The given equations are x - y - 7 = 0 and 2x + y - 2 = 0
Equation of any line which passes through the intersection of the given lines is
(x - y - 7) + k (2x + y - 2) = 0 ... (i)
If equation (i) also passes through the origin
i.e., (0, 0), we get
-7 -2k= 0
\(∴\ k={7\over 2}\)
Putting the value of k is equation (i), we get
(x - y - 7 ) - \(7\over2\) (2x + y. - 2) = 0
2x - 2y - 14 - 14x - 7y + 14 = 0
⇒ -12x- 9y = 0
⇒ 4x + 3y = 0

The equations of lines are 2x + 3y + 4 + k (6x - y + 12) = 0 and 7x + 5y - 4 = 0.
Let m1 be slope of the line 2x + 3y + 4 + k (6x - y + 12) = 0.
Then \(m_1=-{2+6k\over 3-k}\)
Let m₂ be slope of the line 7x + 5y - 4 = 0
Then  \(m_2=-{7\over 5}\)
Now m₁m₂ =-1
\(⇒ \left(-{2+6k\over 3-k}\right)\left(-{7\over5}\right)=-1\)
⇒ 14 + 42k = - 15 + 5k
⇒ 42k - 5k = - 15 - 14
⇒ 37k =- 29
⇒ \(k=-{29\over 37}\)

Here the line 2x - 5y = 0 passes through point B(1, 2) and intersect 3x - 5y + 8 = 0 at point A.
So the coordinates of point A will be obtained by solving 2x- 5y = 0 and 3x- 5y + 18 = 0.
∴ coordinates of point A are \(\left(-8,{-16\over 5}\right)\)
Now PQ =\(\sqrt{(1+8)^2+\left(2+{16\over5}\right)^2 }=\sqrt{81+{676\over25}}\)
\(=\sqrt{20258+676\over 25}=\sqrt{2701\over25}\)
\(={\sqrt{2701}\over 5}\)units

Let m₁ be slope of OC then
\(m_1={5-0\over -3-0}=-{5\over3}\)

Let mbe slope ofAB then
m x m₁ =-1
\(∴\ m\times-{5\over 3}=-1\ ∴\ m={3\over5}\)
The equation of line AB is
y - 5 = \(3\over 5\)[x - (- 3)]
⇒ 5y - 25 = 3x + 9
⇒ 3x - 5y + 34 = 0.

\(\left| x \right| +\left| y \right| =1\),
\(\Rightarrow \pm x\pm y=1\)  which forms a square
The locus of the point is a square.

The given equations are 3x + 4y - 7 = 0 and x - y + 2 = 0
Equation of any line passing through the intersection of the given lines is in the form
(3x + 4y - 7) + k (x - y + 2) = 0 ... (i)
⇒ 3x + 4y - 7 + kx - ky + 2k = 0
⇒ (3 + k) x + (4 - k) y - 7 + 2k = 0
Slope= \(-(3+k)\over 4-k\)
∴ \(-{(3+k)\over 4-k}=5\)
⇒ 20 - 5k = -3 - k
⇒ 4k = 23
∴ \(k={23\over 4}\)
Substituting the value of k is equation (i)
\((3x + 4y - 7) +{23\over 4}(x - y + 2) = 0\)
⇒ 12x + 16y - 28 + 23x - 23y + 46 = 0
⇒ 35x - 7y + 18 = 0

The given equation are 2x-5y+9=0 and x+2y+3=0
Equation of any line passing through the point of intersection of the given lines is in the form
(2x-5y+9)+k(x+2y+3) = 0....(i)
\(\Rightarrow\) 2x-5y+9+kx+2ky+3k=0 \(\Rightarrow\) (2+k)x+(2k-5)y+(3k=9)=0
Slope of this line (i) is \(\frac { -(2+k) }{ (2k-5) } ={ m }_{ 1 }\) (Say)
Slope of the given line 3x + 4y + 7 = 0 is =\(\frac { -3 }{ 4 } ={ m }_{ 2 }\) (Say)
If these two lines are parallel to each other
then m₁ = m₂
\(\Rightarrow -\left( \frac { 2+k }{ 2k-5 } \right) =\frac { -3 }{ 4 } \Rightarrow \) 6k-15=8+4k \(\Rightarrow\) 2k = 23
\(\therefore k=\frac { 23 }{ 2 } \)
Now substituting the value of k in Eq. (1)
(2x-5y+9)+\(\frac { 23 }{ 2 } \)(x+2y+3) = 0
\(\Rightarrow\) 4x-10y+18+23x+46y+69=0
\(\Rightarrow\) 27x + 36y + 87 = 0
which is the required equation.

