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