# How to Separate Pair of Straight Lines

In this post, you will learn how to separate pair of straight lines with example.

Let’s begin –

## How to Separate Pair of Straight Lines

(i) Let us consider the homogeneous equation of second degree as

$$ax^2+2hxy+by^2$$ = 0  ……(i)

which represent pair of straight lines passes through the origin.

Now, we divide by $$x^2$$, we get

a + $$2h({y\over x})$$ + $$b{({y\over x})}^2$$ = 0

Let  $$y\over x$$ = m  (say)

then a + 2hm + $$bm^2$$ = 0  …….(ii)

If $$m_1$$ & $$m_2$$ are the roots of equation (ii),

then $$m_1$$ + $$m_2$$ = -$$2h\over b$$, $$m_1m_2$$ = $$a\over b$$

and also tan$$\theta$$ = $$\pm$$$$2\sqrt{h^2-ab}\over {a+b}$$

These lines will be :

(1) Real and different, if $$h^2-ab$$ > 0

(2) Real and Coincident, if $$h^2-ab$$ = 0

(3) Imaginary if $$h^2-ab$$ < 0

(ii) The condition that these lines are :

(1) At Right angles to each other, is a + b = 0

(2) Coincident is $$h^2$$ = ab

(3) Equally inclined to the axes of x is h = 0. i.e. coefficient of xy = 0.

(iii) Homogeneous equation of second degree $$ax^2+2hxy+by^2$$ = 0 always represent a pair of straight lines whose equations are

y = ($$-h \pm \sqrt{h^2-ab}\over b$$)x = y = $$m_1$$x & y = $$m_2$$x and $$m_1$$ + $$m_2$$ = -$$2h\over b$$ ; $$m_1m_2$$ = $$a\over b$$

These straight lines passes through the origin

(iv) Pair of straight lines perpendicular to the line $$ax^2+2hxy+by^2$$ = 0 and through the origin are given by

$$bx^2-2hxy+ay^2$$ = 0

(v) The Product of the perpendiculars drawn from the point ($$x_1,y_1$$) on the lines $$ax^2+2hxy+by^2$$ = 0 is

|$$a{x_1}^2 + 2hx_1y_1 + b{y_1}^2\over {\sqrt{{(a-b)}^2 + 4h^2}}$$|

Note : A homogeneous equation of degree n represent n straight lines passing through origin.

General Equation and Homogeneous Equation of second degree :

(i)  The general equation of second degree $$ax^2+2hxy+by^2+2gx+2fy+c$$ = 0 represents a pair of straight lines, if

$$\Delta$$ = $$abc+2fgh-af^2-bg^2-ch^2$$ = 0

(ii)  The product of the perpendiculars drawn from origin to the lines $$ax^2+2hxy+by^2+2gx+2fy+c$$ = 0 is

|$$c\over \sqrt{{(a-b)}^2 + 4h^2}$$|

Example : Find the separate equation of the following pair of straight lines $$x^2 + 4xy + y^2$$ = 0

Solution : Divide the given equation by $$x^2$$

$$1 + 4{y\over x} + {y^2\over x^2}$$ = 0

Let y/x = m

$$\implies$$ $$1 + 4m + m^2$$ = 0 $$\implies$$ h = 2 and a = 1 and b = 1

Let $$m_1$$ & $$m_2$$ are the roots of equation , then $$m_1$$ + $$m_2$$ = -$$2h\over b$$ = -4 , $$m_1m_2$$ = $$a\over b$$ = 1

$$\implies$$ $$m_1$$ = $$-2 + \sqrt{3}$$ and $$m_2$$ = $$-2 – \sqrt{3}$$

Hence separate pair of straight lines are y = $$m_1$$x and y = $$m_2$$x

$$\implies$$ y = ($$-2 + \sqrt{3}$$)x and y = ($$-2 – \sqrt{3}$$)x

Hope you learnt how to separate pair of straight lines, learn more concepts of straight lines and practice more questions to get ahead in the competition. Good luck!