# Examples on Pairs of Lines Through the origin Set 1

Go back to  'Straight Lines'

Example - 16

Find the straight lines represented by

\begin{align} & (a)\;\;{y^2} - 5xy + 6{x^2} = 0\\ & ({\rm{b}})\;\;3{y^2} - 10xy + 3{x^2} = 0\\ & ({\rm{c}})\;\;{y^2} + xy + {x^2} = 0 \end{align}

Solution: Note that the homogenous nature of these equations tells us that the lines will pass through the origin.

$$\textbf{(a)}\;\;{y^2} - 5xy + 6{x^2} = 0$$

We either factorize this equation straightaway:

$(y - 2x)(y - 3x) = 0$

so that the lines are

$y = 2x{\text{ and }}y = 3x$

OR,

we divide it by $$x^2$$ and substitute \begin{align}m = \frac{y}{x}\end{align} to obtain:

\begin{align} \qquad {m^2} - 5m + 6 = 0\\ \Rightarrow \qquad & m = 2,\,\,3\\ \Rightarrow \qquad & \frac{y}{x} = 2,\,\,3\\ \Rightarrow \qquad & y = 2x\;\;{\rm{or}}\;\;y = 3x \end{align}

Both alternatives are entirely equivalent.

$$\textbf{(b)}$$

\begin{align} \qquad & 3{y^2} - 10xy + 3{x^2} = 0\\ \Rightarrow \qquad & 3{m^2} - 10m + 3 = 0 \qquad \qquad \left( {m = \frac{y}{x}} \right)\\ \Rightarrow \qquad & m = 3,\;\,\frac{1}{3}\\ \Rightarrow \qquad & y = 3x,\;\,y = \frac{x}{3} \end{align}

$$\textbf{(c)}$$

\begin{align} \qquad & {y^2} + xy + {x^2} = 0\\ \Rightarrow \qquad & {m^2} + m + 1 = 0 \qquad \qquad \left( {{\text{Again, }}m = \frac{y}{x}} \right) \end{align}

This has no real roots and thus physically, no lines will exist with the joint equation $${y^2} + xy + {x^2} = 0.$$ We sometimes say that this equation represents imaginary lines.

Note that in the entire plane, only (0, 0) satisfies this equation.