Using co-ordinate geometry, prove that...

MathematicsCentroid, Incenter, Orthocenter, and CircumcenterJEE Advanced 1998Moderate

Using co-ordinate geometry, prove that the three altitudes of any triangle are concurrent.

Visualized Solution (English)

Visualizing the Altitudes

Objective: Prove that the three altitudes of $△ A B C$ are concurrent.
Let the vertices be $A (x_{1}, y_{1})$ , $B (x_{2}, y_{2})$ , and $C (x_{3}, y_{3})$ .
An altitude is a line passing through a vertex and perpendicular to the opposite side.

The Perpendicularity Condition

Slope of side $B C$ ( $m_{B C}$ ) = $\frac{y _{2} - y _{3}}{x _{2} - x _{3}}$
Since altitude is perpendicular to $B C$ , its slope ( $m_{a l t}$ ) = $- \frac{1}{m _{B C}}$
Therefore, $m_{a l t} = - \frac{x _{2} - x _{3}}{y _{2} - y _{3}}$

Equation of the First Altitude

Using point-slope form: $y - y_{1} = - \frac{x _{2} - x _{3}}{y _{2} - y _{3}} (x - x_{1})$
Rearranging the terms: $(x - x_{1}) (x_{2} - x_{3}) + (y - y_{1}) (y_{2} - y_{3}) = 0$
Let this be equation $L_{1} = 0$ .

The Three Altitude Equations

By symmetry, the other two altitudes are:
$L_{2}$ : $(x - x_{2}) (x_{3} - x_{1}) + (y - y_{2}) (y_{3} - y_{1}) = 0$
$L_{3}$ : $(x - x_{3}) (x_{1} - x_{2}) + (y - y_{3}) (y_{1} - y_{2}) = 0$

The Concurrency Condition

Three lines $L_{1}, L_{2}, L_{3}$ are concurrent if there exist constants such that $λ_{1} L_{1} + λ_{2} L_{2} + λ_{3} L_{3} = 0$ .
Here, we will check the sum $L_{1} + L_{2} + L_{3}$ .
If the sum of equations is identically zero, the lines must meet at a point.

Atomic Compute: Summing X-Terms

Sum of $x$ coefficients: $(x_{2} - x_{3}) + (x_{3} - x_{1}) + (x_{1} - x_{2}) = 0$
Sum of $y$ coefficients: $(y_{2} - y_{3}) + (y_{3} - y_{1}) + (y_{1} - y_{2}) = 0$
The variable parts of the sum $L_{1} + L_{2} + L_{3}$ vanish completely.

Atomic Compute: Summing Constants

Sum of constants: $\sum - x_{1} (x_{2} - x_{3}) + \sum - y_{1} (y_{2} - y_{3})$
Expanding: $(- x_{1} x_{2} + x_{1} x_{3} - x_{2} x_{3} + x_{2} x_{1} - x_{3} x_{1} + x_{3} x_{2}) + \dots$
All terms cancel out, so the total constant sum is $0$ .

Conclusion: Concurrency Proven

Since $L_{1} + L_{2} + L_{3} \equiv 0$ , the lines are concurrent.
The point of concurrency is called the Orthocenter ( $H$ ).
Key Takeaway: The geometric property of concurrency is perfectly reflected in the algebraic symmetry of line equations.

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