

Now
\((3 x-1)^2+(3 y)^2=(4 \cos t+2 \sin t)^2+(4 \sin t-2\cos t)^2 \)
\((3 x-1)^2+(3 y)^2=20\)

Hence \(\alpha=20\)

\( L_1: x+3=0\)
\( L_2: x-y=0 \)
\( L_3: 3 x+y=0 \)
\( \Rightarrow A(-3,-3)\) and \( B(-3,9)\)
Obtuse angle bisector of \( L_3\) and \( L_2\)
\( \frac{x-y}{\sqrt{2}}=\frac{3 x+y}{\sqrt{10}} \ldots\)(1)

point of intersection of \(L_1\) with eq(1) will give point \(C\)
\( \Rightarrow C\left(-3, \frac{9-3 \sqrt{5}}{1+\sqrt{5}}\right)\)

using distance formulae, we get \(BC\) and \(AC\).

Thus, \( \frac{(B C)^2}{(A C)^2}=\left(\frac{9-\frac{9-3 \sqrt{5}}{1+\sqrt{5}}}{3+\frac{9-3 \sqrt{5}}{1+\sqrt{5}}}\right)=5\)

Let Slope of third side is m
$$\begin{gathered} \frac{-\frac{1}{2}-\mathrm{m}}{1-\frac{\mathrm{m}}{2}}=\frac{\mathrm{m}+1}{1-\mathrm{m}} \\ \Rightarrow \frac{1+2m}{m-2}=\frac{m+1}{1-m} \\ \Rightarrow -2{ \mathrm{m}}^2+\mathrm{m}+1={\mathrm{m}}^2-\mathrm{m}-2 \\ \Rightarrow 3{ \mathrm{m}}^2-2 \mathrm{m}-3=0 \\ \Rightarrow {\mathrm{m}}_1+{\mathrm{m}}_2=\frac{2}{3} \end{gathered}$$

$$\begin{gathered} Here, x = \frac{11}{2} \& y = \alpha \\ then put in x+y = 11, we get \\ \frac{11}{2}+\alpha =11\Rightarrow \alpha =5.5 \\ Now put in x+2y = 16, we get \\ \frac{11}{2}+2\alpha =16\Rightarrow \alpha =5.2 \\ Atlast put in 2x+3y = 29, we get \\ 2\times \frac{11}{2}+3\alpha =29 \Rightarrow \alpha =6 \\ Now the minimum value of \alpha is 5.2 \\ and maximum value is 6. \\ So the product of minimum and maximum = 6\times 5.2 = 31.2 \end{gathered}$$

$$\begin{gathered} \text{ Let }(h,k)\text{ be the image of }(1,1)\text{ w.r.to }x+y+5=0 \\ \therefore \frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-2(a{x}_1+b{y}_1+c)}{a^2+b^2} \\ \Rightarrow \frac{h-1}{1}=\frac{k-1}{1}=\frac{-2(1+1+5)}{1+1} \\ \Rightarrow \frac{h-1}{1}=\frac{k-1}{1}=-7 \\ h=-6 \& k =-6 \end{gathered}$$

Now, $\text{P}\left(-6,-6\right)\text{ \& O}\left(0,0\right)$

$$\begin{gathered} OP=\sqrt{{\left(0+6\right)}^2+{\left(0+6\right)}^2} \\ =\sqrt{36+36}=\sqrt{72} \\ =6\sqrt{2} \end{gathered}$$

Let Slope of third side is m
$$\begin{gathered} \frac{-\frac{1}{2}-\mathrm{m}}{1-\frac{\mathrm{m}}{2}}=\frac{\mathrm{m}+1}{1-\mathrm{m}} \\ \Rightarrow \frac{1+2m}{m-2}=\frac{m+1}{1-m} \\ \Rightarrow -2{ \mathrm{m}}^2+\mathrm{m}+1={\mathrm{m}}^2-\mathrm{m}-2 \\ \Rightarrow 3{ \mathrm{m}}^2-2 \mathrm{m}-3=0 \\ \Rightarrow {\mathrm{m}}_1+{\mathrm{m}}_2=\frac{2}{3} \end{gathered}$$

$$\begin{gathered} \text{ Let }(h,k)\text{ be the image of }(1,1)\text{ w.r.to }x+y+5=0 \\ \therefore \frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-2(a{x}_1+b{y}_1+c)}{a^2+b^2} \\ \Rightarrow \frac{h-1}{1}=\frac{k-1}{1}=\frac{-2(1+1+5)}{1+1} \\ \Rightarrow \frac{h-1}{1}=\frac{k-1}{1}=-7 \\ h=-6 \& k =-6 \end{gathered}$$

