Angle Between Pairs of Lines Through the origin

Go back to  'Straight Lines'

Consider now that we’ve been given the equation of a pair of straight lines passing through the origin as :

\[a{x^2} + 2hxy + b{y^2} = 0 \qquad \qquad ...(1)\]

We wish to determine the angle between these two lines. Let \(m_1\) and \(m_2\) be the slopes of these two lines. By dividing(1) by x2 and substituting we have

This quadratic in m will have its roots as \(m_1\) and \(m_2.\) Thus,

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

The angle between the two lines, say \(\theta,\) is given by

\[\begin{align}   \tan \theta & = \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|\\   & = \frac{{\sqrt {{{({m_1} + {m_2})}^2} - 4{m_1}{m_2}} }}{{\left| {1 + {m_1}{m_2}} \right|}}\\    & = \frac{{2\sqrt {{h^2} - ab} }}{{\left| {a + b} \right|}} \qquad \qquad \qquad \qquad  (\text{Using }(2))   \end{align}\]

As a consequence of this formula, we see that the lines represented by (1), are:

\[\boxed{\begin{align}& {{\bf{Parallel}}{\text{ (in fact coincident since both pass through the origin)}}}&{{\text{if}}}\qquad{{h^2} = ab} \\ & {{\bf{Perpendicular}}{\rm{                                                                             }}}&{{\text{if  }}}\quad{a + b = 0}  \end{align}}\]

The importance of this condition must be mentioned; it is very widely used and should be committed to memory.

As an example, the locus given by

\[3{y^2} - 8xy - 3{x^2} = 0 \qquad \qquad\qquad \qquad ...(3)\]

represent two perpendicular straight lines since

\[a + b = (3) + ( - 3) = 0\]

Verify this by the explicit factorization of (3).

Learn math from the experts and clarify doubts instantly

  • Instant doubt clearing (live one on one)
  • Learn from India’s best math teachers
  • Completely personalized curriculum