Inter 1st Year Maths 1B Pair of Straight Lines Formulas

Use these Inter 1st Year Maths 1B Formulas PDF Chapter 4 Pair of Straight Lines to solve questions creatively.

Intermediate 1st Year Maths 1B Pair of Straight Lines Formulas

→ If ax² + 2hxy + by² = 0 represents a pair of lines, then the sum of the slopes of lines is \(-\frac{2 h}{b}\) and the product of the slopes of lines is \(\frac{a}{b}\).

→ If ‘θ’ is an angle between the lines represented by ax² + 2hxy + by² = 0
then cos θ = \(\frac{a+b}{\sqrt{(a-b)^{2}+4 h^{2}}}\)
tan θ = \(\frac{2 \sqrt{h^{2}-a b}}{a+b}\)
If ‘θ’ is accute, cos θ = \(\frac{|a+b|}{\sqrt{(a-b)^{2}+4 h^{2}}}\); tan θ = \(\frac{2 \sqrt{h^{2}-a b}}{|a+b|}\)

→ If h² = ab, then ax² + 2hxy + by² = 0 represents coincident or parallel lines.

→ ax² + 2hxy + by² = 0 represents a pair of ⊥^lr lines ⇒ a+ b = 0 i.e., coeft. of x² + coeff. of y² = 0.

→ The equation of pair of lines passing through origin and perpendicular to ax² + 2hxy + by² = 0 is bx² – 2hxy + ay² = 0.

→ The equation of pair of lines passing through (x₁, y₁) and perpendicular to ax² + 2hxy + by² = 0 isb (x – x₁)² – 2h(x – x₁) (y – y₁) + a(y – y₁)² = 0.

→ The equation of pair of lines passing through (x₁, y₁) and parallel to ax² + 2hxy + by² = 0 is a(x-x,)² + 2h(x-x1)(y-y1) + b(y – y₁)² = 0.

→ The equations of bisectors of angles between the lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0, is \(\frac{a_{1} x+b_{1} y+c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\pm \frac{\left(a_{2} x+b_{2} y+c_{2}\right)}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\)

→ The equation to the pair of bisectors of angles between the pair of lines ax² + 2hxy + by² = 0 is h(x² – y²) = (a – b)xy.

→ The product of the perpendicular is from (α, β) to the pair of lines ax² + 2hxy + by² = 0 is \(\frac{\left|a \alpha^{2}+2 h \alpha \beta+b \beta^{2}\right|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)

→ The area of the triangle formed by ax² + 2hxy + by² = 0 and lx + my + n = 0 is \(\frac{n^{2} \sqrt{h^{2}-a b}}{\left|a m^{2}-2 h / m+b\right|^{2} \mid}\)

→ The line ax + by + c = 0 and pair of lines (ax + by)² – 3(bx – ay)² =0 form an equilateral triangle and the area is \(\frac{c^{2}}{\sqrt{3}\left(a^{2}+b^{2}\right)}\) sq.units

→ If ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of lines, then

• abc + 2fgh – af² – bg² – ch² = 0
• h² ≥ ab
• g² ≥ ac
• f² ≥ be

→ If ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of lines and h² > ab, then the point of intersection of the lines is \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)

→ If ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of parallel lines then h² = ab and af² = bg²
The distance between the parallel lines = \(2 \sqrt{\frac{g^{2}-a c}{a(a+b)}}=2 \sqrt{\frac{f^{2}-b c}{b(a+b)}}\)

Pair of Straight Lines:
Let L₁ = 0, L₂ = 0 be the equations of two straight lines. If P(x₁, y₁) is a point on L₁ then it satisfies the equation L₁ = 0. Similarly, if P(x₁, y₁) is a point on L₂ = 0 then it satisfies the equation.

If P(x₁, y₁) lies on L₁ or L₂, then P(x₁,y₁) satisfies the equation L₁L₂= 0.
L₁L₂ = 0 represents the pair of straight lines L₁ = 0 and L₂ = 0 and the joint equation of L₁ = 0 and L₂ = 0 is given by L₁ L₂= 0. ……………(1)
On expanding equation (1) we get and equation of the form ax² + 2hxy + by² + 2 gx + 2 fy + c = 0 which is a second degree (non – homogeneous) equation in x and y.
Definition: If a, b, h are not all zero,then ax² + 2hxy + by² = 0 is the general form of a second degree homogeneous equation in x and y.
Definition: If a, b, h are not all zer, then ax² + 2hxy + by² + 2gx + 2 fy + c = 0 is the general form of a second degree non – homogeneous equation in x and y.

