(4)

We have, $(x^2+2x)(12–k)=2$

$\Rightarrow \lambda x^2+2\lambda x–2=0, k\ne 12$          [Let 12 – k = $\lambda$]

For equal roots, D = 0

$\Rightarrow 4{\lambda }^2+8\lambda =0$

$\Rightarrow 4\lambda (\lambda +2)=0$

$\Rightarrow \lambda =0$     (which is not possible)

or   $\lambda =–2 \Rightarrow 12–k=–2$

$\Rightarrow k=14$

So, $P(k,\frac{k}{2})=(14,7)$

$\therefore \mathrm{Distance}, d=$$\left|\frac{3\times 14+4\times 7+5}{\sqrt{9+16}}\right|$$=15$.

(4)

We have, $x(3\lambda +1)+y(7\lambda +2)=17\lambda +5$

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

The intersection of family of lines is the point P(1, 2) and let $Q\equiv (3,6)$.

Now, $d=PQ=\sqrt{2^2+4^2}=\sqrt{20} \Rightarrow d^2=20$

(3)

Number of terms in ${(x+y)}^{2n–3}=2n–3+1=2n–2$

If x = 1, y = 1

Then, sum of all coefficients = $2^{2n–3}$

So, arithmetic mean of all coefficients = $\frac{2^{2n–3}}{(2n–3)+1}$

$\Rightarrow \frac{2^{2n–3}}{2n–2}=16 \Rightarrow 2^{2n}=256(n–1) \Rightarrow n = 5$

Now, $P(2n–1,n^2–4n)=(9,5)$

So, distance from P(9, 5) to the line x + y = 8 is

$\frac{(9+5–8)}{\sqrt{1+1}}=\frac{6}{\sqrt{2}}=3\sqrt{2} \mathrm{units}$.

(4)

Equation of Altitude AP, which is perpendicular to BC is given by

x = 1          ... (i)

Equation of Altitude BP, which is perpendicular to AC is given by

y – 0 = 1(x + 2) $\Rightarrow$ x – y + 2 = 0          ... (ii)

Hence, $P\equiv (1, 3)$          [From (i) and (ii)]

$\therefore \mathrm{Area} \mathrm{of} △PBC=\frac{1}{2}\times 4\times 3=6 \mathrm{sq}. \mathrm{units}$

(2)

We have, $x=3 \tan (\theta +\frac{\pi }{3})$

$\Rightarrow x=3\left(\frac{\tan \theta +\sqrt{3}}{1–\sqrt{3} \tan \theta }\right)$

$\Rightarrow x–x\sqrt{3} \tan \theta =3 \tan \theta +3\sqrt{3}$

$\Rightarrow \tan \theta =\frac{x–3\sqrt{3}}{3+\sqrt{3}x}$          ... (i)

Also, $y=2 \tan (\theta +\frac{\pi }{6})$

$\Rightarrow y=2\frac{(\tan \theta +{\displaystyle \frac{1}{\sqrt{3}}})}{(1–{\displaystyle \frac{\tan \theta }{\sqrt{3}}})}$

$\Rightarrow y(\sqrt{3}–\tan \theta )=2(\sqrt{3} \tan \theta +1)$          ... (ii)

Solving (i) and (ii), we get

$\Rightarrow 2(\frac{x–3\sqrt{3}}{\sqrt{3}+x}+1)=y(\sqrt{3}–\frac{(x–3\sqrt{3})}{\sqrt{3}(\sqrt{3}+x)})$

$\Rightarrow 2\sqrt{3}(x–3\sqrt{3}+x+\sqrt{3})=y\left(3\right(\sqrt{3}+x)–x+3\sqrt{3})$

$\Rightarrow 4\sqrt{3}x–12=y(2x+6\sqrt{3})$

$\Rightarrow xy–2\sqrt{3}x+3\sqrt{3}y+6=0$

Comparing with $xy+\alpha x+\beta y+\gamma$, we get

$\alpha =–2\sqrt{3}, \beta =3\sqrt{3}, \gamma =6$

$\therefore {\alpha }^2+{\beta }^2+{\gamma }^2=12+27+36=75$.

(2)

Using lines PQ and QR, we get

Point Q = $(\frac{1–c}{m–1},\frac{1–c}{m–1}+1)$

$m_{QH}=\frac{{\displaystyle \frac{1–c}{m–1}}+2}{{\displaystyle \frac{1–c}{m–1}}–3}=\frac{1–c+2m–2}{1–c–3m+3}=–\frac{1}{4} [\because QH\perp PR]$          ... (i)

$\because m_{PH}=\frac{5}{0}\to \infty \Rightarrow \mathrm{Slope} \mathrm{of} \mathrm{line} QR \left(m\right)=0$

Put value of m in equation (i), we get

         $\frac{1–c–2}{1–c+3}=–\frac{1}{4} \Rightarrow c=0$

$\Rightarrow m–c=0$.

