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**Introduction:**

- Two lines
*l*_{1}, and*l*_{2}are said to form a pair of lines and in notation it is denoted by*l*_{1}U*l*_{2}. - If the point of intersection of the two lines is the origin then the pair is called a pair of lines passing through origin.
- If the point of intersection of the two lines is other than the origin then the pair of lines is called a pair of lines not passing through origin.

**Homogeneous Equation:**

- If the index of every variable in an expression is a non negative integer and the sum of the indices of the variable in each term of an expression is the same (say n), then the expression is called homogeneous expression of degree ‘n’ in these variables.
- Further equation obtained by equating above expression with zero is called homogeneous equation of degrre ‘n’ in these variables.
- Example : 4x² – 3xy + 2y² = 0

**Second Degree Homogeneous Equation in x and y:**

- Any equation in the form ax² + 2hxy + by² = 0 is called a second degree homogeneous equation in x and y. where a, h, b are real and not all zero.
**Examples :**- 4x² – 3xy + 2y² = 0 (a = 4, 2h = – 3, b = 2)
- 3x² – 2y² = 0 (a = 3, 2h = 0, b = – 2)
- xy = 0 (a = 0, 2h = 1, b = 0)

**Theorem – I:** ** **

- Show that the joint equation of a pair of straight lines passing through origin is second degree homogeneous equation in x and y. OR Prove that joint equation of a pair of lines passing through origin is in the form ax² + 2hxy + by² = 0
**Proof:**

Let the equations of two lines passing through the origin be a_{1}x + b_{1}y = 0 and a2x + b2y = 0

Now Their joint equation is

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

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

a_{1}a_{2}x² + a_{1}b_{2}xy + a_{2}b_{1}xy + b_{1}b_{2}y² = 0

a_{1}a_{2}x² + ( a_{1}b_{2} + a_{2}b_{1}) xy + b_{1}b_{2}y² = 0

Now put a_{1}a_{2} = a, ( a_{1}b_{2} + a_{2}b_{1}) = 2h and b_{1}b_{2} = b

Thus the combined equation is in the form ax² + 2hxy + by² = 0**.**

which is second degree homogeneous equation in x and y.

**Theorem – II:** ** **

- A homogeneous equation of degree two in x and y i.e. ax² + 2hxy + by² = 0 represents a pair of lines through the origin if h² – ab ≥ 0
**Proof:**

Let the second degree homogeneous equation in x and y be ax² + 2hxy + by² = 0

where a, h, b are real numbers and not all zero.

**Case I :**

Let a = 0 and b = 0, but h ≠ 0.

∴ hxy = 0 as h ≠ 0

x y = 0.

x = 0 or y = 0.

Separate equations are x = 0 and y = 0 which are the equations of the co-ordinate axes.

ax² + 2hxy + by² = 0 represents a pair of lines through origin when a = 0 and b = 0.

**Case II :**

Let a = 0,

Given equation becomes 2hxy + by² = 0

y(2hx + by) = 0

separate equation are y = 0 and 2hx + by = 0

The two factors are linear in x and y and do not contain constant terms. It represents a pair of lines.

The separate equations y = 0 and 2hx + by = 0 which are satisfied separately by the origin.

Hence ax² + 2hxy + by² = 0 represents a pair of lines through origin for a = 0.

The Same thing can be proved by taking b = 0.

**Case III :**

Let a ≠ 0

Now, ax² + 2hxy + by² = 0

Multiply by both side of equation by a

a²x² + 2ahxy + aby² = 0

a²x² + 2(ax)(hy) + aby² = 0

Add and substract (h y)² = h² y²

a²2x² + 2(ax)(hy) + h² y² + (aby² – h² y²) = 0

∴ (a x + h y)² – y² (h² – ab) = 0

(a x + h y)² – (√ h² – ab )² = 0

(ax + hy – √ h² – ab ) (ax + hy + √ h² – ab ) = 0

The two factors are linear in x and do not contain constant terms. and lines will be real if and only if h² – ab ≥ 0**.**

It is a pair of lines. Their separate equations are (ax + hy – √ h² – ab ) = 0 and (ax + hy + √ h² – ab ) = 0

which are separately satisfied by the origin.

Hence ax² + 2hxy + by² = 0 represents a pair of lines through origin, for a ¹ 0

- Combining the cases (I), (II) and (III) we get that every second degree homogeneous equation in x and y in general represents a pair of lines through origin.

**Nature of Lines:**

- If h² – ab ≥ 0, then the lines are real and distinct.
- If h² – ab = 0, then the lines are real and coincident.
- If h² – ab < 0, then the lines are not real and can’t be drawn.

**Theorem – III:**

- If m
_{1}and m_{2}are the slopes of the two lines represented by ax² + 2hxy + by² = 0, show that m_{1}+ m_{2}= -2h/b and m_{1}m_{2}= a/b. Deduce that lines are perpendicular if a + b = 0 and Lines are coincident if h² = ab. **Proof:**

m_{1}and m_{2} are the slopes of the two lines represented by ax² + 2hxy + by² = 0

Equation of first line is y = m_{1} x i.e. m_{1} x – y = 0

Equation of second line is y = m_{2}x i.e. m_{2} x – y = 0

Combined equation is

(m_{1}x – y)( m_{2}x – y) = 0

∴ m_{1}x ( m2x – y) – y ( m_{2}x – y)= 0

∴ m_{1} m_{2} x² – m_{1}xy – m_{2}xy – y²= 0

∴ m_{1} m_{2} x² – (m_{1} + m_{2}) xy – y²= 0

Also ax² + 2hxy + by² = 0 is combined equation of the two lines

The equation ax² + 2hxy + by² = 0 and m_{1} m_{2} x² – (m_{1} + m_{2}) xy – y²= 0 are identical.

Hence their corresponding coefficients are proportional.

**Case – I**

Lines are perpendicular if m_{1} m_{2} = – 1

i.e. if a/b= – 1

i.e. if a = – b

i.e. if a + b = 0

**Case – II**

Lines are coincident if m_{1} = m_{2}

i.e. if (m_{1} – m_{2}) = 0

i.e. (m_{1} – m_{2})²= 0

i.e. if (m_{1} + m_{2})² – 4 m1 m2 = 0.

Substitute for (m_{1} + m_{2} ) and m1 m2

i.e. if (-2h/b)² – 4(a/b) = 0

i.e. if 4h²/b² – 4a/b = 0

Multiplying both sides by b²

i.e. if 4h² – 4ab = 0

i.e. if h² – ab = 0

Science > Mathematics > Pair of Straight Lines > You are Here |

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Very helpful for me and this explanation is not in text.Thank you for this brief senerio.

Thank you so much for the help