Theorem:
If a, b, h are not all zero and h² ≥ ab then ax² + 2hxy + by² = 0 represents a pair of straight lines passing through the origin.
Proof:
Case (i) : Suppose a = 0.
Given equation ax² + 2hxy + by² = 0 reduces to 2hxy + by² = 0 ^ y(2hx + by) = 0 .
Given equation represents two straight lines y = 0 ………..(1) and 2hx + by = 0 ………(2) which pass through the origin.
Case (ii): Suppose a ≠ 0.
Given equation ax² + 2hxy + by² = 0
⇒ a²x² + 2ahxy + aby² = 0
⇒ (ax)² + 2(ax)(hy) + (hy)² – (h² – ab)y² = 0
⇒ (ax + hy)² – (y\(\sqrt{h^{2}-a b}\))² = 0
[ax + y (h + \(\sqrt{h^{2}-a b}\))][ax + y (h – \(\sqrt{h^{2}-a b}\)] = 0

∴ Given equation represents the two lines
ax + hy + y\(\sqrt{h^{2}-a b}\) = 0, ax + hy – y\(\sqrt{h^{2}-a b}\) = 0 which pass through the origin.

Note 1:

• If h² > ab , the two lines are distinct.
• If h² = ab , the two lines are coincident.
• If h² < ab , the two lines are not real but intersect at a real point (the origin).
• If the two lines represented by ax² + 2hxy + by² = 0 are taken as l₁x + m₁y = 0 and l₂x + m₂y = 0 then
  ax² + 2kxy + by² = (4 x + m₁y) (x + m₂y) = 2 x² + (l₁m₂ + l₂m₁) xy + m₁m₂ y²
• Equating the coefficients of x², xy and y² on both sides, we get l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.

Theorem:
If ax² + 2hxy + by² = 0 represent a pair of straight lines, then the sum of slopes of lines is \(\frac{-2 h}{b}\) product of the slopes is \(\frac{a}{b}\).
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ………….(1) and l₂x + m₂y = 0 ………….(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.
Slopes of the lines (1) and (2) are –\(\frac{l_{1}}{m_{1}}\) and \(-\frac{l_{2}}{m_{2}}\).
sum of the slopes = \(\frac{-l_{1}}{m_{1}}+\frac{-l_{2}}{m_{2}}=-\frac{l_{1} m_{2}+l_{2} m_{1}}{m_{1} m_{2}}=-\frac{2 h}{b}\)
Product of the slopes = \(\left(\frac{-l_{1}}{m_{1}}\right)\left(-\frac{l_{2}}{m_{2}}\right)=\frac{l_{1} l_{2}}{m_{1} m_{2}}=\frac{a}{b}\)

Angle Between A Pair of Lines:
Theorem :
If θ is the angle between the lines represented by ax² + 2hxy + by² = 0, then cos θ = ±\(\frac{a+b}{\sqrt{(a-b)^{2}+4 h^{2}}}\)
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁ x + m₁ y = 0 ………..(1) and l₂x + m₂ y = 0 …………..(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.

Let θ be the angle between the lines (1) and (2). Then cos θ = ±\(\frac{l_{1} l_{2}+m_{1} m_{2}}{\sqrt{\left(l_{1}^{2}+m_{1}^{2}\right)\left(l_{2}^{2}+m_{2}^{2}\right)}}\)

Note 1:
If θ is the accute angle between the lines ax² + 2hxy + by² = 0 then cos θ = \(\frac{|a+b|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)
If θ is the accute angle between the lines ax² + 2hxy + by² = 0 then tan θ = ±\(\frac{2 \sqrt{h^{2}-a b}}{a+b}\) and sin θ = \(\frac{2 \sqrt{h^{2}-a b}}{\sqrt{(a-b)^{2}+4 h^{2}}}\)

Conditions For Perpendicular And Coincident Lines:

• If the lines ax² + 2hxy + by² = 0 are perpendicular to each other then θ = π/ 2 and cos θ = 0 ⇒ a + b = 0 i.e., co-efficient of x² + coefficient of y² = 0.
• If the two lines are parallel to each other then 0 = 0.
  ⇒ The two lines are coincident ⇒ h² = ab

Bisectors of Angles:
Theorem:
The equations of bisectors of angles between the lines a₁ x + b₁ y + c₁ = 0, a₂ x + b₂ y + c₂ = 0 are \(\frac{a_{1} x+b_{1} y+c_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}}}\) = ±\(\frac{a_{2} x+b_{2} y+c_{2}}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\)

Pair of Bisectors of Angles:
The equation to the pair bisectors of the angle between the pair of lines ax² + 2hxy + by² = 0 is h(x² – y²) = (a – b)xy (or) \(\frac{x^{2}-y^{2}}{a-b}=\frac{x y}{h}\).
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ……….(1)
and l₂x + m₂y = 0 ………(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.

