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!

Utkarsh BharadwajBest article on this topic!