Here,

3x^{2} - 5xy - 2y^{2} - x + 2y = 0

or, 3x^{2} - 6xy + xy - 2y^{2} - x + 2y = 0

or, 3x(x - 2y) + y(x - 2y) - 1(x - 2y) = 0

or, (x - 2y) (3x + y - 1) = 0

Either: x - 2y = 0

Or: 3x + y - 1 = 0

∴ The required equations are: x - 2y = 0 and 3x + y - 1 = 0_{Ans}

Here,

x^{2} - 3xy + 2y^{2} = 0

or, x^{2} - 2xy - xy + 2y^{2} = 0

or, x(x - 2y) - y(x - 2y) = 0

or, (x - 2y) (x - y) = 0

Either: x - 2y = 0

Or: x - y = 0

∴ The required seperate equations are:x - 2y = 0 andx - y = 0 _{Ans}

Here,

The given equation is:6x^{2} + 11xy - 6y^{2} = 0..................(1)

The homogeneous equation of second is: ax^{2} + 2hxy + by^{2} = 0.................(2)

Comparing (1) and (2)

a = 6

2h = 11 i.e. h = \(\frac {11}2\)

b = -6

Now.

a + b = 0

or, 6 - 6 = 0

∴ 0 = 0

Hence, a + b = 0 is satisfied by the given equation so the equation6x^{2} + 11xy - 6y^{2} = 0 are perpendicular to each other. _{Proved}

Here,

Given equation is:x^{2} - 4xy + 4y^{2} = 0......................(1)

The homogenous equation of second degree is:

ax^{2}+ 2hxy + by^{2} = 0...........................(2)

Comparing (1) and (2)

a = 1

2h = -4 i.e. h = \(\frac {-4}2\) = -2

b = 4

Now,

h^{2} = ab

or, (-2)^{2} = 1× 4

∴ 4 = 4

Hence, h^{2} = ab is satisfied by the given equation so the equationx^{2} - 4xy + 4y^{2} = 0 are coincident to each other. _{Proved}

Here,

ax^{2} + 2hxy + by^{2} = 0

When: h^{2} = ab then straight lines are coincident each other.

When: a + b = 0 then straight lines are perpendicular each other.

Given eq^{n} of line are:

x = 2y

i.e. x - 2y = 0..........................(1)

2x = y

i.e. 2x - y = 0..........................(2)

Combined eq^{n} of (1) and (2) is:

(x - 2y) (2x - y) = 0

or, 2x^{2}- xy - 4xy + 2y^{2} = 0

or, 2x^{2} - 5xy + 2y^{2} = 0

∴ The required eq^{n} is: 2x^{2} - 5xy + 2y^{2} = 0 _{Ans}

Here,

Given eq^{n} is: y^{2} = x^{2}

or, x^{2} - y^{2} = 0

or, (x - y) (x + y) = 0

either: x + y = 0

Or, x - y = 0

Slope of eq^{n} x + y = 0 is: m_{1} = -\(\frac 11\) = -1

Slope of eq^{n} x -y = 0 is: m_{2} = -\(\frac 1{-1}\) = 1

m_{1}× m_{2} = -1× 1 = -1

Hence, they are perpendicular to each other. _{Proved}

Here,

Given eq^{n} is:

6x^{2} - 5xy - 6y^{2} = 0...............................(1)

The eq^{n}of homogenous is:

ax^{2} + 2hxy + by^{2} = 0..........................(2)

Comparing eq^{n} (1) and (2)

a = 6

b = -6

Now,

a + b = 0

or, 6 - 6 = 0

∴ 0 = 0

Hence, the angle between two lines is 90°. _{Proved}

Here,

Given eq^{n} is:

x^{2} - 2xy sec\(\alpha\) + y^{2} = 0...........................(1)

The homogenous eq^{n} is:

ax^{2} + 2hxy + by^{2} = 0..............................(2)

