Let AB be the initial position of the line and AC be its new position.

Slope of the line

$AC=tan{15}^{\circ }=(2-\sqrt{3})$

$\therefore$ Equation of the line AC is

$(y-0)=(2-\sqrt{3})(x-2)ory=(2-\sqrt{3})x-4+2\sqrt{3}or(2-\sqrt{3})x-y-4+2\sqrt{3}=0$

$△$ = $\frac{1}{2}$[4(- 2 + 16) + 3(-16 - 4) + 3(4 + 2)]

                =  $\frac{1}{2}$ [56 - 60 + 18] = 7.

If (x,y) is the foot of the perpendicular from $(x_1,y_1)$ to the line ax + by + c = 0 then

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-(a{x}_1+b{y}_1+c)}{a^2+b^2}$ 

Here $(x_1,y_1)=(2,3);ax+by+c=x+y-3$

$\therefore$ $\frac{x-2}{1}=\frac{y-3}{1}=\frac{-\left(\right(1\times 2)+(1\times 3)+(-3\left)\right)}{1^2+1^2}$ 

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

$\Rightarrow$ $\frac{x-2}{1}=\frac{y-3}{1}=\frac{-2}{2}$

$\Rightarrow$ $\frac{x-2}{1}=\frac{y-3}{1}=-1$ 

$\therefore$ x - 2 = -1 ; y-3 = -1 

$\Rightarrow$ x = -1 + 2 ; y = -1 + 3 

$\Rightarrow$  x = 1 ; y = 2 

$\therefore$ (x,y) = (1,2)

Given line is $(3x+14y+7)+k(5x+7y+6)=0$

$\Rightarrow (3+5k)x+(14+7k)y+(7+6k)=0$

If it is parallel to y- axis, then coefficient of y $=0$

$\Rightarrow 14+7k=0$

$\Rightarrow k=-2$

4a+5b+6c=0
(6x)a+(6y)b+6c=0 $\to$ [Multiply given set of lines ax+by+c=0 with '6']
Now on comparing 6x=4 and 6y =5
⇒ (x,y) =  $(\frac{2}{3},\frac{5}{6})$
$\therefore$ ax+by+c=0 must passes through $(\frac{2}{3},\frac{5}{6})$

Pair of straight lines represented by a second degree equation with coefficient of $x^2$  as a and coefficient of $y^2$  as b are perpendicular if a+b = 0
Here a + b = 0, so they are perpendicular to each other

Pair of straight lines represented by a second degree equation with coefficient of $x^2$  as a and coefficient of $y^2$  as b are perpendicular if a+b = 0. Herere a + b = 0 for every p and q. 

Making the equation of curve homogeneous with the help of line x + y =1,we get

$x^2+y^2-2y(x+y)+\lambda (x+y{)}^2=0$

$\Rightarrow x^2(1+\lambda )+y^2(-1+\lambda )-2yx=0$

Therefore the lines be perpendicular, if A + B = 0.

$\Rightarrow 1+\lambda -1+\lambda =0\Rightarrow \lambda =0$

According to the given condition

$\sqrt{(x-1{)}^2+y^2}-\sqrt{(x+1{)}^2+y^2}$ = $\pm$1

On squaring both sides, we get

$2x^2+2y^2+1=2\sqrt{(x-1{)}^2+y^2}\sqrt{(x+1{)}^2+y^2}$

Again on squaring, we get $12x^2-4y^2=3$.