Question

Find the conditions that the straight lines $y={{m}_{1}}x+{{c}_{1}},y={{m}_{2}}x+{{c}_{2}}$ and $y={{m}_{3}}x+{{c}_{3}}$ may meet in a point.

Answer 1

Hint: The lines cannot meet if they are parallel, which means that the slopes of all the lines must not be equal. So, we get a relation between the slopes of the lines that, ${{m}_{1}}\ne {{m}_{2}},{{m}_{1}}\ne {{m}_{3}}$ and ${{m}_{2}}\ne {{m}_{3}}$. We will first find the point of line 1 and 2 by equating 1 and 2 of the lines as, ${{m}_{1}}x+{{c}_{1}}={{m}_{2}}x+{{c}_{2}}$. And finally, we will put the obtained point of intersection of the line 1 and 2 in the equation of line 3 and solve accordingly to get the coordinates of the point of intersection of all the three lines.

Complete step-by-step answer:
It is given in the question that we have to find the conditions that the straight lines $y={{m}_{1}}x+{{c}_{1}},y={{m}_{2}}x+{{c}_{2}}$ and $y={{m}_{3}}x+{{c}_{3}}$ may meet in a point. To solve this, let us assume the equations of the first, second and third lines as equations 1, 2 and 3 respectively. So, we will get as,
$\begin{align}
  & {{y}_{1}}={{m}_{1}}x+{{c}_{1}}\ldots \ldots \ldots \left( 1 \right) \\
 & {{y}_{2}}={{m}_{2}}x+{{c}_{2}}\ldots \ldots \ldots \left( 2 \right) \\
 & {{y}_{3}}={{m}_{3}}x+{{c}_{3}}\ldots \ldots \ldots \left( 3 \right) \\
\end{align}$
We know that the lines cannot meet if they are parallel. So, it means that their slopes must be unequal. We can represent it as,
$\begin{align}
  & {{m}_{1}}\ne {{m}_{2}} \\
 & {{m}_{1}}\ne {{m}_{3}} \\
 & {{m}_{2}}\ne {{m}_{3}} \\
\end{align}$
From this, we can get a relation as follows,
$\begin{align}
  & {{m}_{1}}-{{m}_{2}}\ne 0 \\
 & {{m}_{1}}-{{m}_{3}}\ne 0 \\
 & {{m}_{2}}-{{m}_{3}}\ne 0 \\
\end{align}$
Now, we have to find the point of intersection of these lines. If the point of intersection of lines 1 and 2 lies on line 3, then the three lines meet at a point. So, the point of intersection of lines 1 and 2 are,
$\begin{align}
  & y={{m}_{1}}x+{{c}_{1}} \\
 & y={{m}_{2}}x+{{c}_{2}} \\
\end{align}$
On equating equation 1 with equation 2, we will get as,
${{m}_{1}}x+{{c}_{1}}={{m}_{2}}x+{{c}_{2}}$
On transposing ${{m}_{2}}x$ from RHS to LHS, we will get,
${{m}_{1}}x-{{m}_{2}}x+{{c}_{1}}={{c}_{2}}$
Now, we will transpose ${{c}_{1}}$ from LHS to RHS. So, we get,
$\begin{align}
  & {{m}_{1}}x-{{m}_{2}}x={{c}_{2}}-{{c}_{1}} \\
 & \Rightarrow x\left( {{m}_{1}}-{{m}_{2}} \right)={{c}_{2}}-{{c}_{1}} \\
 & \Rightarrow x=\dfrac{{{c}_{2}}-{{c}_{1}}}{{{m}_{1}}-{{m}_{2}}}=\dfrac{{{c}_{1}}-{{c}_{2}}}{{{m}_{2}}-{{m}_{1}}} \\
\end{align}$
Since we know that ${{m}_{2}}-{{m}_{1}}\ne 0$, so $x=\dfrac{{{c}_{1}}-{{c}_{2}}}{{{m}_{2}}-{{m}_{1}}}$. Now, we will put the value of $x=\dfrac{{{c}_{1}}-{{c}_{2}}}{{{m}_{2}}-{{m}_{1}}}$ in equation 1. So, we get,
$\begin{align}
  & y={{m}_{1}}x+{{c}_{1}} \\