The equations of bisectors of angles between (1) and (2) are \(\frac{l_{1} x+m_{1} y}{\sqrt{l_{1}^{2}+m_{1}^{2}}}-\frac{l_{2} x+m_{2} y}{\sqrt{l_{2}^{2}+m_{2}^{2}}}\) = 0 and
\(\frac{l_{1} x+m_{1} y}{\sqrt{l_{1}^{2}+m_{1}^{2}}}+\frac{l_{2} x+m_{2} y}{\sqrt{l_{2}^{2}+m_{2}^{2}}}\) = 0

The combined equation of the bisectors is

Theorem
The equation to the pair of lines passing through (x₀, y₀) and parallel ax² + 2hxy + by² = 0
is a( x – x₀)² + 2h( x – x₀)(y – y₀) + b( y – y₀)² = 0
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ……(1) and l₂x + m₂ y = 0 …….. (2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.
The equation of line parallel to (1) and passing through (x₀, y₀) is l₂(x – x₀) + m₂(y – y₀) = 0 ………(3)
The equation of line parallel to (2) and passing through (x₀, y₀) is l₂(x – x₀) + m₂(y – y₀) = 0 ………(4)

The combined equation of (3), (4) is
[l₁( x – x₀) + m₁( y – y₀)][l₂( x – x₀) + m₂(y – y₀)] = 0
⇒ l₁l₂(x – x₀)² + (l₁m₂ + l₂m₁)(x – x₀)(y – y₀) + m₁m₂(y – y₀)² = 0
⇒ a( x – x₀)² + 2h( x – x₀)( y – y₀) + b( y – y₀)2² = 0

Theorem:
The equation to the pair of lines passing through the origin and perpendicular to ax² + 2hxy + by² = 0 is bx² – 2hxy + ay² = 0 .
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ………(1) and l₂x + m₂y = 0 ……..(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.
The equation of the line perpendicular to (1) and passing through the origin is m₁x – l₁y = 0 ……….(3)
The equation of the line perpendicular to (2) and passing through the origin is m₂ x – l₂ y = 0 — (4)
The combined equation of (3) and (4) is
⇒ (m₁x – l₁ y)(m₂ x – l₂ y) = 0
⇒ m₁m₂x – (l₁m₂ + l₂m₁ )ny + l₁l₂ y = 0
bx² – 2hxy + ay² = 0

Theorem:
The equation to the lines passing through (x₀, y₀) and Perpendicular to ax² + 2hxy + by² = 0 is b(x – x₀)² – 2h(x – x₀)(y – y₀) + a(y – y₀)² = 0

Area of the triangle:
Theorem:
The area of triangle formed by the lines ax² + 2hxy + by² = 0 and lx + my + n = 0 is \(\frac{n^{2} \sqrt{h^{2}-a b}}{\left|a m^{2}-2 h \ell m+b \ell^{2}\right|}\)
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 ………(1) and l₂x + m₂y = 0 ……..(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.

The given straight line is lx + my + n = 0 ………(3)
Clearly (1) and (2) intersect at the origin.
Let A be the point of intersection of (1) and (3). Then

Theorem:
The product of the perpendiculars from (α, β) to the pair of lines ax² + 2hxy + by² = 0 is \(\frac{\left|a \alpha^{2}+2 h \alpha \beta+b \beta^{2}\right|}{\sqrt{(a-b)^{2}+4 h^{2}}}\)
Proof:
Let ax² + 2hxy + by² = 0 represent the lines l₁x + m₁y = 0 — (1) and l₂x + m₂ y = 0 …………..(2).
Then l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b.
The lengths of perpendiculars from (α, β) to
the line (1) is p = \(\frac{\left|l_{1} \alpha+m_{1} \beta\right|}{\sqrt{l_{1}^{2}+m_{1}^{2}}}\)
and to the line (2) is q = \(\frac{\left|l_{2} \alpha+m_{2} \beta\right|}{\sqrt{l_{2}^{2}+m_{2}^{2}}}\)