Comparing eq^{n} (1) and (2)

a = 1

h = - sec\(\alpha\)

b = 1

Now,

tan\(\theta\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or, tan\(\theta\) =± \(\frac {2\sqrt {(-sec\alpha)^2 - 1 × 1}}{1 + 1}\)

or, tan\(\theta\) =± \(\frac {2\sqrt {sec^2\alpha - 1}}2\)

or, tan\(\theta\) =± \(\sqrt {sec^2\alpha - 1}\)

or, tan\(\theta\) =± \(\sqrt {tan^2\alpha}\)

or, tan\(\theta\) =± tan\(\alpha\)

∴ \(\theta\) = \(\alpha\) _{Ans}

Here,

Given eq^{n} are:

x cos\(\alpha\) + y sin\(\alpha\) = 0...................(1)

x sin\(\alpha\) + y cos\(\alpha\) = 0...................(2)

Combined eq^{n} of (1) and (2)

(x cos\(\alpha\) + y sin\(\alpha\)) (x sin\(\alpha\) + y cos\(\alpha\)) = 0

or, x^{2} cos\(\alpha\).sin\(\alpha\) +xy cos^{2}\(\alpha\) + xy sin^{2}\(\alpha\) + y^{2} sin\(\alpha\).cos\(\alpha\) = 0

or, x^{2} cos\(\alpha\).sin\(\alpha\) + y^{2} sin\(\alpha\).cos\(\alpha\) + xy (cos^{2}\(\alpha\) + sin^{2}\(\alpha\)) = 0

or, sin\(\alpha\).cos\(\alpha\) (x^{2} + y^{2}) + xy× 1 = 0

∴(x^{2} + y^{2}) sin\(\alpha\).cos\(\alpha\) + xy = 0 _{Ans}

Given equation is:

x^{2} + 4xy + y^{2} = 0.............................(1)

The homogeneous equation of second degree is:

ax^{2} + 2hxy + by^{2} = 0.....................(2)

Comparing (1) and (2)

a = 1

h =2

b = 1

Now,

tan\(\theta\) = ± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or, tan\(\theta\) = ± \(\frac {2\sqrt {2^2 - 1 × 1}}{1 + 1}\)

or, tan\(\theta\) = ± \(\frac {2\sqrt {4 - 1}}2\)

or, tan\(\theta\) = ± \(\sqrt 3\)

∴ obtuse angle (\(\theta\) = tan^{-1} (-\(\sqrt 3\))

∴ \(\theta\) = (90 + 30)° = 120°_{Ans}

Here,

Given equation is:

3x^{2} + 2y^{2} - 5xy = 0

The homogeneous equation of second degree is:

ax^{2} + 2hxy + by^{2} = 0.....................(2)

Comparing (1) and (2)

a = 1

h =2

b = 1

Now,

tan\(\theta\) = \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or, tan\(\theta\) = \(\frac {2\sqrt {(\frac {-5}2)^2 - 3 × 2}}{3 + 2}\)

or, tan\(\theta\) = \(\frac {2\sqrt {\frac {25}4 - \frac 61}}5\)

or, tan\(\theta\) = \(\frac {2\sqrt {\frac {25 - 24}4}}5\)

or, tan\(\theta\) = \(\frac {2 × \frac 12}{\frac 51}\)

or, tan\(\theta\) = \(\frac 15\)

or, \(\theta\) = tan^{-1}(\(\frac 15\))

∴ \(\theta\) = 11.31° _{Ans}

Here,

The given equation is:

12x^{2} - 23xy + 5y^{2} = 0.....................(1)

The homogeneous equation of second degree is:

ax^{2} + 2hxy + by^{2} = 0.....................(2)

Comparing (1) and (2)

a =12

h = -\(\frac {23}2\)

b = 5

Now,

tan\(\theta\) = ±\(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or,tan\(\theta\) = ± \(\frac {2\sqrt {(\frac {-23}2)^2 - 12 × 5}}{12 + 5}\)

or,tan\(\theta\) = ± \(\frac {2\sqrt {\frac {529}4 - 60}}{17}\)

or,tan\(\theta\) = ± \(\frac {2\sqrt {\frac {529 - 240}4}}{17}\)

or,tan\(\theta\) = ± \(\frac {2\sqrt {289}}2\)× \(\frac 1{17}\)

or,tan\(\theta\) = ± \(\frac {17}{17}\)

