# Pair of Straight Lines – Homogeneous Second Degree Equation

## Pair of Straight Lines – Homogeneous Second Degree Equation

An equation (whose RHS is zero) in which the sum of the power of x and y in every term is the same say n, is called a Homogeneous equation of nth degree in x and y.

Thus, ax² + 2hxy + by² = 0 is a homogeneous equation of second degree and cx³ + dxy² + ey³ = 0 is a homogeneous equation of third degree in x and y.

Pair of Straight Lines through the Origin: If ax² 2hxy + by² + 2gx + 2fy + c = 0 represent two lines, then the LHS can be resolved into two linear factors.

Let the factors be (l₁x + m₁y + n₁) (l₂x + m₂y + n₂)

Simplifying the equation, we will get (l₁x + m₁y) (l₂x + m₂y), the terms of second degree, must be identical with ax² + 2hxy + by² =0 is identical with (l₁x + m₁y) x (l₂x + m₂y) = 0.

Thus, equations ax² + 2hxy + by² represents two lines l₁x + m₁y = 0, l₂x + m₂y = 0 which are parallel to lines. l₁x + m₁y + n₁ = 0, l₂x + m₂y + n₂ = 0, respectively.

Let us consider a general homogeneous equation of second degree in x and y as

ax² + 2hxy + by² = 0 . . . (1)

dividing both sides by x², we get

b(y/x)² + 2h(y/x) + a = 0 . . .(2)

since (2) is an equation of second degree in y/x it has two roots.

Let the roots be m₁ and m₂. if α and β are the roots of equation ax² + bx + c = 0.

ax² + bx + c = 0

= a (x – α) x (x – β)

Therefore, (2) can be written as

b (y/x – m₁) (y/x – m₂) = 0

Thus equation (1) represents two straight lines y – m₁x = 0 and y – m₂x = 0 both of which pass through the origin. Comparing b (y – m₁x) (y – m₂x) = 0 with equation (1), we have

$$\frac{b{{m}_{1}}{{m}_{2}}}{a}=-\frac{b({{m}_{1}}+{{m}_{2}})}{2h}=\frac{1}{1}$$,

$${{m}_{1}}+{{m}_{2}}=\frac{-2h}{b}$$,

$${{m}_{1}}\times {{m}_{2}}=\frac{a}{b}$$.

Example: If the lines px² – qxy – y² = 0 make angles α and β with the x – axis, then the value of tan (α + β) is

Solution: Let the lines represented by the equation.

px² – qxy – y² = 0

y = m₁ x

and y = m₂ x.

Then, m₁ = tanα and m₂ = tanβ

m₁ + m₂ = -2h/b = -q

m₁ m₂ = a/b = p

$$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$$,

$$\tan (\alpha +\beta )=\frac{{{m}_{1}}+{{m}_{2}}}{1-{{m}_{1}}{{m}_{2}}}$$,

$$\tan (\alpha +\beta )=\frac{-q}{1+p}$$.