Here 6x - 8y + 7 = 0
⇒ -6x+ 8y = 7
\(⇒ -{6x\over \sqrt{(-6)^2+(8)^2}}+{8y\over \sqrt{(-6)^2+(8)^2}}\)
\(={7\over \sqrt{(-6)^2+(8)^2}}\)
\(⇒-{6x\over 10}+{8y\over 10}={7\over 10}\)
\(⇒ -{3\over 5}x+{4\over 5}y={7\over 10}\)
Length of perpendicular from origin =  \(7\over 10\)
Now 8x - 6y + 11= 0 ⇒ - 8x + 6y = 11
\(⇒ -{8x\over \sqrt{(-8)^2+(6)^2}}+{6y\over \sqrt{(-8)^2+(6)^2}}\)
\(={11\over \sqrt{(-8)^2+(6)^2}}\)
\(⇒ -{8x\over 10}+{6y\over 10}={11\over 10}\)
\(⇒ -{4\over 5}x+{3\over 5}y={11\over 10}\)
Length of perpendicular from origin = \(11\over 10\)
Now \({7\over 10}<{11\over 10}\)
Thus 6x- 8y + 7 = 0 is nearer to the origin.

The equations of the given lines are 3x - 5y + 11 = 0 and x + 7y - 1 = 0
Equation of any line that passes through the intersection of the given lines are in the form
(3x - 5y + 11) + k (x + 7y - 1) = 0 ...(i)
\(\Rightarrow\)  (3 + k)x + (-5 + 7k) + 11 - k = 0
Slope of the line is \(\frac{-(3+k)}{(-5+7k)}\)
As the equation (i) is parallel to x-axis
\(\therefore\) Slope is zero
\(\therefore -(\frac{3+k}{-5+7k})=0\)
\(\Rightarrow\) 3 + k = 0
\(\Rightarrow\) k = -3
Now substitute the value of k in equation (i), we get
(3x - 5y + 11) - 3(x + 7y - 1) = 0
\(\Rightarrow\)  3x - 5y + 11 - 3x - 21y + 3 = 0
\(\Rightarrow\) -26y + 14 = 0
\(\Rightarrow\) 13y - 7 = 0.

The equation of lines are
y - x = 0.....(i)
x + y = 0....(ii)
x - k = 0....(iii)

By solving (i) and (ii), we get the coordinates of point C.
\(\therefore\) Coordinate of Care (0, 0).
By solving (ii) and (iii), we get the coordinates of point A.
\(\therefore\) Coordinate of A are (k, - k).
By solving (i) and (iii), we get the coordinates of point B.
\(\therefore\) Coordinates of B are (k, k).
\(\therefore\) Area of \(\Delta ABC=\frac { 1 }{ 2 } \left| \begin{matrix} k & -k & 1 \\ k & k & 1 \\ 0 & 0 & 1 \end{matrix} \right| \)
= \(\frac{1}{2}\)[(k² + k²) + (0 - 0) + (0 - 0)]
= \(\frac{1}{2}\times\) 2k² = k² sq.units.

i) Here x - \(\sqrt{3}\)y + 8 = 0
\(\Rightarrow\) x - \(\sqrt{3}\)y = -8 \(\Rightarrow\)  -x + \(\sqrt{3}\)y = 8
Dividing both sides by \(\sqrt{(-1)^2+(\sqrt{3})^2}=2,\) we have
\({-x\over 2}+{\sqrt{3}\over 2}y={8\over 2}\Rightarrow {-1\over 2}x+{\sqrt{3}\over 2}y=4\)
Put \(\cos\alpha={-1\over 2}\)  and  \(\sin\alpha={\sqrt{3}\over 2}\)
\(\Rightarrow \alpha\) lies on II^nd quadrant
\(\therefore \cos\alpha={-1\over 2}\)
= -cos 60 = cos (180 - 60)
\(\Rightarrow\alpha=120^o\)
\(\therefore\) Equation of line in normal from is
\(x\cos{2\pi\over 3}+y\sin{2\pi\over 3}=4\)
Comparing it with \(x\cos \alpha+y\sin\alpha=p,\) we have
\(\alpha={2\pi\over 3}\) and p = 4
ii) Here y - 2 = 0
\(\Rightarrow\)  y = 2  \(\Rightarrow\)  0x + y = 2
Dividing both sides by \(\sqrt{(0)^2+(1)^2}=1,\) we have
Put \(\cos\alpha=0\) and \(\sin\alpha =1\)
\(\Rightarrow\)  \(\alpha={\pi\over 2}\)
\(\therefore\)   Equation of line in normal form is
\(x\cos{\pi\over 2}+y\sin{\pi\over 2}=2\)
Comparing it with \(x\cos\alpha+y\sin\alpha=p,\) we have
\(\alpha={\pi\over 2}\)  and p = 2
iii) Here x - y = 4
Dividing both sides by \(\sqrt{(1)^2+(-1)^2}=\sqrt{2},\) we have
\({x\over \sqrt{2}}-{y\over \sqrt{2}}={4\over \sqrt{2}}\)
\(\Rightarrow {1\over\sqrt{2}}x-{1\over\sqrt{2}}y=2\sqrt{2}\)
Put \(\cos\alpha={1\over\sqrt{2}}\)  and \(\sin\alpha={-1\over \sqrt{2}}\)
\(\Rightarrow\alpha\) lies in IV^th quadrant.
\(\therefore \cos\alpha={1\over \sqrt{2}}=\cos(2\pi-{\pi\over 4})\)
\(\Rightarrow\alpha={7\pi\over 4}\)
\(\therefore\)  Equation of line \(x\cos\alpha+y\sin\alpha=p,\) we have
\(\alpha={7\pi\over 4}\)  and  \(p=2\sqrt{2}\)