## Pairs of straight lines

Subject: Optional Mathematics

#### Overview

General equation of the second degree :

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0.

##### Pairs of straight lines

General equation of second degree

The general equation of first degree in x and y always represents a straight line.

Let A1x + B1y + C1 = 0.......................(i)

and A2x + B2y + C2 = 0.......................(ii)

be the equations of two straight lines.

Now, Combining these equations we get,

(A1x + B1y + C1) (A2x + B2y + C2) = 0........................(iii)

The coordinates of any point, which satisfy the equation (iii), will also satisfy either equation (i) or equation (ii).

Similarly, the co-ordinates of any point, which satisfy any one of the equation (i) or (ii) will also satisfy the equation (iii).

Therefore, equation (iii) represents two separate straight lines (i) and (ii). In other words, equation (iii) represents a pair of the straight lines given by (i) and (ii). So, (iii) is the equation of a pair of lines.

Now, Expanding the left hand side of equation (iii) we get,

A1A2x2 + (A1B2 + A2B1)xy + B1B2y2 + (A1C2 + A2C1)x + (B1C2 + B2C1)y + C1C2 = 0

If we put A1A2 = a,A1B2 + A2B1 = 2h, B1B2 = b, A1C2 + A2C1 = 2g, B1C2 + B2C1 = 2f and C1C2 = c, then the above equation becomes,

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0..........................(iv)

This equation is called the general equation of the second degree in x and y. Thus we see that the equation of a pair of lines is a second degree equation. But the converse of this statement is not always true. It means every second degree equations in x and y may not represent a pair of straight lines. Equations of second degree will represent a pair of straight lines only if the left hand side can be resolved into two linear factors.

Consider an equation y2 - 3xy + 2x2 = 0. This equation is equivalent to (y - x) (y - 2x) = 0.

So, the equation y2 - 3xy + 2x2 = 0 represents the two straight lines y - x = 0 and y - 2x = 0.

Similarly, the equation xy = 0 represents the two straight lines x = 0 and y = 0.

Again,

Consider an equation x2 - 5x + 6 = 0 represents two straight lines x - 2 = 0 and x - 3 = 0.

And the equation x2 - y2 = 0 represents the two straight lines x + y = 0 and x - y = 0.

Condition that the general equation of second degree may represent a line pair

The general equation of second degree in x and y is

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

or, ax2 + (2hy + 2g)x + (by2 + 2fy + c) = 0

This is quadratic equation in x.

So, x = $\frac {-(2hy + 2g) ± \sqrt {(2hy + 2g)^2 - 4a(by^2 + 2fy + c)}}{2a}$

or, x = $\frac {-(hy + g) ± \sqrt {(hy + g)^2 - a(by^2 + 2fy + c)}}{a}$

These two equations will be linear if

(hy + g)2 - a (by2 + 2fy + c) is a perfect square.

or, h2y2 + 2ghy + g2 - aby2 - 2afy - ac is a perfect square.

i.e. (h2 - ab) y2 + (2gh - 2af) y + (g2 - ac) is a perfect square.

i.e. (2gh - 2af)2 - 4 (h2 - ab) (g2 - ac) = 0

i.e. g2h2 - 2ghaf + a2f2 - g2h2 + h2ac + abg2 - a2bc = 0

i.e. a (af2 + bg2 + ch2 - 2fgh - abc) = 0

i.e. abc + 2fgh - af2 - bg2 - ch2 = 0

Hence, the general equation of second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 may represent a line pair if abc + 2fgh - af2 - bg2 - ch2 = 0.

Any equation in the form of ax2 + bx + c = 0 is called quadratic equation in x.

Multiplying both sides of this equation by 4a we get,

4a (ax2 + bx + c) = 0

or, 4a2x2 + 4abx + 4ac = 0

or, (2ax)2 + 2 . 2ax . b + b2 - b2 + 4ac = 0

or, (2ax + b)2 = b2 - 4ac

or, (2ax + b)2 = ($\sqrt {b^2 - 4ac}$)2

or, 2ax + b = ± $\sqrt {b^2 - 4ac}$

or, 2ax = - b ± $\sqrt {b^2 - 4ac}$

or, x = $\frac {- b ± \sqrt {b^2 - 4ac}}{2a}$

Let α = $\frac {- b + \sqrt {b^2 - 4ac}}{2a}$ and β = $\frac {- b - \sqrt {b^2 - 4ac}}{2a}$.

