Given equation of pair of lines is
12x² - 10xy + 2y² + 11x - 5y + \(\lambda\) = 0 ....(1)
The standard equation of pair of lines is
ax² + 2hxy + by² + 2gx + 2fy + c = 0...(2)
Comparing (1) and (2) we get,
\(a=12,\left| \begin{matrix} 2h=-10 \\ h=-5 \end{matrix} \right| b=2\left| \begin{matrix} 2g=11 \\ g=\frac { 11 }{ 2 } \end{matrix} \right| \left| \begin{matrix} 2f=-5 \\ f=\frac { -5 }{ 2 } \end{matrix} \right| c=\lambda \)
The condition for equation (2) to represent pair of lines is abc + 2fgh - af² - bg² - ch² = 0
\(\Rightarrow \left( 12 \right) \left( 2 \right) \left( \lambda \right) +2\left( \frac { -5 }{ 2 } \right) \left( \frac { 11 }{ 2 } \right) \left( -5 \right) -12\left( \frac { 25 }{ 4 } \right) -2\left( \frac { 121 }{ 4 } \right) -\lambda \left( 25 \right) =0\)
\(\Rightarrow 24\lambda +\frac { 275 }{ 2 } -75-\frac { 121 }{ 2 } -25\lambda =0\)
\(\Rightarrow \frac { 275 }{ 2 } -75-\frac { 121 }{ 2 } =\lambda \)
\(\Rightarrow \lambda =\frac { 275-150-121 }{ 2 } =\frac { 4 }{ 2 } \)
\(\Rightarrow \lambda =2\)

Given \(\alpha\) = 30° and p = 12
Equation of the straight line in normal form is x cos\(\alpha\)+y sin \(\alpha\) = p
⇒ x cos 30°+ y sin 30° = 12

⇒ \(x(\frac{\sqrt{3}}{2})+y(\frac{1}{2})=12\)
\(\frac{\sqrt{3}x+y}{2}=12\)
√3x + y = 24

Given equation of pair of lines is 2x² + 3xy - 2y² + 3x + y + 1 = 0
The condition to represent pair of perpendicular lines is a + b = 0
Here a = 2, 2h = 3, b = -2, 2g = 3, 2f = 1 and c = 1
h = \(\frac { 3 }{ 2 } \), \(\Rightarrow\) a + b = 2 - 2 = 0
Hence the given equation represents a pair of perpendicular lines.

Equation of the line in intercept form is \(\frac{x}{a}+\frac{y}{b}=1\)
⇒ \(\frac{x}{a}+\frac{y}{b}-1=0\)
p=Length of perpendicular from (0, 0) to (1)
= \(\frac { \left| 0+0-1 \right| }{ \sqrt { { \left( \frac { 1 }{ a } \right) }^{ 2 }+{ \left( \frac { 1 }{ b } \right) }^{ 2 } } } \)
= \(\frac { 1 }{ \sqrt { \frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } } } \)
⇒ p\(\frac { 1 }{ \sqrt { \frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } } } \) = 1
Squaring on both sides we get,
\({ p }^{ 2 }\left( \frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } \right) =1\)
⇒ \(\frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } =\frac { 1 }{ { p }^{ 2 } } \)
Hence proved

Any line parallel to 12x + 5y + 2 = 0 will be of the form 12x + 5y + k = 0.....(1)
Given distance from (1, -1) to the line 12x + 5y + k = 0 is 1.
\(\therefore\)  \(1=\pm \frac { 12(1)+5(-1)+k }{ \sqrt { { 12 }^{ 2 }+{ 5 }^{ 2 } } } \) 
\(\because\) distance = \(\pm \frac { a{ x }_{ 1 }+b{ y }_{ 1 }+c }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)
\(\Rightarrow 1=\pm \left( \frac { 7+k }{ \sqrt { 144+25 } } \right) =\pm \left( \frac { -7-k }{ 13 } \right) \)
\(\Rightarrow 1=\frac { 7+k }{ 13 } or\quad 1=\frac { +7+k }{ 13 } \)
\(\Rightarrow\) 13 = 7 + k or 13 = -7 - k
\(\Rightarrow\) 6 = k or k = -20
\(\therefore\)  k = 6 or k = -20
\(\therefore\) The required lines are 12x + 5y + 6 = 0 and 12x + 5y - 20 = 0.

