Straight Lines

Slope = 1

$\frac{dy}{dx}=x=1$

$P\left(1,\frac{1}{2}\right)$

Equation of tangent at P : y - $\frac{1}{2}$ = 1 (x – 1)

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

dist= $\frac{\frac{1}{2}}{\sqrt[]{2}}=\frac{1}{2\sqrt[]{2}}$  

Equation of given line x – y + 1 = 0 .(i),

equation of per perpendicular line PP’ is

-x – y + $\lambda =0$  

As line passing through (3, 5)

$\therefore -3-5+\lambda =0$   

$\lambda =8$  

Equation of line PP’ is –x – y + 8 = 0 .(ii)

Solving (i) and (ii)

$\begin{array}{l}-2y=-9\\ y=\frac{9}{2}\end{array}\}Q=\left(\frac{7}{2},\frac{9}{2}\right)$

$y=\frac{9}{2}$

$\therefore x-\frac{9}{2}+1=0$

$P\text{'}=\left(4,4\right)$           

$\frac{5+y_1}{2}=\frac{9}{2}$       &nb

$\frac{dy}{dx}=\frac{ax-by+a}{bx+cy+a}$

$=bxdy+cydy+ady=axdx-bydx+adx$

$=\frac{c{y}^2}{2}+ay-\frac{a{x}^2}{2}-ax+bxy=k$

$a{x}^2+a{y}^2+2ax-2ay=k$

$\Rightarrow x^2+y^2+2x-2y=\lambda$

Short distance of (11,6)

= $\sqrt[]{12^2+5^2}-5$ $\text{exception 25:}$

= 13 – 5

= 8

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

$\therefore \angle B=90°$ circum centre = $O\left(\frac{1}{2},\frac{11}{2}\right)$

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

Equation of OD is y = 2x + $\frac{9}{2}.$ This line passes through

$\left(0,\frac{\alpha }{2}\right)\therefore \alpha =9$

$?f\left(x\right)=ln\left(x+\sqrt[]{1+x^2}\right)\therefore f\left(-x\right)=-f\left(x\right)$ .

Hence f(x) is an odd function.

$Nowg\left(t\right)={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}cos\left(\frac{\pi }{4}t+f\left(x\right)\right)dx}$

Put t = 0, g(0) = ${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}cos\left(f\left(x\right)\right)dx=2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}cos\left(f\left(x\right)\right)dx}...............\left(i\right)$

where cos(f(x)) is an even function.

Now again put t = 1,

$\therefore g\left(1\right)=\frac{1}{\sqrt[]{2}}\times g\left(0\right),from\left(i\right)$

$\therefore g\left(0\right)=\sqrt[]{2}g\left(1\right)$

$\Rightarrow \frac{6}{3}=\frac{2}{-t}\Rightarrow t=-1$

 ->R (-1,0)

    $P{R}^2+R{Q}^2=20+5=25$          

ar (ABC) = 4ar (DEF)

$=4\times \frac{1}{2}\times \left|2\left(2-5\right)+1\left(5-3\right)+7\left(3-2\right)\right|=2\left|-6+2+7\right|=6$

$f\left(x\right)=lo{g}_{\sqrt[]{5}}\left(3+\sqrt[]{2}\left(cosx-sinx\right)\right)$

$-\sqrt[]{2}\le cosx-sinx\le \sqrt[]{2}$

$\Rightarrow 0\le f\left(x\right)\le 2$

$\left|\begin{array}{ccc}\hfill \alpha \hfill & \hfill \beta \hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill 6\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill 1\hfill \end{array}\right|=24$

$4\alpha -2\beta -8=\pm 24$

$4\alpha -2\beta =24+8,4\alpha -2\beta =-24+8$

$2\left(2\alpha -\beta \right)=322\alpha -\beta =-8$

Distance of origin

$D=\sqrt[]{{\alpha }^2+{\left(2\alpha +8\right)}^2}$

$\alpha =-\frac{16}{5}$

$D=\sqrt[]{{\left(\frac{-16}{5}\right)}^2+{\left(\frac{8}{5}\right)}^2}$

if 2 - = 16

$D=\frac{16}{\sqrt[]{5}}$

Equation of ${\perp }^r$ bisector of

$AB:y-3=\frac{t}{3}\left(x-t\right)$

For C put x = 0 so C $\left(0,3-\frac{t^2}{3}\right)$

$h=\frac{t}{2}\&K=\frac{6-\frac{t^2}{3}}{2}$

$2k=6-\frac{1}{3}\times 4h^2$          

$2x^2+3y-9=0$