(4)

Slope of new line $=2–\sqrt{3}$

The new angle with x-axis is $\alpha –\frac{\alpha }{2}=\frac{\alpha }{2}$

$\therefore \tan \left(\frac{\alpha }{2}\right)=2–\sqrt{3}=\tan \left(\frac{\pi }{12}\right) \Rightarrow \frac{\alpha }{2}=\frac{\pi }{12} \Rightarrow \alpha =\frac{\pi }{6}$

The new line passes through (a, 0) and has slope $2–\sqrt{3}$.

So equation of new line is

$(y–0)=(2–\sqrt{3})(x–a) \Rightarrow y–(2–\sqrt{3})x=–(2–\sqrt{3})a$

Now, distance of this line from origin is $\frac{1}{\sqrt{2}}$

$\Rightarrow$$\left|\frac{0+(2–\sqrt{3})a}{\sqrt{{(2–\sqrt{3})}^2+1^2}}\right|$$=\frac{1}{\sqrt{2}} \Rightarrow \frac{(2–\sqrt{3})\left|a\right|}{\sqrt{8–4\sqrt{3}}}=\frac{1}{\sqrt{2}}$

On squaring both sides, we get $\frac{{(2–\sqrt{3})}^2{a}^2}{8–4\sqrt{3}}=\frac{1}{2}$

$\Rightarrow 2(7–4\sqrt{3})a^2=8–4\sqrt{3}$

$\Rightarrow (7–4\sqrt{3})a^2=4–2\sqrt{3}$

$\Rightarrow a^2=\frac{4–2\sqrt{3}}{7–4\sqrt{3}}\times \frac{7+4\sqrt{3}}{7+4\sqrt{3}}$

           $=\frac{28+16\sqrt{3}–14\sqrt{3}–24}{49–48}$

       $a^2=4+2\sqrt{3}$

Now, $3a^{2 }{\tan }^{2 }\alpha –2\sqrt{3}=3(4+2\sqrt{3}) {\tan }^2\frac{\pi }{6}–2\sqrt{3}$

                                                 $=(12+6\sqrt{3})·\frac{1}{3}–2\sqrt{3}$

                                                 $=4+2\sqrt{3}–2\sqrt{3}=4$.

(1)

Slope of OB = $\frac{\sqrt{3}+1}{1–\sqrt{3}}=\tan 105°$

Now, $\alpha +45°=105° \Rightarrow \alpha =60°$

$\therefore$   Coordinates of A are (a cos 60°, a sin 60°)

i.e., $(\frac{a}{2},\frac{\sqrt{3}a}{2})$

Now, A lies on diagonal AC.

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

$\Rightarrow \frac{a}{2}[\sqrt{3}–1–3–\sqrt{3}]+8\sqrt{3}=0$

$\Rightarrow \frac{a}{2}[–4]+8\sqrt{3}=0$

$\Rightarrow a=4\sqrt{3}$

$\Rightarrow a^2=48$.

(4)

Let P, Q and R be the mirror images of points (1, 3), (3, 1) and (2, 4) in the line x + 2y – 2 = 0.

Coordinates of P is given by

$\frac{x–1}{1}=\frac{y–3}{2}=–2\left(\frac{1+6–2}{5}\right) i.e., P(–1,–1)$

Similarly, $Q(\frac{9}{5},\frac{–7}{5}) \mathrm{and} R(\frac{–6}{5},\frac{–12}{5})$

Centroid of $△PQR=(\frac{–2}{15},–\frac{8}{5})$

$\therefore \alpha =\frac{–2}{15} \mathrm{and} \beta =–\frac{–8}{5}$

Thus, $15(\alpha –\beta )=15(\frac{–2}{15}+\frac{8}{5})=22$.

(4)

Let P(h, k) be the point which divides AB internally in the ratio 2 : 1.

$\therefore h=\frac{2\beta +\alpha }{3} \mathrm{and} k=\frac{–4+\alpha +2}{3}$

$\Rightarrow \alpha =3k+2$

$\therefore 2\beta =3h–\alpha =3h–3k–2$

So, AB = 8

$\Rightarrow {(\alpha –\beta )}^2+{(\alpha +4)}^2=64$

$\Rightarrow (3k+2–{\left(\frac{3h–3k–2}{2}\right))}^2+{(3k+2+4)}^2=64$

$\Rightarrow \frac{{(9k–3h+6)}^2}{4}+{(3k+6)}^2=64$

$\Rightarrow 9[{(3k–h+2)}^2+4{(k+2)}^2]=64\times 4$

$\Rightarrow 9(9k^2+h^2+4–6kh+12k–4h+4k^2+16+16k)=256$

$\Rightarrow 9(13k^2+h^2–6kh+28k–4h)=76$

$\Rightarrow 9(x^2+13y^2–6xy–4x+28y)=76$

Comparing the equation with

$\left[9\right(x^2+\alpha y^2+\beta xy+\gamma x+28y)–76]=0$

$\Rightarrow \alpha =13, \beta =–6, \gamma =–4$

$\therefore \alpha –\beta –\gamma =13+6+4=23$