∴tan\(\theta\) = ± 1

Taking +ve sign,

tan\(\theta\) = 1

or, tan\(\theta\) = tan 45°

∴ \(\theta\) = 45°

Taking -ve sign,

tan\(\theta\) = -1

or, tan\(\theta\) = tan (180 - 35)

or, tan\(\theta\) = tan 135°

∴ \(\theta\) = 135°

Hence, the obtuse angle between the pair of equation is: 135°. _{Ans}

Here,

The homogenous equation of second degree is:

ax^{2} + 2hxy +by^{2} = 0................................(1)

The given equation is:

2x^{2} - 5xy + 2y^{2} = 0...................................(2)

Comparing (1) and (2)

a = 2

2h = -5 i.e. h = -\(\frac 52\)

b = 2

If \(\theta\) be the angle between pair of lines:

tan\(\theta\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or, tan\(\theta\) =± \(\frac {2\sqrt {(\frac {-5}2)^2 - 2 × 2}}{2 + 2}\)

or, tan\(\theta\) =± \(\frac {2\sqrt {\frac {25}4 - \frac 41}}4\)

or, tan\(\theta\) =± \(\frac {\sqrt {\frac {25 - 16}4}}2\)

or, tan\(\theta\) =± \(\sqrt {\frac 94}\)× \(\frac 12\)

or, tan\(\theta\) =± \(\frac 32\)× \(\frac 12\)

or, tan\(\theta\) =± \(\frac 34\)

∴ \(\theta\) = tan^{-1}(± \(\frac 34\))

Hence, the acute angle (\(\theta\)) =tan^{-1}(± \(\frac 34\)) _{Ans}

Here,

Given equation is:

6x^{2} + 5xy - 3x - 2y - 6y^{2} = 0

or, 6x^{2} + 9xy - 4xy - 6y^{2} - 3x - 2y = 0

or, 3x (2x + 3y) - 2y (2x + 3y) - 1 (2x + 3y) = 0

or, (2x + 3y) (3x - 2y - 1) = 0

Either: 2x + 3y = 0..........................(1)

Or: 3x - 2y - 1 = 0............................(2)

Slope of equation (1), m_{1} = -\(\frac 23\)

Slope of equation (2), m_{2} = \(\frac 32\)

Again,

m_{1}× m_{2} = \(\frac {-2}3\)× \(\frac 32\) = -1

The product of two slopes = -1

Hence, these equations are perpendicular each other. _{Hence, Proved}

Here,

Let: the two equation of the lines through origin be:

a_{1}x + b_{1}y = 0..............................(1)

a_{2}x + b_{2}y = 0..............................(2)

Combined equation of (1) and (2) is:

(a_{1}x + b_{1}y) (a_{2}x + b_{2}y) = 0

or, a_{1}a_{2}x^{2} + a_{1}b_{2}xy + a_{2}b_{1}xy + b_{1}b_{2}y^{2} = 0

or,a_{1}a_{2}x^{2} + (a_{1}b_{2}+ a_{2}b_{1})_{}xy + b_{1}b_{2}y^{2} = 0

Let:

a_{1}a_{2} = a

b_{1}b_{2} = b

(a_{1}b_{2}+ a_{2}b_{1}) = 2h

Now,

ax^{2} + 2hxy + by^{2} = 0

Thus,ax^{2} + 2hxy + by^{2} = 0 is homogenous equation of second degree. _{Ans}

Here,

The given equation is: ax^{2} + 2hxy + by^{2} = 0 if a≠ 0, then:

The given equation is multiplied by 'a' on both sides:

a^{2}x^{2} + 2ahxy + aby^{2} = 0

or (ax)^{2} + 2 (ax) (hy) + (hy)^{2} - h^{2}y^{2} + aby^{2} = 0

or, (ax + hy)^{2} - (h^{2} - ab) y^{2} = 0

or, (ax + hy)^{2} - {(\(\sqrt {(h^2 - ab)y})}^{2} = 0

or, (ax + hy + \(\sqrt {(h^2 - ab)y}\)) (ax + hy - \(\sqrt {(h^2 - ab)y}\)) = 0

Either:(ax + hy + \(\sqrt {(h^2 - ab)y}\)) = 0.................(1)

Or:(ax + hy - \(\sqrt {(h^2 - ab)y}\)) = 0..........................(2)

Equation (1) and (2) are satisfied (0, 0) so both straight lines will passes through origin.

