Question

A triangle has corners A, B, and C located at `(6 ,1 )`, `(2 ,4 )`, and `(3 , 2 )`, respectively. What are the endpoints and length of the altitude going through corner C?

Answer 1

End points of altitude CF are `(3,2) and (54/15) , (14/5)`
Length of altitude CF = 1

Explanation:

Equation of side AB is
`(y-y1)/(y2-y1)=(x-x1)/(x2-x1)`
`(y-1)/(4-1)=(x-6)/(2-6)`
`3x+4y=22` (Equation 1)
Slope of AB m1`=(y2-y1)/(x2-x1)=(4-1)/(2-6)=-(3/4)`
Slope of altitude CF
`m2=-1/(m1)=-1/(-3/4)=4/3`
Equation of altitude CF
`(y-y3)=m2(x-x3)`
`(y-2)=(4/3)(x-3)`
`4x-3y=6` (Equation 2)
By solving Equation 1 & 2, we get point F
`12x+16y=88`
`12x-9y=18`
Subtracting, `25y=70`
`y=(14/5)`
Substituting value of y in Equation 1, we get `x=54/15`
Coordinates of `F (54/15,14/5)`
Length of altitude CF
`CF=sqrt(((14/5)-2)^2+((54/15)-3)^2)`
`CF=sqrt((4/5)^2+(9/15)^2)=sqrt((144+81)/225)=1`