Then α and β are called roots of the quadratic equation ax2 + bx + c = 0.

Now,

\begin{align*} \text{Sum of the roots (α + β)} &= \frac {- b + \sqrt {b^2 - 4ac}}{2a} + \frac {- b - \sqrt {b^2 - 4ac}}{2a}\\ &= - \frac ba\\ &= -\frac {coefficient\;of\;x}{coefficient\;of\;x^2}\\ \end{align*}

\begin{align*} \text{Sum of the roots (αβ)} &= (\frac {- b + \sqrt {b^2 - 4ac}}{2a}) (\frac {- b - \sqrt {b^2 - 4ac}}{2a})\\ &= \frac ca\\ &= \frac {constant\;term}{coefficient\;of\;x^2}\\ \end{align*}

Homogeneous equation of second degree

An equation in x and y in which the sum of the power of x and y in every term is the same, is called homogenous equation. If this sum is two, then the equation is called a homogenous equation of second degree. The equation ax2 + 2hxy + by2 = 0 is the general homogenous equation of the second degree.

A homogeneous equation of the second degree represents a pair of straight lines which pass through the origin.

Proof: Consider the homogenous equation of the second degree.

ax2 + 2hxy + by2 = 0...........................(i)

or, by2 + 2hxy + ax2 = 0

If b ≠ 0, the equation can be written as y2 + $\frac {2h}b$xy + $\frac ab$x2 = 0

or, ($\frac yx$)2 + $\frac {2h}b$($\frac yx$) + $\frac ab$ = 0

This is quadratic equation in $\frac yx$. So it has two roots. Let these roots be m1 and m2.

Then,

$\frac yx$ = m1 and $\frac yx$ = m2

or, y = m1x and y = m2x.

These two equations are the equations of straight lines passing through the origin.

If b = 0, then equation (i) becomes

ax2 + 2hxy = 0

or, x(ax + 2hy) = 0 which represents two straight lines x = 0 and ax + 2hy = 0.

These two lines pass through the origin.

Hence, the homogenous equation of second degree always represents a pair of straight lines passing through the origin.

Angle between the line pair represented by ax2 + 2hxy + by2 = 0

Homogeneous equation of second degree is:

ax2 + 2hxy + by2 = 0

or, by2 + 2hxy + ax2 = 0

or, y2 + $\frac{2h}{b}$($\frac yx$) + $\frac ab$ = 0.........................(i)

This is quadratic equation in $\frac yx$. So, it has two roots. Let these two roots be m1 and m2.

Then,

$\frac yx$ = m1 and $\frac yx$ = m2

or, y = m1x and y = m2x which are two seperate equations represente by the given equation ax2 + 2hxy + by2 = 0.

Now,

m1 + m2 = $\frac {-2h}b$ and m1m2 = $\frac ab$

Now,

\begin{align*} m_1- m_2 &= \sqrt {(m_1 + m_2)^2 - 4m_1m_2}\\ &= \sqrt {\frac{4h^2}{b^2} - 4\frac ab}\\ &= \sqrt {\frac {4h^2 - 4ab}{b^2}}\\ &= \frac 2b \sqrt {h^2 - ab}\\ \end{align*}

Let $\theta$ be the angle between the lines y = m1x and y = m2x. Then,

\begin{align*} tan\theta &= ± \frac {m_1 - m_2}{1 + m_1m_2}\\ &= ± \frac {\frac 2b \sqrt {h^2 - ab}}{1 + \frac ab}\\ &= ± \frac {2\sqrt {h^2 - ab}}{a + b}\\ \end{align*}

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

Second Method:

We have ax2 + 2hxy + by2 = 0............................................(i)

Let the seperate equations represented by this equation be y = m1x and y = m2x.

