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!