The family of equations of straight lines is of the form (a₁ x + b₁ y + c₁) + \(\lambda\)(a₂ x + b₂ y + c₂) = 0
(i) That is (4x - y + 3) + \(\lambda\)(5x + 2y + 7) = 0....(1)
Since the required equation passes through the point (-1, 2), the point satisfies equation (1),
\(\therefore\)  (-4 - 2 + 3) + \(\lambda\)(-5 + 4 + 7) = 0
\(\Rightarrow\) (-3) + \(\lambda\)(6) = 0
\(\Rightarrow\) 6\(\lambda\) = 3
\(\Rightarrow\)\(\lambda\) = 2
Substituting \(\lambda\) = 2 in (1) we get,
(4x - y + 3) + 2(5x + 2y + 7) = 0
\(\Rightarrow\)4x - y + 3 + 10x + 4y + 14​​​​​​​ = 0
\(\Rightarrow\) 14x + 3y + 17​​​​​​​ = 0
(ii) Any line perpendicular to x - 2y + 1 = 0 will be of the form 2x + y + k = 0
Since this line passes through (-1,-1)
2 (-1) -1 + k = 0
\(\Rightarrow\) 2 - 1 + k = 0
\(\Rightarrow\) -3 + k = 0
\(\Rightarrow\) k = 3
\(\therefore\)  Equation of the required line is 2x + y + 3 = 0
(ii) Parallel to x - y + 5 = 0....(1)
Given lines are   4x - y + 3 = 0 and...(2)
5x + 2y + 7.....(3)
\(\begin{matrix} (2)\times 2\Rightarrow 8x-2y+6=0 \\(3)\Rightarrow \ \ \ \ \ 58x+2y+7=0 \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_\_ \_ \_ \_ \_ \_ \_ \_ \_\_ \_ \_ \_\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 13x\ \ \ +13=0 \end{matrix}\\\Rightarrow x=-1\)
Substituting x = -1 in (2) we get,
-4x - y + 3 = 0
\(\Rightarrow\) -1 - y = 0
\(\Rightarrow\) -1 = +y
\(\Rightarrow\) y = -1
\(\therefore\) (-1, -1) is the point of intersection of the given lines.
Slope of the required line is \(m=-\frac { Co-efficient\quad of\quad x }{ Co-efficient\quad of\quad y } =\frac { -1 }{ -1 } =1\)
\(\therefore\) Equation of the line passing through (-1, -1) with slope 1 is
y + 1 = 1(x + 1)        [ y - y₁ = m(x - x₁) ]
\(\Rightarrow\) y + 1 = x + 1  
\(\Rightarrow\)  x - y = 0 
\(\Rightarrow\) x - y = 0
(iii) Any line perpendicular to x - 2y + 1 = 0 will be of the form 2x + y + k = 0
Since this line passes through (-1, -1)
2 (-1) -1 + k = 0
\(\Rightarrow\) 2 - 1 + k = 0
\(\Rightarrow\) -3 + k = 0
\(\Rightarrow\) k = 3
\(\therefore\)  Equation of the required line is 2x + y + 3 = 0

Given (0,0) is the vertex of y² = 4x and P (h, k) be the mid-point on the locus and the co-ordinates of R be (a, b)
Now P is the mid-point of OR
ஃ (h, k) = \((\frac{0+a}{2},\frac{0+b}{2})=(\frac{a}{2},\frac{b}{2})\)
⇒ h = \(\frac{a}{2}\) and k = \(\frac{b}{2}\)
⇒ a = 2h and b = 2k

Here a and b are two variable which are to be eliminated.
Since (a, b) lies on y² = 4x
⇒ b² = 4a
⇒ (2k)² = 4(2h)
⇒ 4k² = 8h
⇒ k² = 2h
ஃ Locus of (h, k) in y² = 2x

Given that P (-3,1) lie on the locus of x² - 5x + ky = 0.
⇒ (-3)²-5 (-3)+k(1) = 0
⇒ 9 +15 + k = 0
⇒ k = -24
Also, it is given that (2, b) lie on the locus of x² - 5x + ky = 0.
⇒ 2² - 5 (2) + kb = 0
⇒ 4 - 10 - 24 (b) = 0
⇒ - 6 - 24b = 0
⇒ -24b = 6
⇒ b = \(\frac{-6}{24}=\frac{-1}{4}\)

(i) two units from the X-axis
(h, k) be any point on the required path

Any line parallel to x - axis will be of the form y = c.
By the given condition, the distance from x - axis is 2 units ⇒ c = 2.
ஃ Locus of the point P is y = 2.
(ii) Any line parallel to y-axis will be of the formy = c.

By the given condition, the distance from y-axis is 3 units ⇒ c = 2
ஃ Locus of the point P is x = 3.

Let the vertex B is (-4, 7) and the equation of the diagonal AC is 5x -y + 7 = 0
In rhombus, the diagonals are perpendicular to each other.

\(\therefore\)  Equation of the diagonal BD is x + 5y + k = 0 which is \(\bot \) to AC
Since BD passes through the point B (-4, 7) we get
-4 + 5 (7) + k = 0
\(\Rightarrow\)  31 + k = 0
\(\Rightarrow\)  k = -31
\(\therefore\)   Equation of the all another diagonal is x + 5y - 31 = 0.