Straight Lines

Slope of AH = $\frac{a+2}{1}$ slope of BC = $-\frac{1}{p}\therefore p=a+2$ $\therefore C\left(\frac{18p-30}{p+1},\frac{15p-33}{p+1}\right)$  

slope of HC =  $\frac{16p-p^2-31}{16p-32}$  

slope of BC * slope of HC = -1 Þ p = 3 or 5

hence p = 3 is only possible value.

Let point P : (h, k)

Therefore according to question,  ${\left(h-1\right)}^2+{\left(k-2\right)}^2+{\left(h+2\right)}^2+{\left(k-1\right)}^2=14$

$\therefore$ locus of P(h, k) is $x^2+y^2+x-3y-2=0$  

Now intersection with x – axis are  $x^2+x-2=0\Rightarrow x=-2,1$  

Now intersection with y – axis are  $y^2-3y-2=0\Rightarrow y=\frac{3\pm \sqrt[]{17}}{2}$  

Therefore are of the quadrilateral ABCD is =  $\frac{1}{2}\left(\left|x_1\right|+\left|x_2\right|\right)\left(\left|y_1\right|+\left|y_2\right|\right)=\frac{1}{2}*3*\sqrt[]{17}=\frac{3\sqrt[]{17}}{2}$  

If the two lines are perpendicular then a = 2

D = 0       $\Rightarrow$    c = -3

(D = abc + 2fgh - af² – bg² - ch²)

$E\left(\frac{-a{x}_1+b{x}_2+c{x}_3}{-a+b+c},\frac{-a{y}_1+b{y}_2+c{y}_3}{-a+b+c}\right)E\left(\frac{-4*0+4*3+0*5}{-4+3+5},\frac{-4*3+3*0+5*0}{-4+3+5}\right)$

The two altitudes are

$\left(x-\sqrt[]{3}y+1\right)\left(x+\sqrt[]{3}y-5\right)=0$            

Point of int. of the 2 altitudes is  $\left(2,\sqrt[]{3}\right)$  

Let slope of 3^rd altitude be 'm'

then   $\left|\frac{m-\frac{1}{\sqrt[]{3}}}{1+\frac{1}{\sqrt[]{3}}}\right|=\sqrt[]{3}\Rightarrow \frac{\sqrt[]{3}m-1}{\sqrt[]{3}+m}=\pm \sqrt[]{3}$

$\sqrt[]{3}m-1=\pm 3\pm \sqrt[]{3}m$            

  $m=\infty ,=\frac{-2}{2\sqrt[]{3}}=\frac{-1}{\sqrt[]{3}}$          

$\therefore$ The third altitude is x = 2

(k-3)/ (h-2) * (k-0)/ (h-0) = -1
⇒ k (2k – 3) = -2 (h – 2)h
⇒ 2h² + 2k² – 4h – 3k = 0
2x² + 2y² – 4x – 3y = 0
(0,0)

$P\left(x_1,y_1\right)$

$Q\left(x_2,y_2\right)$

$x_1+x_2=\frac{r}{2},x_1{x}_2=\frac{P}{2}$

$y_1+y_2=s,y_1,y_2=-9$

$\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$

$2\left(x^2+y^2\right)-rx-2sy+p-2q=0$

r = 11, s = 7, p – 2q = -22

According to question,

  $\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{4}$         

$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}........(i)$           

Equation of required line is $\frac{x}{a}+\frac{y}{b}=1$  

Obviously B (2, 2) satisfying condition (i)

3x² + 3x²y - 3xy² + dy³ = 0
3x² (x + y) – 3y² (x - dy/3) = 0
x - dy/3 = x + y for getting two perpendicular straight lines
d = -3