Again,

If a = 0

The equation ax^{2}+ 2hxy + by^{2} = 0 will be

2hxy + by^{2} = 0

or, y(2hx + by) = 0

Either: y = 0.......................(3)

Or: 2hx + by = 0..............(4)

Equation (3) and (4) are satisfied by the co-ordinates (0, 0).

Hence, the second degree homogeneous equation: ax^{2} + 2hxy + by^{2} = 0 always represents a pair of straight lines through the origin. _{Proved}

Here,

Given equationax^{2} + 2hxy + by^{2}= 0 represents two straight lines that passes through the origin.

Let: y = m_{1}x and y = m_{2}x are the two lines.

y -m_{1}x = 0............................(1)

y - m_{2}x = 0............................(2)

Product of the equation (1) and (2) is:

(y - m_{1}x) (y - m_{2}x) = 0

or, m_{1}m_{2}x^{2} - m_{1}xy - m_{2}xy + y^{2} = 0

or, m_{1}m_{2}x^{2} - (m_{1 }+m_{2})xy + y^{2} = 0...................(3)

Given equationax^{2} + 2hxy + by^{2}= 0

\(\frac ab\)x^{2} + \(\frac {2h}b\) xy + y^{2} = 0...................(4)

Comparing equation (3) and (4), we get:

m_{1} + m_{2} = -\(\frac {2h}b\) and m_{1}m_{2} = \(\frac ab\)

Let \(\theta\) be the angle between the lines:

tan\(\theta\) =± \(\frac {m_1 - m_2}{1 + m_1m_2}\)

or, tan\(\theta\) =± \(\frac {\sqrt {(m_1 + m_2)^2 - 4m_1m_2}}{1 + m_1m_2}\) [\(\because\) a - b = \(\sqrt {(a + b)^2 - 4ab}\)]

or, tan\(\theta\) =± \(\frac {\sqrt {\frac {4h^2}{b^2} - \frac {4a}b}}{1 + \frac ab}\)

or, tan\(\theta\) =± \(\frac {\sqrt {\frac {4h^2 - 4ab}{b^2}}}{\frac {b + a}b}\)

or, tan\(\theta\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)× \(\frac bb\)

or, tan\(\theta\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

∴\(\theta\) = tan^{-1}(± \(\frac {2\sqrt {h^2 - ab}}{a + b}\))_{ Ans}

Here,

x^{2} - 5xy + 4y^{2} = 0

or, x^{2} - xy - 4xy + 4y^{2} = 0

or, x(x - y) - 4y(x - y) = 0

or, (x - y) (x - 4y) = 0

Either: x - y = 0.................(1)

Or: x - 4y = 0......................(2)

The eq^{n} (1) changes into parallel form is:

x - y + k_{1} = 0......................(3)

The point (1, 1) passes through eq^{n} (1)

1 - 1 + k_{1} = 0

∴ k_{1} = 0

Putting the value of k_{1} in eq^{n} (3)

x - y + 0 = 0

x - y = 0............................(4)

The eq^{n}(2) changes in parallel form is:

x - 4y + k_{2} = 0..................(5)

The point (1, 1) passes through eq^{n} (5)

1 - 4× 1 + k_{2} = 0

or, -3 + k_{2} = 0

∴ k_{2} = 3

Putting the value of k_{2} in eq^{n} (5)

x - 4y + 3 = 0.......................(6)

The eq^{n} of pair of line is:

