Let two straight lines be represented by

L_{1} = a_{1}x + b_{1}y + c_{1} = 0 and

L_{2} = a_{2}x + b_{2}y + c_{2} = 0

The combined equation for L_{1} and L_{2} is given by

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

**The general equation of second degree in x and y** is given by

S ≡ ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0

If the locus of the equation S = 0 contains a straight line L, then

S can be written as the product of two linear factors in x and y.

And hence the equation S = 0 represents a pair of straight lines.

We can write S ≡ (l_{1}x + m_{1}y + n_{1})(l_{2}x + m_{2}y + n_{2})

**The homogeneous equation of second degree in x and y,**

H ≡ ax^{2} + 2hxy + by^{2} = 0 represents a pair of straight lines
passing through the origin iff h^{2} ≥ ab.

We have:

i) if h^{2} = ab, the lines are coincident

ii) if h^{2} > ab, the two lines are real and different

iii) if h^{2} < ab, the lines are imaginary having real point of intersection.

**Example: ** Does the equation x^{2} + xy + y^{2} = 0 represent a pair of lines ?

**Sol:** No.

**Explanation: ** Comparing the given equation with ax^{2} + 2hxy + by^{2} = 0, we get

a = b = 1, h =

Now, h^{2} – ab = – 1 < 0, i.e., h^{2} < ab.

So the given equation does not represent a pair of lines.

If H ≡ ax^{2} + 2hxy + by^{2} = 0 represents a pair of straight lines passing through origin,

then it can be written as

(a) H ≡ (l_{1}x + m_{1}y) (l_{2}x + m_{2}y)

or

(b) H ≡ (y – m_{1}x) (y – m_{2}x)

(a) If H = 0 represents a pair of straight lines

l_{1}x + m_{1}y = 0 and l_{2}x + m_{2}y = 0, then

H ≡ (l_{1}x + m_{1}y) (l_{2}x + m_{2}y) = 0

⇒ l_{1}l_{2}x^{2} + (l_{1}m_{2} + l_{2}m_{1}) xy + m_{1}m_{2}y^{2} = 0

Comparing with H = 0, we get

l_{1}l_{2} = a; m_{1}m_{2} = b; l_{1}m_{2} + l_{2}m_{1} = 2h

(b) If H = 0 represents a pair of straight lines

y = m_{1}x and y = m_{2}x, then

H ≡ (y – m_{1}x) (y – m_{2}x) = 0

⇒ y^{2} – (m_{1} + m_{2}) xy + m_{1}m_{2}x^{2} = 0

⇒ m_{1}m_{2}x^{2} – (m_{1} + m_{2}) xy + y^{2} = 0

Comparing with H = 0,

i.e, we have,

m_{1}m_{2} = and m_{1} + m_{2} =

i.e, if H = 0 represents a pair of straight lines, then

(i) the sum of the slopes of the lines is

(ii) the product of the slopes of the lines is

**Note: ** The two lines represented by H = 0 (b ≠ 0) are non-vertical to each other.

The angle(θ) between two lines represented by H = 0 is given by

or

or

Actually, cos θ gives the acute angle between the lines if a + b > 0

and the obtuse angle if a + b < 0.

So the acute angle is given by (note the modulus)

⇒

Note that the two lines are

**co-incident**if

**h**

^{2}= abThe two lines are

**perpendicular**if

**cos θ = 0 or a + b = 0**

The equations of the ** bisectors of angles** between the lines L_{1} ≡ a_{1}x + b_{1}y + c_{1} = 0 and L_{2} ≡ a_{2}x + b_{2}y + c_{2}= 0 are given by

**Note:**

(i) If the angle between one of the given lines and one of the angle bisectors is less than , then that angle bisector is acute angle bisector and the other is obtuse angle bisector.

(ii) The bisectors of the angles between any two intersecting lines are always perpendicular to each other.

(iii) Conditions for equation of bisectors of angles

Condition | Equation of the acute angle bisector |
Equation of the obtuse angle bisector |
---|---|---|

For the combined equation (of the pair of straight lines) H = 0, the equations of bisectors of angles is

**h(x ^{2} – y^{2}) = (a – b)xy**

**Note:**

(i) A pair of lines L_{1} L_{2} = 0 is said to be equally inclined to a line L = 0

if the lines L_{1} = 0, L_{2} = 0 subtend the same angle with the line L = 0. Refer adjacent fig. (i)

(ii) Every pair of lines is equally inclined to either of its angle bisectors.

