Question

A triangle has vertices at A $=(0,0)$, $B=(14,0)$ and $C=(5,12)$. What are the coordinates of the orthocenter ?

• A. $(5,\frac{15}{4})$
• B. $(4,5)$
• C. $(\frac{13}{4},5)$
• D. None of these

Answer 1

The correct option is A $(5,\frac{15}{4})$

The easiest altitude to find is one from C to AB, since that is simply the line $x=5$.

The next easiset to find is the one from B to AC, since AC can be calculated as $y=\frac{12}{5}x$.

A line perpendiculat to AC is of the form $y=-\frac{5}{12}x+b$, for some b, and as the line goes through $(14,0)$, the equation of the altitude is

$y=-\frac{5}{12}x+\frac{35}{6}$

Finally the intersection of this line and the line $x=5$ is $(5,\frac{15}{4})$, which is thus the location of the orthocenter.