Question

A triangle has corners A, B, and C located at `(3 ,5 )`, `(2 ,9 )`, and `(4 , 8 )`, respectively. What are the endpoints and length of the altitude going through corner C?

Answer 1

Endpoints `(4,8)` and `( 40/17, 129/17)` and length `7/sqrt{17}`.

Explanation:

I am apparently an expert in answering two year old questions. Let's continue.

The altitude through C is the perpendicular to AB through C.

There are a few ways to do this one. We can calculate the slope of AB as `-4,` then the slope of the perpendicular is `1/4` and we can find the meet of the perpendicular through C and the line through A and B. Let's try another way.

Let's call the foot of the perpendicular `F(x,y)`. We know the dot product of the direction vector CF with the direction vector AB is zero if they're perpendicular:

`(B-A) cdot (F - C) = 0`

`(1-,4) cdot (x-4,y-8) = 0`

`x - 4 - 4y + 32 = 0`

`x - 4y = -28`

That's one equation. The other equation says `F(x,y)` is on the line through A and B:

`(y - 5)(2-3)=(x-3)(9-5)`

`5 - y = 4(x-3)`

`y = 17 - 4x`

They meet when

`x - 4(17 - 4x) = -28`

`x - 68 + 16 x = -28`

`17 x = 40`

`x = 40/17`

`y = 17 - 4 (40/17) = 129/17`

The length CF of the altitude is

`h = \sqrt{ (40/17-4)^2 + (129/17 - 8)^2} = 7 /sqrt{17}`

Let's check this by calculating the area using the shoelace formula and then solving for the altitude. A(3,5),B(2,9),C(4,8)

`a = \frac 1 2 | 3(9)-2(5) + 2(8)-9(4) + 4(5)-3(8)| = 7/2`

`AB=sqrt{ (3-2)^2+(9-5)^2 } = sqrt{17}`

`a = \frac 1 2 b h`

`7/2 = 1/2 h sqrt{17}`

`h = 7/sqrt{17} quad quad quad sqrt`