Angle between the lines is given by tanθ=\(\frac { \pm 2\sqrt { { h }^{ 2 }-ab } }{ a+b } \)
In this problem, tanθ =\(\frac { \pm 2\sqrt { 16{ a }^{ 2 }{ b }^{ 2 }-({ a }^{ 2 }3{ b }^{ 2 })({ b }^{ 2 }-3{ a }^{ 2 }) } }{ { a }^{ 2 }-3{ b }^{ 2 }+{ b }^{ 2 }-3{ a }^{ 2 } } \)
=\(\frac { \pm 2\sqrt { 16{ a }^{ 2 }{ b }^{ 2 }-{ a }^{ 2 }{ b }^{ 2 }+3{ b }^{ 4 }+3{ a }^{ 4 }-9{ a }^{ 2 }{ b }^{ 2 } } }{ -2{ a }^{ 2 }-2{ b }^{ 2 } } \)
=\(\frac { \pm 2\sqrt { 3{ a }^{ 4 }+3{ b }^{ 4 }+6{ a }^{ 2 }{ b }^{ 2 } } }{ -2({ a }^{ 2 }+{ b }^{ 2 }) } =\pm \sqrt { 3 } \)
tanθ = 60° [If we take the acute angle]

Let m be the slope of the line joining A (a, 2a) and B (-2,3).Then m₁ = \(\frac { 2a-3 }{ a+2 } \)
Let m₂ be the slope of the line 4 x + 3y + 5=0. Then m₂=-\(\frac { 4 }{ 3 } \)
Since given lines are perpendicular. Therefore,
m₁m₂=-1 ⇒ \(\frac { 2a-3 }{ a+2 } \times -\frac { 4 }{ 3 } \)=-1 ⇒ 3a+6 ⇒ a=18/5.

The combined equation of straight lines
(x - 2y - 3) (x + y + 5) = 0
x² + xy + 5x - 2xy - 2y²- 10y - 3x - 3y - 15 = 0
x²- 2y²- xy + 2x -13y - 15 = 0

k = 2, 2x - y + 2 = 0, 6x - 2y + 1 = 0, \(\theta={\tan}^{-1}\left({1\over 7} \right)\)

(3x +y) (x - 3y) = 0 \(\Rightarrow\) 3x²- 8xy - 3y² = 0

5x + 3y + 8 = 0

5x + 3y + 5 = 0

Given pair of lines is 2x²- 5xy - 7y² = 0
Here a = 2, 2h = -5 and b = -7
\(h=\frac{-5}{2}\)
If \(\theta\) is the acute angle between the lines, then 
\(\tan\theta=\pm\frac{2\sqrt{h^2-ab}}{a+b}\)
\(=\pm\frac{2\sqrt{\frac{25}{4}-2(-7)}}{2-7}\)
\(=\pm\frac{2\sqrt{\frac{25}{4}+14)}}{-5}=\pm\frac{2\sqrt{81}}{2(5)}=\pm\frac{\sqrt{81}}{5}=\frac{9}{5}\)
\(\Rightarrow\tan\theta=\frac{9}{5}\)

Equation ofthe bisectors ofthe angles between the lines axx2 (1 - m2) = xl . m² + 2hxy + by² = 0 is
\(\frac{x^2-y^2}{a-b}=\frac{xy}{h}\) ....(1)
Given equation is x² + 4xy - y² = 0
\(\Rightarrow \) a = 1, 2h = 4, b = 1
h = 2
Substituting these values in (1) we get,
\(\frac{x^2-y^2}{1-(-1)}=\frac{xy}{2}\Rightarrow\frac{x^2-y^2}{2}=\frac{xy}{2}\) 
\(\Rightarrow\)x² - y² = xy
Given y = mx is one of the bisector....(2)
Substituting y = mx in (2) we get,
\(\Rightarrow\) x²- m²x² = x.mx
\(\Rightarrow\) x²(1-m²) = x².m
\(\Rightarrow\)1-m² = m
\(\Rightarrow\)m² + m - 1 = 0
\(\Rightarrow m=\frac{-1\pm\sqrt{1-4(1)(-1)}}{2}=\frac{-1\pm\sqrt{1+4}}{2}=\frac{-1\pm\sqrt5}{2}\)