Now, $\text{P}\left(-6,-6\right)\text{ \& O}\left(0,0\right)$

$$\begin{gathered} OP=\sqrt{{\left(0+6\right)}^2+{\left(0+6\right)}^2} \\ =\sqrt{36+36}=\sqrt{72} \\ =6\sqrt{2} \end{gathered}$$

The mirror line is \(x+2 y-4=0\)
image of \(\mathrm{A}(1,3)\) is \(\frac{x-1}{1}=\frac{y-3}{2}=-2\left(\frac{1+6-4}{5}\right)\)
image of \(\mathrm{B}(3,1)\) is \(\frac{x-3}{1}=\frac{y-1}{2}=-2\left(\frac{3+2-4}{5}\right)\)
image of \((2,4)\) is \(\frac{x-2}{1}=\frac{y-4}{2}=-2\left(\frac{2+8-4}{5}\right)\)

So \(D=\left(-\frac{1}{5}, \frac{3}{5}\right), E=\left(\frac{13}{5}, \frac{1}{5}\right), F=\left(\frac{-2}{5}, \frac{-4}{5}\right)\)
Centroid \(=\left(\frac{\frac{-1}{5}+\frac{13}{5} -\frac{2}{5}}{3}, \frac{\frac{3}{5}+\frac{1}{5}-\frac{4}{5}}{3}\right)\)
\(=\left(\frac{10}{15}, 0\right)=\left(\frac{2}{3}, 0\right)\)

Let 'G' be the centroid of triangle formed by (1,3), (3,1) and (2,4).

$G\equiv \left(2,\frac{8}{3}\right)$
Image of 'G', with respect to $x+2y-2=0$

$\frac{\alpha -2}{1}=\frac{\beta -\frac{8}{3}}{2}=-2\frac{\left(2+\frac{16}{3}-2\right)}{1+4}=\frac{-2}{5}\left(\frac{16}{3}\right)$

$$\begin{gathered} \Rightarrow \alpha =\frac{-2}{15}, \beta =\frac{-24}{15} \\ \text{Now}, 15(\alpha -\beta )=15\left(\frac{-2}{15}+\frac{24}{15}\right)=22 \end{gathered}$$

$$\begin{gathered} \text{Since,} \cos x.\cos \left(\frac{\pi }{3}-x\right).\cos \left(\frac{\pi }{3}+x\right) \\ =\frac{1}{4}\cos 3x \\ f\left(x\right)=6+16\left(\frac{1}{4}\cos 3x\right)\sin 3x·\cos 6x \\ =6+4\cos 3x\sin 3x\cos 6x \\ =6+\sin 12x \\ \text{Since, Range of }\sin \text{12x is }[-1,1] \\ \text{ Range of }f\left(x\right)\text{ is }[5,7] \\ (\alpha ,\beta )\equiv (5,7) \\ \text{ distance }=\left|\frac{15+28+12}{5}\right|=11 \end{gathered}$$

$$\begin{gathered} \text{Given,} \\ 3x-4y-\alpha =0, \\ 8x-11y-33=0 \\ 2x – 3y + \lambda = 0 \\ \text{Here, these equations are concurrent.} \\ \text{then ∆}=0 \\ \Rightarrow \left|\begin{array}{ccc}3& -4& -\alpha \\ 8& -11& -33\\ 2& -3& \lambda \end{array}\right|=0 \\ \Rightarrow 2\alpha -\lambda =33 .........\left(i\right) \\ \text{Now for image of point }\left(1,2\right)\text{:} \\ \frac{x-1}{2}=\frac{y-2}{-3}=\frac{-2·\left(2-6+\lambda \right)}{4+9}=\frac{-2\lambda +8}{13} \\ \mathrm{Now}, x=\frac{29-4\lambda }{13}=\frac{57}{13}\text{ (given) } \\ \Rightarrow \lambda =-7 \\ \text{put in eqn }\left(i\right), \text{we get } \\ \alpha =13 \\ \left|\alpha \lambda \right|=91 \end{gathered}$$