∴ The product of perpendiculars is

Pair of Lines-Second Degree General Equation:
Theorem:
If the equation S ≡ ax² + 2hxy + by² + 2 gx + 2 fy + c = 0 represents a pair of straight lines then
(i) Δ ≡ abc + 2fgh – af² – bg² – ch² =0 and
(ii) h² ≥ ab, g² ≥ ac, f² ≥ bc
Proof:
Let the equation S = 0 represent the two lines l₁x + m₁ y + n₁ = 0 and l₂ x + m₂ y + n₂ = 0. Then
ax2 + 2hxy + by2 + 2 gx + 2 fy + c
≡ (l₁x + m₁y + n₁)(l₂ x + m₂ y + n₂) = 0

Equating the co-efficients of like terms, we get
l₁l₂ = a, l₁m₂ + l₂m₁ = 2h , m₁m₂ = b, and l₁n₂ + l₂n₁ = 2g , m₁n₂ + m₂n₁ = 2 f , n₁n₂ = c

(i) Consider the product(2h)(2g)(2f)
= (l₁m₂ + l₂m₁)(l₁n₂ + l₂n₁)(m₁n₂ + m₂n₁)
= l₁l₂ (m₁²n₂ + m₂²n₁²) + m₁m₂ (l₁²n₂² + l₂²n₁²) + n₁n₂ (l₁²m₂² + l₂²m₁²) + 2l₁l₂m₁m₂n₁n₂
= l₁l₂[(m₁n₂ + m₂n₁) – 2m₁m₂n₁n₂] + m₁m₂[(l₁n₂ + l₂n₁) – 2l₁l₂n₁n₂] + n₁n₂[(l₁m₂ + l₂m₁) – 2l₁l₂m₁m₂] + 2l₁l₂m₁m₂n₁n₂
= a(4 f² – 2bc) + b(4g² – 2ac) + c(4h² – 2ab)
8 fgh = 4[af² + bg² + ch² – abc]
abc + 2 fgh – af² – bg² – ch² = 0

(ii) h² – ab = \(\left(\frac{l_{1} m_{2}+l_{2} m_{1}}{2}\right)^{2}\) – l1l2m1m2 = \(\frac{\left(l_{1} m_{2}+l_{2} m_{1}\right)^{2}-4-l_{1} l_{2} m_{1} m_{2}}{4}\)
= \(\frac{\left(l_{1} m_{2}-l_{2} m_{1}\right)^{2}}{4}\) ≥ 0
Similarly we can prove g² > ac and f² ≥ bc

Note :
If A = abc + 2 fgh – af² – bg² – ch² = 0 , h² ≥ ab, g² ≥ ac and f² ≥ bc, then the equation S ≡ ax² + 2hxy + by² + 2 gx + 2 fy + c = 0 represents a pair of straight lines

Conditions For Parallel Lines-Distance Between Them:
Theorem:
If S = ax² + 2hxy + by² + 2 gx + 2 fy + c = 0 represents a pair of parallel lines then h² = ab and bg² = af². Also the distance between the two parallel lines is 2\(\sqrt{\frac{g^{2}-a c}{a(a+b)}}\) (or) 2\(\sqrt{\frac{f^{2}-b c}{b(a+b)}}\)
Proof:
Let the parallel lines represented by S = 0 be
lx + my + n₁ = 0 ……….(1) lx + my + n₂ = 0 ………..(2)
ax² + 2hxy + 2gx + 2 fy + c
= (lx + my + n₁)(lx + my + n₂)

Equating the like terms
l² = a ………(3)
2lm = 2h …………(4)
m² = b ………..(5)
l(n₁ + n₂) = 2g …….(6)
m(n₁ + n₂) = 2 f ….(7)
n₁n₂ = c …….(8)

From (3) and (5), l²m² = ab and from (4) h² = ab .

