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

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Last updated date: 22nd Jul 2024
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Answer
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Hint: We first take the coordinates of the point in which the straight lines $y={{m}_{1}}x+{{c}_{1}}$, $y={{m}_{2}}x+{{c}_{2}}$, $y={{m}_{3}}x+{{c}_{3}}$ meet each other. We put the values and then using the condition of the coefficient matrix of the equations’ having determinant value 0, we find the required condition.

Complete step by step answer:
We assume the point $\left( h,k \right)$ in which the straight lines $y={{m}_{1}}x+{{c}_{1}}$, $y={{m}_{2}}x+{{c}_{2}}$, $y={{m}_{3}}x+{{c}_{3}}$ meet each other.
It means that the point $\left( h,k \right)$ lies on every line of $y={{m}_{1}}x+{{c}_{1}}$, $y={{m}_{2}}x+{{c}_{2}}$, $y={{m}_{3}}x+{{c}_{3}}$.
The point satisfies the equations.
Therefore, $k={{m}_{1}}h+{{c}_{1}}$, $k={{m}_{2}}h+{{c}_{2}}$, $k={{m}_{3}}h+{{c}_{3}}$.
We convert them into equations into side of the equality.
$k-{{m}_{1}}h-{{c}_{1}}=0$, $k-{{m}_{2}}h-{{c}_{2}}=0$, $k-{{m}_{3}}h-{{c}_{3}}=0$.
The three lines represents the same point with area being 0 when the coefficient matrix of the equations has determinant value 0 which means the matrix is singular matrix.
Therefore, $\left| \begin{matrix}
   1 & -{{m}_{1}} & -{{c}_{1}} \\
   1 & -{{m}_{2}} & -{{c}_{2}} \\
   1 & -{{m}_{3}} & -{{c}_{3}} \\
\end{matrix} \right|=0$.
We now need to simplify the determinant by expanding through the first column.
$\left| \begin{matrix}
   1 & -{{m}_{1}} & -{{c}_{1}} \\
   1 & -{{m}_{2}} & -{{c}_{2}} \\
   1 & -{{m}_{3}} & -{{c}_{3}} \\
\end{matrix} \right|=\left| \begin{matrix}
   1 & {{m}_{1}} & {{c}_{1}} \\
   1 & {{m}_{2}} & {{c}_{2}} \\
   1 & {{m}_{3}} & {{c}_{3}} \\
\end{matrix} \right|$
$\left| \begin{matrix}
   1 & {{m}_{1}} & {{c}_{1}} \\
   1 & {{m}_{2}} & {{c}_{2}} \\
   1 & {{m}_{3}} & {{c}_{3}} \\
\end{matrix} \right|={{m}_{2}}{{c}_{3}}-{{m}_{3}}{{c}_{2}}+{{m}_{3}}{{c}_{1}}-{{m}_{1}}{{c}_{3}}+{{m}_{1}}{{c}_{2}}-{{m}_{2}}{{c}_{1}}$
Now we simplify the equation ${{m}_{2}}{{c}_{3}}-{{m}_{3}}{{c}_{2}}+{{m}_{3}}{{c}_{1}}-{{m}_{1}}{{c}_{3}}+{{m}_{1}}{{c}_{2}}-{{m}_{2}}{{c}_{1}}=0$. We take common terms out.
$\begin{align}
  & {{m}_{2}}{{c}_{3}}-{{m}_{3}}{{c}_{2}}+{{m}_{3}}{{c}_{1}}-{{m}_{1}}{{c}_{3}}+{{m}_{1}}{{c}_{2}}-{{m}_{2}}{{c}_{1}}=0 \\
 & \Rightarrow {{m}_{1}}\left( {{c}_{2}}-{{c}_{3}} \right)+{{m}_{2}}\left( {{c}_{3}}-{{c}_{1}} \right)+{{m}_{3}}\left( {{c}_{1}}-{{c}_{2}} \right)=0 \\
\end{align}$
Therefore, the conditions that the straight lines $y={{m}_{1}}x+{{c}_{1}}$, $y={{m}_{2}}x+{{c}_{2}}$, $y={{m}_{3}}x+{{c}_{3}}$ may meet in a point 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: Three or more distinct lines are said to be concurrent, if they pass through the same point. The point of intersection of any two lines, which lie on the third line is called the point of concurrence. Thus, if three lines are concurrent the point of intersection of two lines lies on the third line.