$(\sqrt{3}+1)x+(\sqrt{3}-1)y=0$

$y=\left(\frac{\sqrt{3}+1}{1-\sqrt{3}}\right)x$

$m=\frac{\sqrt{3}+1}{1-\sqrt{3}}$

$m=\tan {105}^{°}$

$A(a\cos {60}^o,a\sin {60}^o)$

$A\left(\frac{a}{3},\frac{a\sqrt{3}}{2}\right)$

eqn. of AC

$(\sqrt{3}-1)x-(\sqrt{3}+1)y+8\sqrt{3}=0$

$A\left(\frac{a}{2},\frac{a\sqrt{3}}{2}\right)$ lies on AC

$(\sqrt{3}-1)\frac{a}{2}-(\sqrt{3}+1)\frac{a\sqrt{3}}{2}+8\sqrt{3}=0$

$\sqrt{3}a-a-3a-2\sqrt{3}+16\sqrt{3}=0$

$4a=-16\sqrt{3}$

$a=4\sqrt{3}$

$a^2=16\times 3=48$

Sol: $\tan \frac{\alpha}{2} = 2 - \sqrt{3} = \tan \frac{\pi}{12}$

$\alpha = \frac{\pi}{6}$

Equation of new line is $y - 0 = \tan \frac{\alpha}{2} (x - a)$

dis tance from $(0,0)=\frac{\mid -a\tan \frac{\alpha }{2}\mid }{\sqrt{1+{\tan }^2\frac{\alpha }{2}}}$$(0,0) = \frac{\left| -a \tan \frac{\alpha}{2} \right|}{\sqrt{1 + \tan^2 \frac{\alpha}{2}}} = \frac{1}{\sqrt{2}}$

$\Rightarrow$ $\frac{a\tan \frac{\alpha }{2}}{\sec \frac{\alpha }{2}}=\frac{1}{\sqrt{2}}$

$\frac{a \tan \frac{\alpha}{2}}{\sec \frac{\alpha}{2}} = \frac{1}{\sqrt{2}} \Rightarrow a = \sqrt{3} + 1$ $3a^2{\tan }^2\alpha -2\sqrt{3}=4$

Equation of the line passing through (3,8) and perpendicular to $x+3y-7=0$ is $3x-y-1=0$. The intersection point of both the lines is (1,2).

Now let the image of $A$(3,8) be $A\text{'}\left(x_1,y_1\right)$ then point (1,2) will be the mid point of $AA\text{'}$.

$\Rightarrow \frac{x_1+3}{2}=1\Rightarrow x_1=-1$ and $\frac{y_1+8}{2}=2$

$\Rightarrow y_1=-4.$

Hence, the image is $(-1,-4)$.

The equation is

$$x(x+2)(12-k)=2$$

which can be written as a quadratic in $x$:

$$(12-k)(x^2+2x)-2=0$$ $$(12-k)x^2 + 2(12-k)x - 2 = 0$$

For equal roots, its discriminant must be zero.

Let

$$a = 12-k,\quad b = 2(12-k),\quad c = -2$$

Then

$$\Delta = b^2 - 4ac = [2(12-k)]^2 - 4(12-k)(-2)$$ $$\Delta = 4(12-k)^2 + 8(12-k) = 4(12-k)(14-k)$$

Set $\Delta = 0$:

$$4(12-k)(14-k) = 0 \Rightarrow k = 12 \text{ or } k = 14$$

For $k = 12$, the original equation becomes $0 = 2$, which has no roots, so discard.
Thus $k = 14$.

The point is $(k, k/2) = (14, 7)$.
Distance from this point to the line $3x + 4y + 5 = 0$ is:

$$d=\frac{\mathrm{\mid }3(14)+4(7)+5\mathrm{\mid }}{\sqrt{3^2+4^2}}=\frac{\mathrm{\mid }42+28+5\mathrm{\mid }}{\sqrt{9+16}}=\frac{75}{5}=15$$