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!