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Pair of straight lines

Let two straight lines be represented by
L1 = a1x + b1y + c1 = 0 and
L2 = a2x + b2y + c2 = 0
The combined equation for L1 and L2 is given by
(a1x + b1y + c1)(a2x + b2y + c2) = 0

The general equation of second degree in x and y is given by
S ≡ ax2 + 2hxy + by2 + 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 ≡ (l1x + m1y + n1)(l2x + m2y + n2)

The homogeneous equation of second degree in x and y,

H ≡ ax2 + 2hxy + by2 = 0 represents a pair of straight lines passing through the origin iff h2 ≥ ab.
We have:
i) if h2 = ab, the lines are coincident
ii) if h2 > ab, the two lines are real and different
iii) if h2 < ab, the lines are imaginary having real point of intersection.

Example: Does the equation x2 + xy + y2 = 0 represent a pair of lines ?
Sol: No.
Explanation: Comparing the given equation with ax2 + 2hxy + by2 = 0, we get
a = b = 1, h =
Now, h2 – ab = – 1 < 0, i.e., h2 < ab.
So the given equation does not represent a pair of lines.

Sum and product of the slopes

If H ≡ ax2 + 2hxy + by2 = 0 represents a pair of straight lines passing through origin,
then it can be written as
(a) H ≡ (l1x + m1y) (l2x + m2y)
or
(b) H ≡ (y – m1x) (y – m2x)

(a) If H = 0 represents a pair of straight lines
    l1x + m1y = 0 and l2x + m2y = 0, then
    H ≡ (l1x + m1y) (l2x + m2y) = 0
    ⇒ l1l2x2 + (l1m2 + l2m1) xy + m1m2y2 = 0
    Comparing with H = 0, we get
    l1l2 = a;   m1m2 = b;   l1m2 + l2m1 = 2h

(b) If H = 0 represents a pair of straight lines
    y = m1x and y = m2x, then
    H ≡ (y – m1x) (y – m2x) = 0
    ⇒ y2 – (m1 + m2) xy + m1m2x2 = 0
    ⇒ m1m2x2 – (m1 + m2) xy + y2 = 0
    Comparing with H = 0,
  i.e, we have,
    m1m2 = and m1 + m2 =
    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.

Angle between a pair of lines

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)

⇒

Proof

Note that the two lines are co-incident if h2 = ab
The two lines are perpendicular if cos θ = 0 or a + b = 0

Bisectors of angles

The equations of the bisectors of angles between the lines L1 ≡ a1x + b1y + c1 = 0 and L2 ≡ a2x + b2y + c2= 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
Pair of bisectors of angles

For the combined equation (of the pair of straight lines) H = 0, the equations of bisectors of angles is
h(x2 – y2) = (a – b)xy

Proof

Note:

(i) A pair of lines L1 L2 = 0 is said to be equally inclined to a line L = 0
if the lines L1 = 0, L2 = 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 L1L2 = 0 and L3L4 = 0 are said to be
equally inclined to each other if the angle between the lines L1 = 0 and L3 = 0 is
equal to the angle between the lines L2 = 0 and L4 = 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.

Formulae based on pair of lines

Product of the perpendiculars

The product of the perpendiculars from a point (α, β)
to the pair of lines ax2 + 2hxy + by2 = 0 is given by

Area of a Triangle

The area of the triangle formed by ax2 + 2hxy + by2 = 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

Proof

Conditions to represent pair of lines

If S ≡ ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of straight lines, then
i) abc + 2fgh – af2 – bg2 – ch2 = 0 and
ii) h2 ≥ ab, g2 ≥ ac and f2 ≥ bc
If any of the above two conditions fails, then S = 0 does not represent a pair of lines.

Proof

If abc + 2fgh – af 2 – bg2 – ch2 = 0, then
ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 need not represent a pair of lines.
For examples, consider x2 + 4xy + 4y2 + 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 – bg2 – ch2 = 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 h2 = ab
ii) perpendicular if a + b = 0
iii) intersecting if h2 > ab

Distance between parallel lines

If S = 0 represents a pair of parallel straight lines.
i) h2 = ab
ii) af2 = bg2
iii) the distance between them

Proof

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 h2 > 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.

Homogenisation

Generally, the locus of a second degree equation
ax2 + 2hxy + by2 + 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
ax2 + 2hxy + by2 + (2gx).1 + (2fy).1 + c.12 = 0
Replacing '1' by , we get a homogeneous equation in x, y as under.
ax2 + 2hxy + by2 + 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 x2 + coefficient of y2 = 0
i.e.,