Question

A triangle has corners A, B, and C located at `(1 ,8 )`, `(6 ,3 )`, and `(7 ,4 )`, respectively. What are the endpoints and length of the altitude going through corner C?

Answer 1

The Altitude is
`bar(CB)=bar(BC)`
It length is `bar(CB)= sqrt(2)`
and the end point `CB` are:
`B(6,3) " and " C(7,4)`

Explanation:

By inspection `bar(AB)` is perpendicular to `bar(BC)`.
You can show this by taking the dot product of vectors:
`vec(v_bar(AB))*vec(v_bar(BC))`
`vec(v_bar(AB)) = [(1, 1)]*[(-5), (5)]= -5+5 = 0 =vec(v_bar(AB))*vec(v_bar(BC))costheta`
Q.E.D

Or if you prefer you can usd distance formula to find the length of the triangles, and then check is they make up a Pythagorean triples.
Let's do it:
`bar(AB) = sqrt((1-6)^2+(8-3)^2) = 5sqrt(2)`
`bar(BC) = sqrt((7-6)^2+(4-3)^2) = sqrt(2)`
`bar(CA) = sqrt((7-1)^2+(4-8)^2) = sqrt(52)`

Now `bar(AB), bar(BC), bar(CA)` are Pythagorean triplets iff:
`bar(AB)^2 + bar(BC^2) = bar(CA)^2`
`25*2 + 2 = 52`
`:. bar(AB)^2, bar(BC^2)" and "bar(CA)^2`
Q.E.D again

What does this mean? Well if `AB` perpendicular `bar(BC)`
Then the altitude is:
`bar(CB)=bar(BC) sqrt(2)`
and the end point `BC` are:
`B(6,3) " and " C(7,4)`

Note that in geometry: an altitude of a triangle is that line segment from the vertex that perpendicularly cut the opposite side to the vertex (base).