# Point of Intersection of Two Lines

The point of intersection of two lines $${a_1}x + {b_1}y + {c_1} = 0$$ and $${a_2}x + {b_2}y + {c_2} = 0$$ is given by \[\left( {\frac{{{b_1}{c_2} – {b_2}{c_1}}}{{{a_1}{b_2} – {a_2}{b_1}}},\frac{{{a_2}{c_1} – {a_1}{c_2}}}{{{a_1}{b_2} – {a_2}{b_1}}}} \right)\] where $${a_1}{b_2} – {a_2}{b_1} \ne 0$$.

To prove this formula we have the given equations of straight lines:

\[\begin{gathered} {a_1}x + {b_1}y + {c_1} = 0\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ {a_2}x + {b_2}y + {c_2} = 0\,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right) \\ \end{gathered} \]

We solve the above equations using the simultaneous method.

Multiplying equation (i) by $${b_2}$$, we have

\[{a_1}{b_2}x + {b_1}{b_2}y + {b_2}{c_1} = 0\,\,\,\,{\text{ – – – }}\left( {{\text{iii}}} \right)\]

Multiplying equation (ii) by $${b_1}$$, we have

\[{a_2}{b_1}x + {b_1}{b_2}y + {b_1}{c_2} = 0\,\,\,\,{\text{ – – – }}\left( {{\text{iv}}} \right)\]

Now subtracting (iv) from equation (iii), we get

\[ \Rightarrow x = \frac{{{b_1}{c_2} – {b_2}{c_1}}}{{{a_1}{b_2} – {a_2}{b_1}}}\]

Multiplying equation (i) by $${a_2}$$, we have

\[{a_1}{a_2}x + {b_1}{a_2}y + {b_2}{a_1} = 0\,\,\,\,{\text{ – – – }}\left( {\text{v}} \right)\]

Multiplying equation (ii) by $${a_1}$$, we have

\[{a_2}{a_1}x + {a_1}{b_2}y + {a_1}{c_2} = 0\,\,\,\,{\text{ – – – }}\left( {{\text{vi}}} \right)\]

Now subtracting (vi) from equation (v), we get

\[ \Rightarrow y = \frac{{{a_2}{c_1} – {a_1}{c_2}}}{{{a_1}{b_2} – {a_2}{b_1}}}\]

This shows that the point of intersection is \[\left( {\frac{{{b_1}{c_2} – {b_2}{c_1}}}{{{a_1}{b_2} – {a_2}{b_1}}},\frac{{{a_2}{c_1} – {a_1}{c_2}}}{{{a_1}{b_2} – {a_2}{b_1}}}} \right)\]

Note: If $${a_1}{b_2} – {a_2}{b_1} = 0$$, then lines (i) and (ii) will have no common point and therefore, these will be parallel lines.

Now

\[\begin{gathered} {a_1}{b_2} – {a_2}{b_1} = 0 \\ \Rightarrow \frac{{{a_1}}}{{{b_1}}} = \frac{{{a_2}}}{{{b_2}}} \\ \end{gathered} \]

This is the condition for lines (i) and (ii) to be parallel lines.