# 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).

Straight Lines
Straight Lines
grade 11 | Questions Set 1
Straight Lines
Straight Lines
grade 11 | Questions Set 2