Now,

The combined equation of theses equation is:

(y - m1x) (y - m2x) = 0

or, y2 - (m1 + m2)xy + m1m2x2 = 0

or, m1m2x2- (m1 + m2)xy + y2 = 0.......................................(ii)

Comparing (i) and (ii) we have,

$\frac {m_1m_2}{a}$ = $\frac {- (m_1 + m_2)}{2h}$ = $\frac 1b$

Taking 1st and last, m1m2 = $\frac ab$

Taking 2nd and last, m1 + m2 = -$\frac {2h}b$

Now,

\begin{align*} m_1 - m_2 &= \sqrt {(m_1 + m_2)^2 - 4m_1m_2}\\ &= \sqrt {\frac {4h^2}{b^2} - 4\frac ab}\\ &= \sqrt {\frac {4h^2 - 4ab}{b^2}}\\ &= \frac 2b \sqrt {h^2 - ab}\\ \end{align*}

Let $\theta$ be the angle between the lines y = m1x and y = m2x.

Then,

\begin{align*} tan\theta &=± \frac {m_1 - m_2}{1 + m_1m_2}\\ &= ± \frac {\frac 2b \sqrt {h^2 - ab}}{1 + \frac ab}\\ &= ± \frac {2\sqrt {h^2 - ab}}{a + b}\\ \end{align*}

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

Condition that the straight lines given by the equation ax2 + 2hxy + by2 = 0 may be (1) Perpendicular and (2) Coincident.

1. If a + b = 0 the value of tan$\theta$ is∞ and hence $\theta$ is 90°. Hence two straight lines represented by the equation ax2 + 2hxy + by2 = 0 are perpendicular to each other if coefficient of x2 + coefficient of y2 = 0.
For example: the equations x2 - y2 = 0 and 6x2 + 11xy - 6y2= 0 both represent pairs of straight lines at right angles.
Similarly, whatever be the value of h, the equation x2 + 2hxy - y2 = 0 represents a pair of straight lines at right angles.
2. If h2 = ab, the value of tan$\theta$ is zero and hence $\theta$ is zero. But both the straight lines represented by ax2 + 2hxy + by2 = 0 pass through the origin. So, the two lines are coincident.
Hence, two straight lines represented by the equation ax2 + 2hxy + by2 = 0 are coincident ifh2 = ab.
This may be seen directly from the original equation ax2 + 2hxy + by2 = 0. If h2 = ab i.e. h = $\sqrt {ab}$, then the original equation becomes,
ax2 + 2$\sqrt {ab}$xy + by2 = 0
or, ($\sqrt a$x)2 + 2 $\sqrt a$ $\sqrt b$ xy + ($\sqrt b$y)2 = 0
or, ($\sqrt a$x + $\sqrt b$y)2 = 0 which gives two coincident straight lines
i.e.$\sqrt a$x + $\sqrt b$y = 0 and$\sqrt a$x + $\sqrt b$y = 0.

If the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of lines, then ax2 + 2hxy + by2 = 0 represents a pair of lines through the origin parallel to the above pair.

Proof:

Let ax2+ 2hxy + by2 + 2gx + 2fy + c = 0............................................................(i)

represent a pair of straight lines. Then the left-hand side can be resolved into two linear factors. Let these factors be A1x + B1y + C1and A2x + B2y + C2 = 0.

Now

Combining equation of these equations is:
(A1x + B1y + C1) (A2x + B2y + C2) = 0

or, A1A2x2 + (A1B2 + A2B1)xy + B1B2y2 + (A1C2 + A2C1)x + (B1C2 + B2C1)y + C1C2 = 0..............................................(ii)

Equating the coefficients of like terms in equation (i) and (ii) we have,

A1A2 = a2, A1B2 + A2B1 = 2h, B1B2 = b2, A1C2 + A2C1 = 2g, B1C2 + B2C1 = 2f, C1C2 = c2

Now,

Equation of the straight line parallel to A1x + B1y + C1 = 0 and passing through the origin is:

A1x+ B1y = 0.........................................(iii)

Again,

Equation of straight line parallel to A2x + B2y + C2 = 0 and passing through the origin is:

A2x + B2y = 0.........................................(iv)

Now,

Combining (iii) and (iv) we have,

(A1x + B1y) (A2x + B2y) = 0

or, A1A2x2 + (A1B2 + A2B1)xy + B1B2y2 = 0

or, ax2 + 2hxy + by2 = 0. This completes the proof.

