Altitudes of Triangle are Concurrent

The altitudes of a triangle are concurrent.


altitudes-triangle

Let A\left(  {{x_1},{y_1}} \right), B\left(  {{x_2},{y_2}} \right) and C\left( {{x_3},{y_3}}  \right) be the vertices of the triangle ABC. If {m_1} is the slope of AB, then using two point formula to find slope of line

{m_1}  = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\,\,\,\,{\text{ -  -  -  }}\left( {\text{i}} \right)


Since altitude CD is perpendicular to the side AB, so its slope m is given as using condition of perpendicular slope is

m  =  - \frac{1}{m} =  - \frac{{{x_2} - {x_1}}}{{{y_2} -  {y_1}}}\,\,\,\,{\text{ -  -  - }}\left( {{\text{ii}}} \right)


Equation of altitude CD passing through C\left( {{x_3},{y_3}} \right) with slope m is

\begin{gathered} y - {y_3} = m\left( {x - {x_3}} \right) \\ \Rightarrow y - {y_3} =  - \frac{{{x_2} - {x_1}}}{{{y_2} -  {y_1}}}\left( {x - {x_3}} \right) \\ \Rightarrow \left( {{y_2} - {y_1}}  \right)\left( {y - {y_3}} \right) =  -  \left( {{x_2} - {x_1}} \right)\left( {x - {x_3}} \right) \\ \Rightarrow \left( {{x_2} - {x_1}}  \right)\left( {x - {x_3}} \right) + \left( {{y_2} - {y_1}} \right)\left( {y -  {y_3}} \right) = 0 \\ \Rightarrow \left( {{x_2} - {x_1}} \right)x  - \left( {{x_2} - {x_1}} \right){x_3} + \left( {{y_2} - {y_1}} \right)y -  \left( {{y_2} - {y_1}} \right){y_3} = 0 \\ \Rightarrow \left( {{x_2} - {x_1}} \right)x  + \left( {{y_2} - {y_1}} \right)y - \left( {{x_2} - {x_1}} \right){x_3} -  \left( {{y_2} - {y_1}} \right){y_3} = 0\,\,\,{\text{ -  -  -  }}\left( {{\text{iii}}} \right) \\ \end{gathered}


For the equation of altitude AB, we just replace {x_1} by {x_2}, {x_2} by {x_3} and {x_3} by {x_1} in (iii) (i.e. {x_1} \to {x_2},\,\,{x_2} \to {x_3},\,\,{x_3} \to  {x_1}), so

  \Rightarrow \left( {{x_3} - {x_2}} \right)x + \left( {{y_3} - {y_2}} \right)y -  \left( {{x_3} - {x_2}} \right){x_1} - \left( {{y_3} - {y_2}} \right){y_1} =  0\,\,\,{\text{ -  -  - }}\left( {{\text{iv}}} \right)


For the equation of altitude BF, we just replace {x_1} by {x_2}, {x_2} by {x_3} and {x_3} by {x_1} in (iv) (i.e. {x_1} \to {x_2},\,\,{x_2} \to {x_3},\,\,{x_3} \to  {x_1}), so

  \Rightarrow \left( {{x_1} - {x_3}} \right)x + \left( {{y_1} - {y_3}} \right)y -  \left( {{x_1} - {x_3}} \right){x_2} - \left( {{y_1} - {y_3}} \right){y_2} =  0\,\,\,{\text{ -  -  - }}\left( {\text{v}} \right)


To see whether the altitudes (iii), (iv) and (v) are concurrent, consider the determinant.

\left|  {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{ -  \left( {{x_2} - {x_1}} \right){x_3} - \left( {{y_2} - {y_1}} \right){y_3}} \\ {{x_3} - {x_2}}&{{y_3} - {y_2}}&{ -  \left( {{x_3} - {x_2}} \right){x_1} - \left( {{y_3} - {y_2}} \right){y_1}} \\ {{x_1} - {x_3}}&{{y_1} - {y_3}}&{ -  \left( {{x_1} - {x_3}} \right){x_2} - \left( {{y_1} - {y_3}} \right){y_2}} \end{array}}  \right|


{R_3}  + \left( {{R_1} + {R_2}} \right)


 =  \left| {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{ -  \left( {{x_2} - {x_1}} \right){x_3} - \left( {{y_2} - {y_1}} \right){y_3}} \\ {{x_3} - {x_2}}&{{y_3} - {y_2}}&{ -  \left( {{x_3} - {x_2}} \right){x_1} - \left( {{y_3} - {y_2}} \right){y_1}} \\ 0&0&0 \end{array}}  \right| = 0


This shows that the altitudes of the triangle are concurrent.

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