(x - y) (x - 4y + 3) = 0

or, x^{2} - 4xy + 3x - xy - 4y^{2} - 3y = 0

∴x^{2} - 5xy - 4y^{2} + 3x - 3y = 0_{Ans}

Here,

Given equation is:

x^{2} - xy - 2y^{2} = 0

or, x^{2} - 2xy + xy - 2y^{2} = 0

or, x (x - 2y) + y (x - 2y) = 0

or, (x - 2y) (x + y) = 0

Either: x - 2y = 0.........................(1)

Or: x + y = 0.................................(2)

The eq^{n} (1) changes in perpendicular form is:

-2x - y + k_{1} = 0

2x + y - k_{1} = 0..........................(3)

The eq^{n} (3) passes through origin (0, 0):

2× 0 + 0 - k_{1} = 0

∴ k_{1} = 0

Putting the value of k_{1} in eq^{n} (3)

2x + y - 0 = 0

2x + y = 0........................(4)

The eq^{n} (2) change in perpendicular form is:

x - y + k_{2} = 0........................(5)

The eq^{n} (5) passes through origin (0, 0)

0 - 0 + k_{2} = 0

∴ k_{2} = 0

Putting the value of k_{2} in eq^{n} (5)

x - y + 0 = 0

x - y = 0...........................(6)

The equation of the pairs of lines is:

(2x + y) (x - y) = 0

or, 2x^{2} - 2xy + xy - y^{2} = 0

∴2x^{2} - xy - y^{2} = 0_{Ans}

Given eq^{n} of line is:

3x^{2} - 8xy + 5y^{2}^{}= 0

or, 3x^{2} - 3xy - 5xy + 5y^{2}^{}= 0

or, 3x (x - y) - 5y (x - y) = 0

or, (x - y) (3x - 5y) = 0

Either: x - y = 0.......................(1)

Or: 3x - 5y = 0.........................(2)

The eq^{n} (1) changes in perpendicular form is:

x + y + k_{1} = 0...........................(3)

The point (2, 3) passes through eq^{n} (3)

2 + 3 + k_{1} = 0

or, 5 + k_{1} = 0

∴ k_{1} = -5

Putting the value of k_{1} in eq^{n} (3)

x + y - 5 = 0..........................(4)

The eq^{n} (2) change in perpendicular form is:

5x + 3y + k_{2} = 0...................(5)

The point (2, 3) passes through eq^{n} (5)

5× 2 + 3× 3 + k_{2} = 0

or, 10 + 9 + k_{2} = 0

or, 19 + k_{2} = 0

∴ k_{2} = - 19

Putting the value of k_{2} in eq^{n} (5)

5x + 3y - 19 = 0......................(6)

The eq^{n} of pairs of lines is:

(x + y - 5) (5x + 3y - 19) = 0

or, 5x^{2} + 3xy - 19x + 5xy + 3y^{2} - 19y - 25x - 15y + 95 = 0

∴ 5x^{2} + 8xy + 3y^{2} - 44x - 34y + 95 = 0_{Ans}

Here,

x^{2} - xy - 2y^{2} = 0

or, x^{2} - 2xy + xy - 2y^{2} = 0

or, x(x - 2y) + y(x - 2y) = 0

or, (x - 2y) (x + y) = 0

Either: x - 2y = 0..................(1)

Or: x + y = 0...........................(2)

The eq^{n} (1) change in perpendicular form is:

2x + y + k_{1} = 0.......................(3)

The point (3, -1) passes through eq^{n} (3)

2× 3 - 1 + k_{1} = 0

or, 6 - 1 + k_{1} = 0

∴ k_{1}= -5

Putting the value of k_{1} in eq^{n} (3)

2x + y - 5 = 0..........................(4)

The eq^{n} (2) change in perpendicular form is:

x - y + k_{2} = 0.........................(5)

The point (3, -1) passes througheq^{n} (5)

3 + 1 + k_{2} = 0

∴ k_{2} = -4

Putting the value of k_{2} in eq^{n} (5)

x - y - 4 = 0.........................(6)

The eq^{n} of pair of lines is:

(2x + y - 5) (x - y - 4) = 0

or, 2x^{2} - 2xy - 8x + xy - y^{2} - 4y - 5x + 5y + 20 = 0

∴ 2x^{2} - xy- y^{2} - 13x + y + 20 = 0_{Ans}

Here,

2x^{2} + 5xy + 3y^{2} = 0

or, 2x^{2} + 3xy + 2xy + 3y^{2} = 0

or, x(2x + 3y) + y(2x + 3y) = 0

or, (2x + 3y) (x + y) = 0

Either: 2x + 3y = 0

Or: x + y = 0

Given eq^{n} is: 2x^{2} + 5xy + 3y^{2} = 0......................(1)

Homogenous eq^{n} is: ax^{2} + 2hxy + by^{2} = 0...................(2)

Comparing eq^{n} (1) and (2)

a = 2

h = \(\frac 52\)

b = 3

We know that:

tan\(\theta\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or,tan\(\theta\) =± \(\frac {2\sqrt {(\frac 52)^2 - 2 × 3}}{2 + 3}\)

or, tan\(\theta\) =± \(\frac {2\sqrt {\frac {25}4 - 6}}5\)

or,tan\(\theta\) =± \(\frac {2\sqrt {\frac {25 - 24}4}}5\)

or,tan\(\theta\) =± \(\frac {2\sqrt {\frac 14}}5\)

or,tan\(\theta\) =± \(\frac {2 × \frac 12}5\)

or,tan\(\theta\) =± \(\frac 15\)

Taking +ve sign,

\(\theta\) = tan^{-1} (\(\frac 15\)) = 11.31°

Taking -vesign,

\(\theta\) = (180° - 11.31°) = 168.69°

∴ Required eq^{n} are: 2x + 3y = 0 and x + y = 0 and angle between them are: 11.31° and 168.69°. _{Ans}

Given eq^{n} is:x^{2} + 2xy sec\(\theta\) + y^{2} = 0........................(1)

Homogenous eq^{n} is: ax^{2} + 2hxy + by^{2} = 0..........(2)

Comparingeq^{n}(1) and (2)

a = 1

h = sec\(\theta\)

b = 1

If \(\alpha\) be the angle between pairs of lines then,

tan\(\alpha\) = \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or, tan\(\alpha\) = \(\frac {2\sqrt {sec^2\theta - 1 × 1}}{1 + 1}\)

or, tan\(\alpha\) = \(\frac {2\sqrt {sec^2\theta - 1}}2\)

or, tan\(\alpha\) = \(\sqrt {tan^2\theta}\)

or, tan\(\alpha\) = tan\(\theta\)

∴\(\alpha\) = \(\theta\) _{Proved}

Again,

The given eq^{n} is:

y^{2} + 2xy sec\(\theta\) + x^{2} = 0..........................(3)

The quadratic eq^{n} is:

ax^{2} + bx + c = 0.............................(4)

Comparing (3) and (4)

a = 1

b = 2x sec\(\theta\)

c = x^{2}

We know,

x = \(\frac {-b ± \sqrt {b^2 - 4ac}}{2a}\)

or, y = \(\frac {-(2x sec\theta) ± \sqrt {(2x sec\theta)^2 - 4 × 1 × x^2}}{2 × 1}\)

or, y = \(\frac {-2x sec\theta ± \sqrt {4x^2 sec^2\theta - 4x^2}}2\)

or, y = \(\frac {-2x sec\theta ± \sqrt {4x^2 (sec^2\theta - 1)}}2\)

or, y = \(\frac {-2x sec\theta ± 2x\sqrt {tan^2\theta}}2\)

or, y = \(\frac {-2x sec\theta ± 2x tan\theta}2\)

or, y = \(\frac {2(-x sec\theta ± xtan\theta)}2\)

or, y = - xsec\(\theta\)± x tan\(\theta\)

Taking +ve sign:

y = -xsec\(\theta\) + xtan\(\theta\)

Taking -ve sign:

y = -xsec\(\theta\) - xtan\(\theta\)

∴ The required eq^{n} are:y = -xsec\(\theta\) + xtan\(\theta\) andy = -xsec\(\theta\) - xtan\(\theta\) _{Ans}

Given equation is:x^{2} - 2xy cosec\(\theta\) + y^{2} = 0.....................(1)