Note:Angles between the line pair ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 are same as the angles between the pair ax2 + 2hxy + by2 = 0.

The angles are given by,

tan-1 (± $\frac {2\sqrt {h^2 - ab}}{a + b}$)

Hence, the two lines will be perpendicular to each other if a + b = 0 and they will beparallel if h2= ab.

##### Things to remember

General equation of the second degree :

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0.

• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### Pair Of Straight Lines Example -1 l Maths Geometry

Here,

3x2 - 5xy - 2y2 - x + 2y = 0

or, 3x2 - 6xy + xy - 2y2 - 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 = 0Ans

Here,

x2 - 3xy + 2y2 = 0

or, x2 - 2xy - xy + 2y2 = 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:6x2 + 11xy - 6y2 = 0..................(1)

The homogeneous equation of second is: ax2 + 2hxy + by2 = 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 equation6x2 + 11xy - 6y2 = 0 are perpendicular to each other. Proved

Here,

Given equation is:x2 - 4xy + 4y2 = 0......................(1)

The homogenous equation of second degree is:

ax2+ 2hxy + by2 = 0...........................(2)

Comparing (1) and (2)

a = 1

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

b = 4

Now,

h2 = ab

or, (-2)2 = 1× 4

∴ 4 = 4

Hence, h2 = ab is satisfied by the given equation so the equationx2 - 4xy + 4y2 = 0 are coincident to each other. Proved

Here,

ax2 + 2hxy + by2 = 0

When: h2 = ab then straight lines are coincident each other.

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

Given eqn of line are:

x = 2y

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

2x = y

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

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

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

or, 2x2- xy - 4xy + 2y2 = 0

or, 2x2 - 5xy + 2y2 = 0

∴ The required eqn is: 2x2 - 5xy + 2y2 = 0 Ans

Here,

Given eqn is: y2 = x2

or, x2 - y2 = 0

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

either: x + y = 0

Or, x - y = 0

Slope of eqn x + y = 0 is: m1 = -$\frac 11$ = -1

Slope of eqn x -y = 0 is: m2 = -$\frac 1{-1}$ = 1

m1× m2 = -1× 1 = -1

Hence, they are perpendicular to each other. Proved

Here,

Given eqn is:

6x2 - 5xy - 6y2 = 0...............................(1)

The eqnof homogenous is:

ax2 + 2hxy + by2 = 0..........................(2)

Comparing eqn (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 eqn is:

x2 - 2xy sec$\alpha$ + y2 = 0...........................(1)

The homogenous eqn is:

ax2 + 2hxy + by2 = 0..............................(2)

Comparing eqn (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 eqn are:

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

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

Combined eqn of (1) and (2)

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

or, x2 cos$\alpha$.sin$\alpha$ +xy cos2$\alpha$ + xy sin2$\alpha$ + y2 sin$\alpha$.cos$\alpha$ = 0

or, x2 cos$\alpha$.sin$\alpha$ + y2 sin$\alpha$.cos$\alpha$ + xy (cos2$\alpha$ + sin2$\alpha$) = 0

or, sin$\alpha$.cos$\alpha$ (x2 + y2) + xy× 1 = 0

∴(x2 + y2) sin$\alpha$.cos$\alpha$ + xy = 0 Ans

Given equation is:

x2 + 4xy + y2 = 0.............................(1)

The homogeneous equation of second degree is:

ax2 + 2hxy + by2 = 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:

3x2 + 2y2 - 5xy = 0

The homogeneous equation of second degree is:

ax2 + 2hxy + by2 = 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:

12x2 - 23xy + 5y2 = 0.....................(1)

The homogeneous equation of second degree is:

ax2 + 2hxy + by2 = 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:

ax2 + 2hxy +by2 = 0................................(1)

The given equation is:

2x2 - 5xy + 2y2 = 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:

6x2 + 5xy - 3x - 2y - 6y2 = 0

or, 6x2 + 9xy - 4xy - 6y2 - 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), m1 = -$\frac 23$

