Examples on Pairs of Lines Through the origin Set 1

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.