Question
Find the altitudes of triangle $\left(1,~-2\right)$ $\left(-2,~1\right)$ $\left(2,~3\right)$.
Answer
The lengths of the altitudes of a triangle $ ABC $ are:
$$ h_a = \dfrac{ 9 \sqrt{ 5}}{ 5 } ~~,~~ h_b = \dfrac{ 9 \sqrt{ 26}}{ 13 } ~~,~~ h_c = 3 \sqrt{ 2 } $$Explanation
The altitude $ h_a $ can be found using formula:
$$ A = \frac{ a \cdot h_a }{2} $$Where $ A $ is a triange area and $ a $ is a length of a side $ BC $. In this eample we have: $ A = 9 $ and $ d(B,C) = 2 \sqrt{ 5 } $ so:
$$ A = \frac{ a \cdot h_a }{2} => h_a = \frac{2 \cdot A}{ a } = \frac{ 2 \cdot 9}{ 2 \sqrt{ 5 } } = \frac{ 9 \sqrt{ 5}}{ 5 } $$This page was created using
Triangle Calculator