Slope of equation (2), m2 = $\frac 32$

Again,

m1× m2 = $\frac {-2}3$× $\frac 32$ = -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:

a1x + b1y = 0..............................(1)

a2x + b2y = 0..............................(2)

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

(a1x + b1y) (a2x + b2y) = 0

or, a1a2x2 + a1b2xy + a2b1xy + b1b2y2 = 0

or,a1a2x2 + (a1b2+ a2b1)xy + b1b2y2 = 0

Let:

a1a2 = a

b1b2 = b

(a1b2+ a2b1) = 2h

Now,

ax2 + 2hxy + by2 = 0

Thus,ax2 + 2hxy + by2 = 0 is homogenous equation of second degree. Ans

Here,

The given equation is: ax2 + 2hxy + by2 = 0 if a≠ 0, then:

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

a2x2 + 2ahxy + aby2 = 0

or (ax)2 + 2 (ax) (hy) + (hy)2 - h2y2 + aby2 = 0

or, (ax + hy)2 - (h2 - ab) y2 = 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 ax2+ 2hxy + by2 = 0 will be

2hxy + by2 = 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: ax2 + 2hxy + by2 = 0 always represents a pair of straight lines through the origin. Proved

Here,

Given equationax2 + 2hxy + by2= 0 represents two straight lines that passes through the origin.

Let: y = m1x and y = m2x are the two lines.

y -m1x = 0............................(1)

y - m2x = 0............................(2)

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

(y - m1x) (y - m2x) = 0

or, m1m2x2 - m1xy - m2xy + y2 = 0

or, m1m2x2 - (m1 +m2)xy + y2 = 0...................(3)

Given equationax2 + 2hxy + by2= 0

$\frac ab$x2 + $\frac {2h}b$ xy + y2 = 0...................(4)

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

m1 + m2 = -$\frac {2h}b$ and m1m2 = $\frac ab$

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,

x2 - 5xy + 4y2 = 0

or, x2 - xy - 4xy + 4y2 = 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 eqn (1) changes into parallel form is:

x - y + k1 = 0......................(3)

The point (1, 1) passes through eqn (1)

1 - 1 + k1 = 0

∴ k1 = 0

Putting the value of k1 in eqn (3)

x - y + 0 = 0

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

The eqn(2) changes in parallel form is:

x - 4y + k2 = 0..................(5)

The point (1, 1) passes through eqn (5)

1 - 4× 1 + k2 = 0

or, -3 + k2 = 0

∴ k2 = 3

Putting the value of k2 in eqn (5)

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

The eqn of pair of line is:

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

or, x2 - 4xy + 3x - xy - 4y2 - 3y = 0

∴x2 - 5xy - 4y2 + 3x - 3y = 0Ans

Here,

Given equation is:

x2 - xy - 2y2 = 0

or, x2 - 2xy + xy - 2y2 = 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 eqn (1) changes in perpendicular form is:

-2x - y + k1 = 0

2x + y - k1 = 0..........................(3)

The eqn (3) passes through origin (0, 0):

2× 0 + 0 - k1 = 0

∴ k1 = 0

Putting the value of k1 in eqn (3)

2x + y - 0 = 0

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

The eqn (2) change in perpendicular form is:

x - y + k2 = 0........................(5)

The eqn (5) passes through origin (0, 0)

0 - 0 + k2 = 0

∴ k2 = 0

Putting the value of k2 in eqn (5)

x - y + 0 = 0

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

The equation of the pairs of lines is:

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

or, 2x2 - 2xy + xy - y2 = 0

∴2x2 - xy - y2 = 0Ans

Given eqn of line is:

3x2 - 8xy + 5y2= 0

or, 3x2 - 3xy - 5xy + 5y2= 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 eqn (1) changes in perpendicular form is:

x + y + k1 = 0...........................(3)

The point (2, 3) passes through eqn (3)

2 + 3 + k1 = 0

or, 5 + k1 = 0

∴ k1 = -5

Putting the value of k1 in eqn (3)

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

The eqn (2) change in perpendicular form is:

5x + 3y + k2 = 0...................(5)

The point (2, 3) passes through eqn (5)

5× 2 + 3× 3 + k2 = 0

or, 10 + 9 + k2 = 0

or, 19 + k2 = 0

∴ k2 = - 19

Putting the value of k2 in eqn (5)