 & \Rightarrow y={{m}_{1}}\left( \dfrac{{{c}_{1}}-{{c}_{2}}}{{{m}_{2}}-{{m}_{1}}} \right)+{{c}_{1}} \\
 & \Rightarrow y=\dfrac{{{m}_{1}}{{c}_{1}}-{{m}_{1}}{{c}_{2}}}{{{m}_{2}}-{{m}_{1}}}+{{c}_{1}} \\
 & \Rightarrow y=\dfrac{{{m}_{1}}{{c}_{1}}-{{m}_{1}}{{c}_{2}}+{{m}_{2}}{{c}_{1}}-{{m}_{1}}{{c}_{1}}}{{{m}_{2}}-{{m}_{1}}} \\
\end{align}$
On cancelling the similar terms in the RHS, we get,
$y=\dfrac{{{m}_{2}}{{c}_{1}}-{{m}_{1}}{{c}_{2}}}{{{m}_{2}}-{{m}_{1}}}$
Thus, we get the point of intersection of lines 1 and 2 as,
$\begin{align}
  & x=\dfrac{{{c}_{1}}-{{c}_{2}}}{{{m}_{2}}-{{m}_{1}}},y=\dfrac{{{m}_{2}}{{c}_{1}}-{{m}_{1}}{{c}_{2}}}{{{m}_{2}}-{{m}_{1}}} \\
 & \Rightarrow \left\{ \dfrac{{{c}_{1}}-{{c}_{2}}}{{{m}_{2}}-{{m}_{1}}},\dfrac{{{m}_{2}}{{c}_{1}}-{{m}_{1}}{{c}_{2}}}{{{m}_{2}}-{{m}_{1}}} \right\} \\
\end{align}$
Now, as this line lies on line 3, then the coordinates of point x and y must satisfy the equation of line 3 also. So, we can say,
$\begin{align}
  & y={{m}_{3}}x+{{c}_{3}} \\
 & \Rightarrow \dfrac{{{m}_{2}}{{c}_{1}}-{{m}_{1}}{{c}_{2}}}{{{m}_{2}}-{{m}_{1}}}=\left( \dfrac{{{c}_{1}}-{{c}_{2}}}{{{m}_{2}}-{{m}_{1}}} \right){{m}_{3}}+{{c}_{3}} \\
 & \Rightarrow \dfrac{{{m}_{2}}{{c}_{1}}-{{m}_{1}}{{c}_{2}}}{{{m}_{2}}-{{m}_{1}}}=\dfrac{{{m}_{3}}{{c}_{1}}-{{m}_{3}}{{c}_{2}}+{{m}_{2}}{{c}_{3}}-{{m}_{1}}{{c}_{3}}}{{{m}_{2}}-{{m}_{1}}} \\
\end{align}$
On cancelling the similar terms, we get,
${{m}_{2}}{{c}_{1}}-{{m}_{1}}{{c}_{2}}={{m}_{3}}{{c}_{1}}-{{m}_{3}}{{c}_{2}}+{{m}_{2}}{{c}_{3}}-{{m}_{1}}{{c}_{3}}$
On transposing all the terms from RHS to LHS, we get,
$\begin{align}
  & {{m}_{2}}{{c}_{1}}-{{m}_{1}}{{c}_{2}}-{{m}_{3}}{{c}_{1}}+{{m}_{3}}{{c}_{2}}-{{m}_{2}}{{c}_{3}}+{{m}_{1}}{{c}_{3}}=0 \\
 & \Rightarrow {{m}_{2}}{{c}_{1}}-{{m}_{3}}{{c}_{1}}-{{m}_{1}}{{c}_{2}}+{{m}_{3}}{{c}_{2}}-{{m}_{2}}{{c}_{3}}+{{m}_{1}}{{c}_{3}}=0 \\
 & \Rightarrow {{c}_{1}}\left( {{m}_{2}}-{{m}_{3}} \right)+{{c}_{2}}\left( {{m}_{3}}-{{m}_{1}} \right)+{{c}_{3}}\left( {{m}_{1}}-{{m}_{2}} \right)=0 \\
\end{align}$
Thus, the three lines meet at a point if, $\left( {{m}_{2}}-{{m}_{3}} \right){{c}_{1}}+\left( {{m}_{3}}-{{m}_{1}} \right){{c}_{2}}+\left( {{m}_{1}}-{{m}_{2}} \right){{c}_{3}}=0$ and $\left( {{m}_{1}}-{{m}_{2}} \right)\ne 0,\left( {{m}_{3}}-{{m}_{2}} \right)\ne 0,\left( {{m}_{3}}-{{m}_{1}} \right)\ne 0$.
Therefore, the required condition is ${{m}_{1}}\left( {{c}_{2}}-{{c}_{3}} \right)+{{m}_{2}}\left( {{c}_{3}}-{{c}_{1}} \right)+{{m}_{3}}\left( {{c}_{1}}-{{c}_{2}} \right)=0$.

Note: This question includes a lot of long and complex calculations and so because of that many students tend to make mistakes with the use of the plus and minus signs which lead to incorrect answers. Many students also take the point $y=\dfrac{{{c}_{1}}-{{c}_{2}}}{{{m}_{1}}-{{m}_{2}}}$ in equation 3 which leads to a wrong result. So, one should be careful while solving this question.