Homogenous equation is: ax^{2} + 2hxy + by^{2} = 0......................(2)

Comparing eq^{n} (1) and (2)

a = 1

h = -cosec\(\theta\)

b = 1

If \(\alpha\) be the angle between pair of lines then:

tan\(\alpha\) =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or, tan\(\alpha\) =± \(\frac {2\sqrt {(-cosec\theta)^2 - 1 × 1}}{1 + 1}\)

or, tan\(\alpha\) =± \(\frac {2\sqrt {(-cosec\theta)^2 - 1}}2\)

or, tan\(\alpha\) =± \(\frac {2\sqrt {cosec^2\theta - 1}}2\)

or, tan\(\alpha\) =± \(\sqrt {cot^2\theta}\)

∴ tan\(\alpha\) =± cot\(\theta\)

Taking +ve sign,

tan\(\alpha\) = cot\(\theta\)

tan\(\alpha\) = tan(\(\frac p2\) - \(\theta\))

∴\(\alpha\) = (\(\frac p2\) - \(\theta\))_{Hence, Proved}

Here,

Given equation is:

2x^{2} + 7xy + 3y^{2} = 0

or, 2x^{2} + 6xy + xy + 3y^{2} = 0

or, 2x(x + 3y) + y(x + 3y) = 0

or, (x + 3y)(2x + y) = 0

The two equation represented by2x^{2} + 7xy + 3y^{2} = 0 are:

2x + y = 0......................(1)

x + 3y = 0......................(2)

Now,

Slope of equation (1), m_{1} = -\(\frac {x-coefficient}{y-coefficient}\) = -2

Slope of equation (2),m_{2} =-\(\frac {x-coefficient}{y-coefficient}\) = -\(\frac 13\)

If the angle between the lines be \(\theta\) then:

tan\(\theta\) =± \(\frac {m_1 - m_2}{1 + m_1m_2}\)

or, tan\(\theta\) =± \(\frac {-2 + \frac 13}{1 + (-2) (-\frac 13)}\)

or, tan\(\theta\) = ± \(\frac {\frac {-6 + 1}3}{\frac {3 + 2}3}\)

or, tan\(\theta\) = ± \(\frac {-5}3\)× \(\frac 35\)

or, tan\(\theta\) = ± 1

Taking +ve sign,

tan\(\theta\) = tan 45°

∴ \(\theta\) = 45°

Taking -ve sign,

tan\(\theta\) = tan (180 - 45)° = tan 135°

∴ \(\theta\) = 135°

∴ Required angles (\(\theta\)) = 45° and 135°_{Ans}

The given equation is:

2x^{2} + 3xy - 2y^{2} = 0

or, 2x^{2} + 4xy - xy - 2y^{2} = 0

or, 2x(x + 2y) - y(x + 2y) = 0

or, (x + 2y) (2x - y) = 0

∴ Equation are: x + 2y = 0 and 2x - y = 0

Again,

2x^{2} + 3xy - 2y^{2} = 0............................(1)

The homogenous equation of second degree is:

ax^{2} + 2hxy + by^{2} = 0........................(2)

Comparingequation (1) and (2)

a = 2

h = \(\frac 32\)

b = -2

tan\(\theta\) = \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or, tan\(\theta\) = \(\frac {2\sqrt {(\frac 32)^2 - 2 × (-2)}}{2 + (-2)}\)

or, tan\(\theta\) = \(\frac {2\sqrt {\frac 94 + \frac 41}}{2 - 2}\)

or, tan\(\theta\) = \(\frac {2\sqrt {\frac {9 + 16}4}}0\)

or, tan\(\theta\) =∞

∴ \(\theta\) = tan^{-1}∞ = 90°

∴ The required equations are: x + 2y = 0 and 2x - y = 0 and angle between the pairs of straight lines is 90°._{Ans}

Here,

Given equation is:

2x^{2}- 3xy + y^{2} = 0......................(1)

The homogenous equation of second degree is:

ax^{2} + 2hxy + by^{2} = 0..................(2)