(iii) A pair of lines is equally inclined to a line L = 0

iff L = 0 is parallel to one of the angle bisectors.

(iv) Two pairs of lines L_{1}L_{2} = 0 and L_{3}L_{4} = 0 are said to be

equally inclined to each other if the angle between the lines L_{1} = 0 and L_{3} = 0 is

equal to the angle between the lines L_{2} = 0 and L_{4} = 0. Refer adjacent fig. (ii)

(v) Two pairs of concurrent lines are equally inclined to each other

iff both the pairs of lines have the same angle bisectors.

## Product of the perpendiculars

The product of the perpendiculars from a point (α, β)

to the pair of lines ax^{2} + 2hxy + by^{2} = 0 is given by

## Area of a Triangle

The area of the triangle formed by ax^{2} + 2hxy + by^{2} = 0

and lx + my + n = 0 is given by

## Area of equilateral triangle

The line ax + by + c = 0 and the pair of lines (ax + by)^{2} – 3(bx – ay)^{2} = 0 form an equilateral triangle.

Its area is given by sq.units

If S ≡ ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 represents a pair of straight lines, then

i) abc + 2fgh – af^{2} – bg^{2} – ch^{2} = 0 and

ii) h^{2} ≥ ab, g^{2} ≥ ac and f^{2} ≥ bc

If any of the above two conditions fails, then S = 0 does not represent a pair of lines.

If abc + 2fgh – af ^{2} – bg^{2} – ch^{2} = 0, then

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 need not represent a pair of lines.

For examples, consider x^{2} + 4xy + 4y^{2} + 2 = 0.

It does not represent a pair of lines.

In this equation a = 1, b = 4, c = 2, h = 2, g = 0, f = 0

∴ abc + 2fgh – af ^{2} – bg^{2} – ch^{2} =
8 + 0 – 0 – 0 – 8 = 0

f ^{2} – bc = 0 – (4)(2) = – 8 < 0

∴ The given equation is not satisfying f ^{2} ≥ bc.

If S = 0 represents a pair of straight lines, then H = 0 represents a second pair of lines passing through the origin and parallel to the first pair.

The lines represented by S = 0 are

i) parallel if h^{2} = ab

ii) perpendicular if a + b = 0

iii) intersecting if h^{2} > ab

If S = 0 represents a pair of parallel straight lines.

i) h^{2} = ab

ii) af^{2} = bg^{2}

iii) the distance between them

## Point of intersection

Point of intersection of a pair of straight lines represented by S = 0 is given by

If the point of intersection is (0, 0) i.e, origin,

then g = f = c = 0

If h^{2} > ab, then the point of intersection of the pair of lines S = 0 satisfies the 3 equations

ax + hy + g = 0

hx + by + f = 0

gx + fy + c = 0

∴ is a condition for S = 0 to represent a pair of straight lines.

Generally, the locus of a second degree equation

ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c = 0 ------ (i)

is called a second degree curve.

If the graph of the equation (i) contains more than one point,

then (i) can be either a pair of straight lines or a circle or a conic.

Any such curve has at most two points of intersection with a given line
if the given line is not contained in the locus (i).

In this case, we find the combined equation of the pair of lines joining the origin to the points of intersection of (i) with a given straight line as shown in adjacent figures.

Let the straight line be lx + my + n = 0 (n ≠ 0) ------ (ii)

If (ii) intersects (i) in two distinct points A and B,

then the equation of can be taken as

= 1

We homogenise (i) with the help of (ii) as follows:

We write (i) as

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

Replacing '1' by , we get a homogeneous equation in x, y as under.

ax^{2} + 2hxy + by^{2} + 2gx +
2fy + c = 0

⇒ --- (iii)

∴ Equation (iii) represents the combined equation of
and

Note that, the method of finding (iii) is known as "homogenisation" of (i) with the help of (ii).

In this case, the condition for perpendicularity is

coefficient of x^{2} + coefficient of y^{2} = 0

i.e.,