Animated Solution for Mathematics - Straight Lines: Using co-ordinate geometry, prove that the three altitudes of any triangle are concurrent.

Consider a triangle $△ABC$ with vertices $A(x_1,y_1)$, $B(x_2,y_2)$, and $C(x_3,y_3)$ on a coordinate plane. Our objective is to prove that the three altitudes of this triangle are concurrent at a single point, known as the Orthocenter.

To begin, we determine the equation of the altitude dropped from vertex $A$ to the side $BC$. First, we calculate the slope of the base $BC$ using the two-point form:

Since the altitude from $A$ is perpendicular to $BC$, the product of their slopes must be $-1$. Consequently, the slope of the altitude ($m_{alt}$) is the negative reciprocal of $m_{BC}$:

Using the point-slope form $y-y_1=m_{alt}(x-x_1)$, we substitute our slope to obtain:

By cross-multiplying the denominator and rearranging the terms, we arrive at the equation for the first altitude, $L_1$:

We do not need to repeat this derivation for the remaining altitudes. Due to the cyclic symmetry of the triangle's vertices, we can obtain the equations for $L_2$ and $L_3$ by performing a cyclic permutation of the indices ($1\to 2$, $2\to 3$, $3\to 1$):

To prove that these three lines are concurrent, we examine their sum. If the sum of the three equations is identically zero, it implies that any point satisfying two of the equations must necessarily satisfy the third.

Summing the $x$-terms:

The $y$-terms follow the same pattern:

Finally, the constant terms also cancel out in pairs. We are left with the identity:

This identity confirms that the three altitudes are concurrent, as the third line is a linear combination of the first two. This elegant algebraic approach demonstrates the power of cyclic symmetry in solving geometric problems.