# Examples on Concurrency of Three Lines Set 1

**Example – 9**

Prove that the three lines *L*_{1}, *L*_{2} and *L*_{3} whose equations have been mentioned in the preceeding discussion, are concurrent if we can find three constants \({{\lambda }_{1}},{{\lambda }_{2}}\,\,\text{and}\,\,{{\lambda }_{3}}\) such that

\[{{\lambda }_{1}}{{L}_{1}}\,\,+\,\,{{\lambda }_{2}}{{L}_{2}}+{{\lambda }_{3}}{{L}_{3}}=0\]

**Solution:** Assume that *L*_{1} and *L*_{2} intersect at the point *P* whose co-ordinates are (*x*_{0}, *y*_{0}) *P* should satisfy the equations of both *L*_{1 }and *L*_{2}.

\[\begin{align}

& {{L}_{1}}\left( \text{at}\,P \right)\equiv {{a}_{1}}{{x}_{0}}+{{b}_{1}}{{y}_{0}}+{{c}_{1}}=0\qquad \qquad ...\text{ }\left( 1 \right) \\

& {{L}_{2}}\left( \text{at}\,P \right)\equiv {{a}_{2}}{{x}_{0}}+{{b}_{2}}{{y}_{0}}+{{c}_{2}}=0\qquad \qquad ...\text{ }\left( 2 \right) \\

\end{align}\]

Now assume that we can find three non-zero constants \({{\lambda }_{1}},{{\lambda }_{2}}\,\,\text{and}\,\,{{\lambda }_{3}}\) such that \({{\lambda }_{1}}{{L}_{1}}+{{\lambda }_{2}}{{L}_{2}}+{{\lambda }_{3}}{{L}_{3}}=0\) . We will prove that due to this condition, *L*_{3} will definitely have to pass through *P*:

\[\begin{align} &\qquad\quad {{\lambda }_{1}}{{L}_{1}}+{{\lambda }_{2}}{{L}_{2}}+{{\lambda }_{3}}{{L}_{3}}=0 \\ & \Rightarrow\quad {{L}_{3}}=\left( -\frac{{{\lambda }_{1}}}{{{\lambda }_{3}}} \right){{L}_{1}}+\left( -\frac{{{\lambda }_{2}}}{{{\lambda }_{3}}} \right){{L}_{2}} \end{align}\]

If we evaluate the value of *L*_{3} at *P*, we get

\[\begin{align} & {{L}_{3}}\left( \text{at}\,P \right)=\left( -\frac{{{\lambda }_{1}}}{{{\lambda }_{3}}} \right)\times {{L}_{1}}\left( \text{at}\,P \right)+\left( -\frac{{{\lambda }_{2}}}{{{\lambda }_{3}}} \right)\times {{L}_{2}}\left( \text{at}\,P \right) \\ & \qquad\qquad=\left( -\frac{{{\lambda }_{1}}}{{{\lambda }_{3}}} \right)\times 0+\left( -\frac{{{\lambda }_{2}}}{{{\lambda }_{3}}} \right)\times 0\qquad \qquad \left\{ \begin{gathered} \text{By}\,\,\,\left( 1 \right) \\ \text{and}\,\,\left( 2 \right) \end{gathered} \right\} \\ &\qquad\qquad =0 \end{align}\]

Since the value of *L*_{3} is 0 at *P*, the line *L*_{3 }must pass through *P*. Thus, *L*_{1}, *L*_{2} and *L*_{3} are concurrent.

**Example – 10**

Show that the medians of a triangle are concurrent.

**Solution:** Let the triangle have the vertices \(A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right)\,\,\text{and}\,\,C\left( {{x}_{3}},{{y}_{3}} \right)\) as in the figure below:

From the two-point form of the equation of a line, we can write down the equations of *AD*, *BE* and *CF*.

The equation *L*_{1} of the median *AD* is:

\[\begin{align} & \qquad \quad\,\,\,\,{{L}_{1}}\equiv \frac{y-{{y}_{1}}}{x-{{x}_{1}}}=\frac{{{y}_{1}}-\begin{align}\frac{{{y}_{2}}+{{y}_{3}}}{2}\end{align}}{{{x}_{1}}-\begin{align}\frac{{{x}_{2}}+{{x}_{3}}}{2}\end{align}} \\ & \Rightarrow \qquad {{L}_{1}}\equiv \left( 2{{y}_{1}}-\left( {{y}_{2}}+{{y}_{3}} \right) \right)x-\left( 2{{x}_{1}}-\left( {{x}_{2}}+{{x}_{3}} \right) \right)y={{x}_{1}}\left( 2{{y}_{1}}-\left( {{y}_{2}}+{{y}_{3}} \right) \right)-{{y}_{1}}\left( 2{{x}_{1}}-\left( {{x}_{2}}+{{x}_{3}} \right) \right) \end{align}\]

By symmetry, we can write down the corresponding equations *L*_{2} and *L*_{3} of the medians *BE* and *CF*.

Observe carefully that when we subsequently add the three equations *L*_{1}, *L*_{2} and *L*_{3}, their left hand sides sum to 0. Thus, we have found three constants 1, 1 and 1 such that

\[\begin{align} & \qquad\quad \;1\cdot {{L}_{1}}+1\cdot {{L}_{2}}+1\cdot {{L}_{3}}=0 \\

& \Rightarrow \qquad {{L}_{1}},{{L}_{2}}\,\,\text{and}\,{{L}_{3}} \\ & \Rightarrow \qquad \text{The medians of any triangle are concurrent.} \end{align}\]