Question

A triangle has corners A, B, and C located at `(1 ,3 )`, `(7 ,4 )`, and `(5 , 8 )`, respectively. What are the endpoints and length of the altitude going through corner C?

Answer 1

Endpoints are `C(5, 8)` and `(211/37, 140/37)`
Length `l=4.27437`

Explanation:

We need the equation of the line passing thru A(1, 3) and B(7, 4) by the two-point form

`y-y_a=(y_b-y_a)/(x_b-x_a)*(x-x_a)`

`y-3=(4-3)/(7-1)(x-1)`

`y-3=1/6*(x-1)`

`6y-18=x-1`

`x-6y+17=0" "`first equation

Solve for the length `l` using "distance from line to a point" formula:
from the first equation `a=1` and `b=-6` and `c=17`

`l=(ax_c+by_c+c)/(+-sqrt(a^2+b^2))=(1*5-6*8+17)/(+-sqrt(1^2+(-6)^2)`
`l=(5-48+17)/(+-sqrt(37))=(-26)/(-sqrt37)`
`l=4.27437`

We need another equation to solve for the endpoints
Use point `C(x_c, y_c)=C(5, 8)` and slope `m=-6`. This is perpendicular to side c or line segment AB.

`y-y_c=m(x-x_c)`
`y-8=-6(x-5)`
`y-8=-6x+30`
`6x+y=38`

simultaneous solution of the two equations to solve for the other endpoint
`x-6y+17=0" "` first equation
`6x+y=38" "`second equation

`x=6y-17`
`x=6(38-6x)-17`
`x=228-36x-17`
`37x=211`
`x=211/37`

`x=5.70`

`y=38-6x`
`y=38-6(211/37)=(1406-1266)/37=140/37`
`y=140/37`

`y=3.78`

the other endpoint is `(5.70, 3.78)`

God bless...I hope the explanation is useful.