Straight Lines

y – 1 = m(x – 1)

mx – y + 1 – m = 0

$\frac{\left|1-m\right|}{\sqrt[]{m^2+1}}=\frac{4}{\sqrt[]{17}}$

$17{\left(1-m\right)}^2=16\left(m^2+1\right)$

m² – 3m + 1 = 0

$\begin{array}{l}bx+10y-8=0\\ y=\frac{2}{3}x\end{array}\}\Rightarrow \left(b+\frac{20}{3}\right)x=8$

$x=\frac{24}{3b+20},y=\frac{16}{3b+20}$

b = $\frac{2\left(3-2\sqrt[]{2}\right)\left(4+3\sqrt[]{2}\right)}{-2}$

$=\frac{12-8\sqrt[]{2}+9\sqrt[]{2}}{-1}=-\sqrt[]{2}$

b² = 2

$\frac{x^2}{5}+\frac{y^2}{2}=1$

2 = 5 (1 – e²)

$e^2=\frac{3}{5}$

$e$$=$

$D=\left|\begin{array}{ccc}1& -2& 3\\ 2& 1& 1\\ 1& -7& a\end{array}\right|=0\Rightarrow a=8$

also,  $D_1=\left|\begin{array}{ccc}9& -2& 3\\ b& 1& 1\\ 24& -7& 8\end{array}\right|=0\Rightarrow b=3$

hence,  $a-b=8-3=5$

 $\mathrm{m}\mathrm{a}\mathrm{x}\left\{n\right(A),n(B\left)\right\}\le n(A\cup B)\le n\left(U\right)$

$\Rightarrow 76\le 76+63-x\le 100$

$\Rightarrow -63\le -x\le -39$

$\Rightarrow 63\ge x\ge 39$

$L_1:4x+3y+2=0$  

$L_2:3x-4y-11=0$               

Since circle C touches the line L2 at Q intersection point L1 and L2 is (1, -2)

P lies of L₁

$\therefore P\left(x,-\frac{1}{3}\left(2+4x\right)\right)$               

Now,

PQ = 5 ? (x – 1)² + ${\left(\frac{4x+2}{3}-2\right)}^2=25$  

$\Rightarrow x=4,-2$                     

$?$ The circle lies below the axis

y = -6

p (4, -6)

Now distance of P from 5x – 12 y + 51 = 0

$=\left|\frac{20+72+51}{13}\right|=\frac{143}{13}=11$                

C : 4x² + 4y² – 12x + 8y + k = 0

$?\left(1,-\frac{1}{3}\right)$               

Lies on or inside the C then

$4+\frac{4}{9}-12-\frac{8}{3}+k\le 0$

$\Rightarrow k\le \frac{92}{9}$               

Now, circle lies in 4^th quadrant centre

$\equiv \left(\frac{3}{2},-1\right)$               

$\therefore r<1\Rightarrow \sqrt[]{\frac{9}{4}+1-\frac{k}{4}}<1$               

$\Rightarrow \frac{13}{4}-\frac{k}{4}<1$

$\Rightarrow \frac{k}{4}>\frac{9}{4}$

$\Rightarrow k>9$

$\therefore K\in \left(9,\frac{92}{9}\right)$      

28. Kindly consider the following

76. 

If point P be the junction between the lines

2x – 3y + 4 = 0 ______ (1)

3x + 4y – 5 = 0 ______ (2)

Solving (1) and (2) using 3 * (1) – 2 * (2) we get,

6x – 9y + 12 – (6x + 8y – 10) = 0

–17y + 22 = 0

y = $\frac{22}{17}$

And 2x = 3y– 4

=> 2x = 3 * $\frac{22}{17}$ – 4

x = $\frac{33}{17}$ – 2 = $\frac{33-34}{17}$ = $-\frac{1}{17}$

Hence, the co-ordinate of the junction is P $\left(-\frac{1}{17},\frac{22}{17}\right)$

The eqⁿ of the path to be reach is

6x – 7y + 8 = 0 _____ (3)

Then, least distance will be perpendicular path.

So, slope of ⊥ path = $-1$
$\frac{\mathrm{}\mathrm{slope\; of \; line\; (3)}}{}$

$=\frac{-1}{(6/7)}=\frac{-7}{6}$

Hence eqⁿ of shortest/least distance path from P $\left(-\frac{1}{17},\frac{22}{17}\right)$is

$y-\frac{22}{17}=\frac{-7}{6}\left(x+\frac{1}{17}\right).$

$\Rightarrow 6y-\frac{132}{17}=-7x-\frac{7}{17}.$

$\Rightarrow 7x+6y-\frac{132}{17}+\frac{7}{17}=0$

$\Rightarrow 7x+6y-\frac{125}{17}=0.$

119x + 102y – 125 = 0

Kindly go through the solution

Let A have the co-ordinate (x, o)

By laws of reflection

∠PAB = ∠ QAB = θ

And ∠ CAQ + θ = 90°

As normal is ⊥ to surface (x-axis)

⇒ ∠CAQ = 90° - θ

and ∠CAP = 90° + θ

73. The given eqⁿ of limes are.

9x + 6y – 7 = 0 ______ (1)

3x + 2y + b = 0 ______ (2)

Let P (x₀, y₀) be a point equidistant from (1) and (2) so

9x₀ + 6y - 7 = ± 3 (3x₀ + 2y₀ + 6)

When, 9x₀ + 6y₀ – 7 = 3 (3x₀ + 2y₀ + 6)

⇒ 9x₀ + 6y₀ - 7 = 9x₀ + 6y₀ + 18

⇒ - 7 = 18 which in not true

So, 9x₀ + 6y₀ - 7 = -3 (3x₀ + 2y₀ + 6)

⇒ 9x₀ + 6y₀ -7 = -9x₀ -6y₀ -18

⇒ 18x₀ + 12y₀ + 11= 0.

Hence, the required eqn of line through (x₀, y₀) & equidistant  from parallel line 9x + 6y - 7 = 0

and 3x + 2y + 6 = 0 is 18x + 12y + 11 = 0.