Examples on Pairs of Lines Through the origin Set 2

Example - 17

Find the equation of the pair of lines through the origin and perpendicular to the pair of lines \(a{x^2} + 2hxy + b{y^2} = 0.\)

Solution: Let the slopes of the two lines represented by the given equation be \(m_1\) and \(m_2.\) As explained earlier, \(m_1\) and \(m_2\) are the roots of the quadratic

\[b{m^2} + 2hm + a = 0\]

so that

\[\begin{align}{m_1} + {m_2} = \frac{{ - 2h}}{b}, \qquad {m_1}{m_2} = \frac{a}{b} \qquad \qquad ...(1)\end{align}\]

The slopes of the lines whose joint equation we require will simply be \(\begin{align}\frac{{ - 1}}{{{m_1}}}\;{\rm{and}}\;\frac{{ - 1}}{{{m_2}}}\end{align}\) so that their equations will be:

\[\begin{align} \qquad& y = \frac{{ - 1}}{{{m_1}}}x, \qquad \qquad \qquad y = \frac{{ - 1}}{{{m_2}}}x\\   \Rightarrow  \qquad& x + {m_1}y = 0, \qquad \quad \quad \;\;\, x + {m_2}y = 0   \end{align}\]

The required joint equation is

\[\begin{align}  \qquad & (x + {m_1}y)(x + {m_2}y) = 0\\    \Rightarrow \qquad & {x^2} + {m_1}{m_2}{y^2} + ({m_1} + {m_2})xy = 0\\    \Rightarrow \qquad & {x^2} + \frac{a}{b}{y^2} - \frac{{2h}}{b}xy = 0 \qquad \qquad \qquad \qquad \qquad  (Using\;(1))\\     \Rightarrow  \qquad &  b{x^2} - 2hxy + a{y^2} = 0    \end{align}\]

Example - 18

The equation \(a{x^3} + b{x^2}y + c{x^2} + d{y^3} = 0\) is a third degree homogenous equation and hence represents three straight lines passing through the origin. Find the condition so that two of these three lines may be perpendicular.

Solution: We divide the given equation by \(x^3\) and substitute \(\begin{align}\frac{y}{x}=m\end{align}\) to obtain:

\[d{m^3} + c{m^2} + bm + a = 0 \qquad \qquad \qquad  ...(1)\]

This has three roots, say \({m_1},\,{m_2},\,{m_3},\) corresponding to the three straight lines. Since we want two of these lines to be perpendicular, we can assume

\[m_1m_2=-1\]

From (1), we have

\[\begin{align}  &{m_1}\,{m_2}\,{m_3} = \frac{{ - a}}{d}\\    \Rightarrow \qquad & {m_3} = \frac{a}{d}   \end{align}\]

Substituting this value of \(m_3\) back in (1), (since \(m_3\) is a root of (1)), we obtain

\[\begin{align} & \frac{{d{a^3}}}{{{d^3}}} + \frac{{c{a^2}}}{{{d^2}}} + \frac{{ba}}{d} + a = 0\\   \Rightarrow \qquad & {a^2} + ac + bd + {d^2} = 0   \end{align}\]