Given equation is 9x² + 12xy + 4y² + 6x + 4y - 3 = 0
\(\Rightarrow (3x+2y)^2+2(3x+2y)-3=0\)
\(\Rightarrow\) y²+ 2y - 3 = 0 where y = 3x + 2y
\(\Rightarrow\) (y+3) (y-1) = 0
\(\Rightarrow\) (3x + 2y + 3) (3x + 2y - 1) = 0[y = 3x + 2y]
Hence the separate equation are 3x + 2y + 3 = 0 and 3x + 2y - 1 = 0
\(\Rightarrow\) a = 3, b = 2, c₁ = 3 and c₂ = -1
Now, Distance between parallel lines \(=\left|\frac{c_1-c_2}{\sqrt{a^2+b^2}}\right|\)
\(=\left|\frac{3-(-1)}{\sqrt{3^2+2^2}}\right|=\left|\frac{4}{\sqrt{9+4}}\right|=\left|\frac{4}{\sqrt{13}}\right|\)
\(=\frac{4}{\sqrt{13}}\)

The equations of parallel sides of the square are x + y - 1 = 0 and x + y + 2 = 0  
\(\therefore\) Length of the side of the square = Distance between parallel sides
\(=\left|\frac{2-(-1)}{\sqrt{1^2+1^2}}\right|\) [ Distance between parallel sides \(=\left|\frac{C_1-C_2}{\sqrt{a^2+b^2}}\right|\)]
\(=\left|\frac{3}{\sqrt2}\right|=\frac{3}{\sqrt2}\)
Hence, Area of the square = (side)²
\(=\left(\frac{3}{\sqrt2}\right)^2=\frac{9}{2}\) sq.units

Given line (k-3)x-(4-k²)y+(k²-7k+6) = 0....(1)
Since the given line passes through the origin
(0, 0) must satisfy the line(1)
\(\Rightarrow\) (k-3)0-(4-k²)0+(k²-7k+6) = 0
\(\Rightarrow\) k²- 7k + 6) = 0
\(\Rightarrow\) (k-1)(k- 6) = 0
\(\Rightarrow\) k = 1,6.
Hence the values of k are 1 and 6.

Given slope = tan 135^o
m = tan(180^o- 45^o) = -tan 45^o = -1
Also, slope \(=\frac{y_2-y_1}{x_2-x_1}\left[ \begin{matrix} (x_1,y_1)\ is\ (3,4)\\(x_2,y_2)\ is\ (x,5) \end{matrix} \right] \)....(1)
\(\Rightarrow m=\frac{5-4}{x-3}\)
\(\Rightarrow m=\frac{1}{x-3}\)....(2)
From (1) and(2), -1 = \(\frac{1}{x-3}\)
\(\Rightarrow\) -x + 3 = 1
\(\Rightarrow \) -x = 1 - 3 = -2
\(\Rightarrow \) x = 2

Let P(x₁, y₁) be the moving point and A(a, 0) B(-a, 0) are the fixed points
Given PA² + PB² = 2C²
\(\Rightarrow\) (x₁ - a)²+ (y₁ - 0)² + (x₁ + a)² + (y₁ - 0)² = 2c² [using distance formula]

\(\Rightarrow2x^2_1+2y^2_1+2a^2=2c^2\) 
\(\Rightarrow x^2_1+y^2_1+a^2=c^2\)
\(\Rightarrow x^2_1+y^2_1=c^2-a^2\)
\(\therefore\) Locus of (x₁, y₁) is x²+ y² = c²- a²