Comparing equation (1) and (2)

a = 2

h = -\(\frac 32\)

b =1

Now,

tan\(\theta\) = ± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or, tan\(\theta\) = ± \(\frac {2\sqrt {(\frac {-3}2)^2 - 2 × 1}}{2 + 1}\)

or, tan\(\theta\) = ± \(\frac {2\sqrt {\frac 94 - \frac 21}}3\)

or, tan\(\theta\) = ± \(\frac {2\sqrt {\frac {9 - 8}4}}3\)

or, tan\(\theta\) = ± \(\frac {2 × \frac 12}3\)

or, tan\(\theta\) = ± \(\frac 13\)

For the obtuse angle,

tan\(\theta\) = -\(\frac 13\)

\(\theta\) = tan^{-1} (\(\frac 13\))

Now,

2x^{2} - 3xy + y^{2} = 0

or, 2x^{2} - 2xy - xy + y^{2} = 0

or, 2x(x - y) - y(x - y) = 0

or, (x - y) (2x - y) = 0

∴ The required pairs of lines are: x - y = 0 and 2x - y = 0. _{Ans}

Here,

6x^{2} + 5xy - 3x + 2y - 6y^{2} = 0

or, 6x^{2} + 5xy - 6y^{2} - 3x + 2y = 0

or, 6x^{2} + 9xy - 4xy - 6y^{2} - 3x + 2y = 0

or, 3x(2x + 3y) - 2y(2x + 3y) - 1(3x - 2y) = 0

or, (2x + 3y - 1) (3x - 2y) = 0

Either: 2x + 3y - 1 = 0...........(1)

Or: 3x - 2y = 0.........................(2)

Slope of equation (1), m_{1} = \(\frac {-3}{-2}\) = \(\frac 32\)

Slope of equation (2),m_{2} = -\(\frac 23\)

∴ m_{1}× m_{2} = \(\frac 32\)× \(\frac {-2}3\) = -1

Hence, the product of two slopes = -1 so these equations are perpendicular to each other. _{Proved}

The given equation is:

x^{2} - 2xy cosec\(\theta\) + y^{2} = 0

or, (y)^{2} - (2x cosec\(\theta\)) y + (x)^{2} = 0..................(1)

The quadratic equation is:

ax^{2} + bx + c = 0......................(2)

Comparing (1) and (2)

a = 1

b = -2x cosec\(\theta\)

c = x^{2}

x = \(\frac {-b ± \sqrt {b^2 - 4ac}}{2a}\)

y = -(-2x cosec\(\theta\))± \(\frac {\sqrt {(-2x cosec\theta)^2 - 4(1) (x^2)}}{2× 1}\)

y = \(\frac {2x cosec\theta ± \sqrt {4x^2 cosec^2\theta - 4x^2}}2\)

y = \(\frac {2x cosec\theta ± 2x \sqrt {cosec^2\theta - 1}}2\)

y = \(\frac {2(x cosec\theta ± x cot\theta)}2\)

y = x cosec\(\theta\)± x cot\(\theta\)

The pair of lines are:

y - x (cosec\(\theta\) - cot\(\theta\)) = 0 and

y - x (cosec\(\theta\) + cot\(\theta\)) = 0

Comparing x^{2} -2xy cosec\(\theta\) + y^{2} = 0 and ax^{2} + 2hxy + by^{2} = 0

a = 1

h = - cosec\(\theta\)

b = 1

We know that:

tanA =± \(\frac {2\sqrt {h^2 - ab}}{a + b}\)

or, tanA = ± \(\frac {2\sqrt {(-cosec\theta)^2 - 1 × 1}}{1 + 1}\)

or, tanA =± \(\frac {2\sqrt {cosec^2\theta - 1}}2\)

or, tanA =± cot\(\theta\)

∴ A = tan^{-1} (± cot\(\theta\))

∴ The required equations are: y - x (cosec\(\theta\) - cot\(\theta\)) = 0 and y - x (cosec\(\theta\) + cot\(\theta\)) = 0 and angle (A) = tan^{-1} (± cot\(\theta\)). _{Ans}