Lecture 3 [17 Feb, 2020] MATH 127 (Prof Farid)

 

MATH -127 (Coordinate Geometry)
Chapter-2
(Pair of Straight Lines)

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General equation of second degree,

  $ax^2 + 2hxy + by^2 + 2 hx + 2fy + c = 0 \tag{1}\label{eq:1}$  

If we multiply two linear functions we get one non-linear function.

To remove the first-degree term from (1), we must shift the origin to  $(\alpha, \beta)$   where,

  $$\begin{align*} & \alpha = \frac{bg - fh }{h^2 - ab} = \frac{fh - bg}{ab - h^2}\\ & \beta = \frac{af - gh}{h^2 - ab} = \frac{gh - af }{ab - h^2} \end{align*}$$  

and to remove the term containing $xy$, we must rotate the axes through,  $\theta = \frac{1}{2}\tan^{-1} \frac{2h}{a - b}$  

Homogenous quadratic equation (where the total power of variables are same in every term):  $ax^2 + 2hxy + by^2 = 0 \tag{2}\label{eq:2}$  

we know, if $m_1$ and $m_2$ are two roots of an equation, 

  $x^2 - ( m_1 + m_2 )x + m_1 m_2 = 0$  

Now, take $b\ne0$ and divide throughout of (2) by $bx^2$, 

  $\implies \frac{a}{b}+ \frac{2hy}{6x} + \frac{y^2 }{x^2 } = 0$ is a quadratic equation in $\frac{y}{x} \tag{3}\label{eq:3}$

we know, for every quadratic equation,  $AX^2 + BX + C = 0$  

Here,  $A = 1, ~~~B = \frac{2h}{b}, ~~~ C = \frac{a}{b},~~~ X = \frac{y}{x}$  

If  $m_1$, $m_2$ are the roots of (3), then we get, 

  $m_1 + m_2 = - \frac{2h}{b} ~~~\textrm{and}~~~ m_1 m_2 = \frac{a}{b} \tag{4}\label{eq:4}$  

$\therefore$ From (2) we get, 

  $$\begin{align*} &ax^2 + 2hxy + by^2 = \left(\frac{y}{x} - m_1\right) \left(\frac{y}{x} - m_2\right) = 0 \\ \implies & y - m_1 x = 0 ~~~ \textrm{and}~~~ y - m_2 x = 0 \end{align*}$$  

$\therefore$ Equation (2) represents two straight lines that pass through the origin.

        ♦ Generalization: 

an equation of the form,  $a_o y^n + a_1 y^{n - 1}x^1 + a_2 y^{n - 2}x^2 + \cdots + a_n x^n = 0$   is a homogenous equation in $xy$  of degree $'n'$

  $P_n (x) = a_o + a_1 x + \cdots + a_n x^n$  

if $\theta$ be the angle between  $y - m_1 x = 0 ~~~and ~~~ y - m_2 x = 0$   then,

  $$\begin{align*} \tan\theta & = \frac{m_1 - m_2}{1 + m_1 m_2} \\ & = \frac{\sqrt{(m_1 - m_2 )^2}}{1 + m_1 m_2} \\ & = \frac{\sqrt{(m_1 + m_2 )^2 - 4 m_1 m_2}}{1 + m_1 m_2}\\ & = \frac{2 \sqrt{h^2 - ab}}{a + b}\\ \therefore ~\theta & = \tan^{-1} \frac{2 \sqrt{h^2 - ab}}{a + b} \end{align*}$$  

        ♦ Corollary:

1) if the lines  $y - m_1 x = 0 ~~~\textrm{and}~~~ y - m_2 x = 0$ are parallel:

  $$\begin{align*} & \theta = \frac{2 \sqrt{h^2 - ab}}{a + b}= 0 \\ \therefore ~ & h^2 = ab \end{align*}$$  

2)if the line represented by equation (2) are perpendicular:

  $$\begin{align*} & \tan 90^{\circ} = \frac{2 \sqrt{h^2 - ab}}{a + b} = \frac{1}{0} \\ \therefore ~~ & a + b = 0 \end{align*}$$  

•Under which condition equation (1) represents a pair of straight lines:

  $$\begin{align*} & \left| \begin{array}{ccc} a & h & g \\ h & b & f \\ g & f & c \end{array} \right| = 0 \\ \implies & abc + 2fgh - af^2 - b g^2 - ch^2 = 0 \end{align*}$$  

•angle between the lines represented by (1)  $lx + my + n = 0 \textrm{and} ~~~ l_1 x + m_1 y + n_1 = 0$  

They will not pass through the origin. In case of homogenous, they will pass through the origin.

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    Question 1: Prove that the equation $x^2 + 6 xy + 9 y^2 + 12 y - 5 = 0$  represents a pair of straight lines.

Solution: Given that  $x^2 + 6 xy + 9 y^2 + 12 y - 5 = 0 \tag{1.1}\label{eq:1.1}$  

comparing equation (1.1) with the standard equation we get, 

  $ax^2 + 2 hxy + by^2 + 2gx + 2 fy + c = 0$ $a = 1, ~~~ b = 9, ~~~ c = -5, ~~~ h = 3, ~~~ g = 2, ~~~ f = 6$  

the equation will represent a pair of  straight lines if, $$\begin{align*} & abc + 2 fgh - af^2 - bg^2 - ch^2 = 0\\ \\ \textrm{So now,} ~~ & 1\times9\times(-5) + 2\times6.2\times3-1\times36-9\times4+5\times9 \\ = & -45 + 72 - 36 - 36 + 45 \\ = & 0 \end{align*}$$                                                                                                     

$\therefore$ The condition is satisfied.

    Question 2: Find the values of k, so that the equation $3x^2 + 10 xy + 9 y^2 - kx - 26 y + 21 = 0$ represents a pair of straight lines and also find the angle between them.

Solution: The given equation is,$3x^2 + 10 xy + 9 y^2 - kx - 26 y + 21 = 0 \tag{2.1}\label{eq:2.1}$                                                                                                            

comparing equation (2.1) with the standard equation,

  $$\begin{align*} & ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \\ \textrm{we get,}\\ & a = 3, ~~~ b = 8, ~~~ c = 21, ~~~ h = 5, ~~~ g = - \frac{k}{2}, ~~~ f = - 13. \end{align*}$$ $$\begin{align*} \textrm{ now, } & \\ & 3 \times8\times21+2\times(-13)\times(- \frac{k}{2})\times5 -3 \times (- 13)^2 - 8 (\frac{- k}{2})^2 - 21 \times5^2 = 0 \\ & 2 k^2 - 65k + 528 = 0 \end{align*}$$