Point of Intersection of Pair of Lines:
Theorem:
The point of intersection of the pair of lines represented by
a² + 2hxy + by² + 2gx + 2fy + c = 0 when h² > ab is \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)
Proof:
Let the point of intersection of the given pair of lines be (x₁, y₁).
Transfer the origin to (x₁, y₁) without changing the direction of the axes.
Let (X, Y) represent the new coordinates of (x, y). Then x = X + x₁ and y = Y + y₁.
Now the given equation referred to new axes will be
a( X + x₁)2 + 2h( X + x₁)(Y + y₁) +b(Y + y₁)2 + 2 g (X + x₁) + 2 f (Y + y₁) + c = 0
⇒ aX² + 2hXY + bY² + 2 X (ax₁ + hy₁ + g) + 2Y(hx₁ + by₁ + f) +(ax₁² + 2hx₁y₁ + by² + 2 gx₁ + 2 fy₁ + c) = 0

Since this equation represents a pair of lines passing through the origin it should be a homogeneous second degree equation in X and Y. Hence the first degree terms and the constant term must be zero. Therefore,
ax₁ + hy₁ + g = 0
hx₁ + by₁ + f = 0
ax₁² + 2hx₁ y₁ + by₁² + 2 gx₁ + 2 fyx + c = 0
But (3) can be rearranged as
x₁(ax₁ + hy + g) + y (hx₁ + byx + f) + (gx₁ + fq + c) = 0
⇒ gx₁ + fy₁ + c = 0 ………..(4)

Solving (1) and (2) for x₁ and y₁

Hence the point of intersection of the given pair of lines is \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)

Theorem:
If the pair of lines ax² + 2hxy + by² = 0 and the pair of lines ax² + 2hxy + by² + 2gx + 2fy + c = 0 form a rhombus then (a – b) fg+h(f² – g²) = 0.
Proof:
The pair of lines ax² + 2hxy + by² = 0 …………(1) is parallel to the lines ax² + 2hxy + by² + 2gx + 2fy + c = 0 ……….. (2)

Now the equation
ax² + 2hxy + by² + 2gx + 2 fy + c + λ(ax² + 2hxy + by²) = 0

Represents a curve passing through the points of intersection of (1) and (2).
Substituting λ = -1, in (3) we obtain 2gx + 2fy + c = 0 …(4)
Equation (4) is a straight line passing through A and B and it is the diagonal \(\overline{A B}\)

The point of intersection of (2) is C = \(\left(\frac{h f-b g}{a b-h^{2}}, \frac{g h-a f}{a b-h^{2}}\right)\)
⇒ Slope of \(\overline{O C}=\frac{g h-a f}{h f-b g}\)
In a rhombus the diagonals are perpendicular ⇒ (Slope of \(\overline{O C}\)) (Slope of \(\overline{A B}\)) = -1
⇒ \(\left(\frac{g h-a f}{h f-b g}\right)\left(-\frac{g}{f}\right)\) = -1
⇒ g²h – afg = hf² – bfg
⇒ (a – b)fg + h(f² – g²) = 0
\(\frac{g^{2}-f^{2}}{a-b}=\frac{f g}{h}\)

Theorem:
If ax² + 2hxy + by² = 0 be two sides of a parallelogram and px + qy = 1 is one diagonal, then the other diagonal is y(bp – hq) = x(aq – hp)
proof:
Let P(x₁, y₁) and Q(x₂, y₂) be the points where the digonal

px + qy = 1 meets the pair of lines.
\(\overline{O R}\) and \(\overline{P Q}\) biset each other at M(α, β)
∴ α = \(\frac{x_{1}+x_{2}}{2}\) and β = \(\frac{y_{1}+y_{2}}{2}\)

Eliminating y from ax² + 2hxy+by² = 0
and px + qy = 1 ………..(2)
ax² + 2hx\(\left(\frac{1-p x}{q}\right)\) + b\(\left(\frac{1-p x}{q}\right)^{2}\) = 0
⇒ x² (aq² – 2hpq + bp²) + 2 x(hp – bp) + b = 0

The roots of this quadratic equation are x₁ and x₂ where
x₁ + x₂ = \(-\frac{2(h q-b p)}{a q^{2}-2 h p q-b p^{2}}\)
⇒ α = \(\frac{(b p-h q)}{\left(a q^{2}-2 h p q+b p^{2}\right)}\)

Similarly by eliminating x from (1) and (2) a quadratic equation in y is obtained and y₁,
y₂ are its roots where
y₁ + y₂ = \(-\frac{2(h p-a q)}{a q^{2}-2 h p q-n p^{2}}\) ⇒ β = \(\frac{(a q-h p)}{\left(a q^{2}-2 h p q+b p^{2}\right)}\)
Now the equation to the join of O(0, 0) and M(α, β) is (y – 0)(0 – α) = (x – 0)(0 – β)
⇒ αy = βx
Substituting the values of α and β, the equation of the diagonal OR
is y(bp – hq) = x(aq – hp).