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.