# A triangle has corners A, B, and C located at (4 ,5 ), (3 ,6 ), and (8 ,4 ), respectively. What are the endpoints and length of the altitude going through corner C?

##### 1 Answer
Aug 14, 2018

$N \left(\frac{13}{2} , \frac{5}{2}\right) \mathmr{and} C \left(8 , 4\right)$ are the endpoints of altitudes
going through corner $C$
The length of the altitude $= \therefore C N = \frac{3}{\sqrt{2}} \approx 2.12$

#### Explanation:

Let $\triangle A B C \text{ be the triangle with corners at}$

$A \left(4 , 5\right) , B \left(3 , 6\right) \mathmr{and} C \left(8 , 4\right)$

Let $\overline{A L} , \overline{B M} \mathmr{and} \overline{C N}$ be the altitudes of sides $\overline{B C} , \overline{A C} \mathmr{and} \overline{A B}$ respectively.

Let $\left(x , y\right)$ be the intersection of three altitudes

Slope of $\overline{A B} = \frac{5 - 6}{4 - 3} = - \frac{1}{1} = - 1$

$\overline{A B} \bot \overline{C N} \implies$slope of $\overline{C N} = 1$ ,

$\overline{C N}$ passes through $C \left(8 , 4\right)$

$\therefore$The equn. of $\overline{C N}$ is $: y - 4 = 1 \left(x - 8\right)$

$\implies y - 4 = x - 8$

$\implies y = x - 8 + 4$

i.e. color(red)(y=x-4.....to (1)

Now, Slope of $\overline{A B} = - 1$ and $\overline{A B}$ passes through
$A \left(4 , 5\right)$

So, eqn. of $\overline{A B}$ is: $y - 5 = - 1 \left(x - 4\right)$

$\implies y - 5 = - x + 4$

=>color(red)(y=9-x...to(2)

from $\left(1\right) \mathmr{and} \left(2\right)$

$x - 4 = 9 - x$

=>x+x=9+4=>2x=13=>color(blue)(x=13/2=6.5

From $\left(1\right) ,$

y=13/2-4=>color(blue)(x=5/2=2.5

$\implies N \left(\frac{13}{2} , \frac{5}{2}\right) \mathmr{and} C \left(8 , 4\right)$ are the endpoints of altitudes
going through corner $C$

Using Distance formula,

$C N = \sqrt{{\left(\frac{13}{2} - 8\right)}^{2} + {\left(\frac{5}{2} - 4\right)}^{2}} = \sqrt{\frac{9}{4} + \frac{9}{4}}$
$\therefore C N = \sqrt{\frac{18}{4}} = \sqrt{\frac{9}{2}}$

$\therefore C N = \frac{3}{\sqrt{2}} \approx 2.12$