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

The eqn of pairs of lines is:

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

or, 5x2 + 3xy - 19x + 5xy + 3y2 - 19y - 25x - 15y + 95 = 0

∴ 5x2 + 8xy + 3y2 - 44x - 34y + 95 = 0Ans

Here,

x2 - xy - 2y2 = 0

or, x2 - 2xy + xy - 2y2 = 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 eqn (1) change in perpendicular form is:

2x + y + k1 = 0.......................(3)

The point (3, -1) passes through eqn (3)

2× 3 - 1 + k1 = 0

or, 6 - 1 + k1 = 0

∴ k1= -5

Putting the value of k1 in eqn (3)

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

The eqn (2) change in perpendicular form is:

x - y + k2 = 0.........................(5)

The point (3, -1) passes througheqn (5)

3 + 1 + k2 = 0

∴ k2 = -4

Putting the value of k2 in eqn (5)

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

The eqn of pair of lines is:

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

or, 2x2 - 2xy - 8x + xy - y2 - 4y - 5x + 5y + 20 = 0

∴ 2x2 - xy- y2 - 13x + y + 20 = 0Ans

Here,

2x2 + 5xy + 3y2 = 0

or, 2x2 + 3xy + 2xy + 3y2 = 0

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

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

Either: 2x + 3y = 0

Or: x + y = 0

Given eqn is: 2x2 + 5xy + 3y2 = 0......................(1)

Homogenous eqn is: ax2 + 2hxy + by2 = 0...................(2)

Comparing eqn (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 eqn are: 2x + 3y = 0 and x + y = 0 and angle between them are: 11.31° and 168.69°. Ans

Given eqn is:x2 + 2xy sec$\theta$ + y2 = 0........................(1)

Homogenous eqn is: ax2 + 2hxy + by2 = 0..........(2)

Comparingeqn(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 eqn is:

y2 + 2xy sec$\theta$ + x2 = 0..........................(3)

ax2 + bx + c = 0.............................(4)

Comparing (3) and (4)

a = 1

b = 2x sec$\theta$

c = x2

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 eqn are:y = -xsec$\theta$ + xtan$\theta$ andy = -xsec$\theta$ - xtan$\theta$ Ans

Given equation is:x2 - 2xy cosec$\theta$ + y2 = 0.....................(1)

Homogenous equation is: ax2 + 2hxy + by2 = 0......................(2)

Comparing eqn (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:

2x2 + 7xy + 3y2 = 0

or, 2x2 + 6xy + xy + 3y2 = 0

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

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

The two equation represented by2x2 + 7xy + 3y2 = 0 are:

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

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

Now,

Slope of equation (1), m1 = -$\frac {x-coefficient}{y-coefficient}$ = -2

Slope of equation (2),m2 =-$\frac {x-coefficient}{y-coefficient}$ = -$\frac 13$

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:

2x2 + 3xy - 2y2 = 0

or, 2x2 + 4xy - xy - 2y2 = 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,

2x2 + 3xy - 2y2 = 0............................(1)

The homogenous equation of second degree is:

ax2 + 2hxy + by2 = 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:

2x2- 3xy + y2 = 0......................(1)

The homogenous equation of second degree is:

ax2 + 2hxy + by2 = 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,

2x2 - 3xy + y2 = 0

or, 2x2 - 2xy - xy + y2 = 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,

6x2 + 5xy - 3x + 2y - 6y2 = 0

or, 6x2 + 5xy - 6y2 - 3x + 2y = 0

or, 6x2 + 9xy - 4xy - 6y2 - 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), m1 = $\frac {-3}{-2}$ = $\frac 32$

Slope of equation (2),m2 = -$\frac 23$

∴ m1× m2 = $\frac 32$× $\frac {-2}3$ = -1

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

The given equation is:

x2 - 2xy cosec$\theta$ + y2 = 0

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

ax2 + bx + c = 0......................(2)

Comparing (1) and (2)

a = 1

b = -2x cosec$\theta$

c = x2

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 x2 -2xy cosec$\theta$ + y2 = 0 and ax2 + 2hxy + by2 = 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