# Equation of a Line Through the Intersection of Two Other Lines

Consider the two straight lines
$\begin{gathered} {a_1}x + {b_1}y + {c_1} = 0\,\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ {a_2}x + {b_2}y + {c_2} = 0\,\,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right) \\ \end{gathered}$

For any nonzero constant $k$, the equation of the form
${a_1}x + {b_1}y + {c_1} + k\left( {{a_2}x + {b_2}y + {c_2}} \right) = 0\,\,\,\,{\text{ – – – }}\left( {{\text{iii}}} \right)$

being linear in $x$ and $y$ is an equation of a straight line.

If $\left( {{x_1},{y_1}} \right)$ is the point of intersection of line (i) and (ii), then it must satisfy both equations (i) and (ii), i.e.
$\begin{gathered} {a_1}{x_1} + {b_1}{y_1} + {c_1} = 0\,\,\,\,\,{\text{ – – – }}\left( {{\text{iv}}} \right) \\ {a_2}{x_1} + {b_2}{y_1} + {c_2} = 0\,\,\,\,\,{\text{ – – – }}\left( {\text{v}} \right) \\ \end{gathered}$

Next we check whether the point $\left( {{x_1},{y_1}} \right)$ lies on (iii) or not. For this we replace $x$ by ${x_1}$ and $y$ by ${y_1}$ in equation (iii), and we have
${a_1}{x_1} + {b_1}{y_1} + {c_1} + k\left( {{a_2}{x_1} + {b_2}{y_1} + {c_2}} \right) = 0\,\,\,\,{\text{ – – – }}\left( {{\text{vi}}} \right)$

Using equation (iv) and (v) in equation (vi), we get the following result:
$\begin{gathered} 0 + k\left( 0 \right) = 0 \\ 0 = 0 \\ \end{gathered}$

This shows that (vi) is true for all $k$ and for $x = {x_1},\,y = {y_1}$. Thus, the point $\left( {{x_1},{y_1}} \right)$ lies on (vi) for all $k$. In fact (vi) represents the equation of the line through the point of intersection of the lines (i) and (ii). Since $k$ is any real number, equation (vi) shows that there will be an infinite number of lines through the point of intersection of the lines (i) and (ii).

In order to find the particular line, we must have the particular value of $k$, and this particular value of $k$can be found with the help of some given conditions.

Example: Find the equation of a line through the point $\left( {1,3} \right)$ and the point of intersection of lines $2x – 3y + 4 = 0$ and $4x + y – 1 = 0$.

The equation of a straight line through the point of intersection of lines $2x – 3y + 4 = 0$ and $4x + y – 1 = 0$ is given as
$2x – 3y + 4 + k\left( {4x + y – 1} \right) = 0$

Since the required line passes through the point $\left( {1,3} \right)$, this point must satisfy the equation, i.e.
$\begin{gathered} 2\left( 1 \right) – 3\left( 3 \right) + 4 + k\left( {4\left( 1 \right) + 3 – 1} \right) = 0 \\ – 3 + 6k = 0\,\,\, \Rightarrow k = \frac{1}{2} \\ \end{gathered}$

Putting this value of $k$ in the required equation of a straight line, we have
$\begin{gathered} 2x – 3y + 4 + \frac{1}{2}\left( {4x + y – 1} \right) = 0 \\ \Rightarrow 4x – 6y + 8 + 4x + y – 1 = 0 \\ \Rightarrow 8x – 5y + 7 = 0 \\ \end{gathered}$

This is the required equation of the straight line.