Question

A triangle has corners A, B, and C located at `(2 ,7 )`, `(5 ,3 )`, and `(9 , 4 )`, respectively. What are the endpoints and length of the altitude going through corner C?

Answer 1

endpoints are:
`C=(9,4)`
`H=(5.96, 1.72)`
length of `CH=3.8`

Explanation:

`A=(2,7)`
`B=(5,3)`
`C=(9,4)`
Endpoints of height passing through C are C itself and some point lying on segment (side) AB.
To find that point we need to solve system of equations corresponding to the height and segment AB.
To find an equation of a line we must have either 2 points (from which we can get the slope) or 1 point and the slope.

We have 2 points of line AB, the slope is:

`m=(rise)/(run)=(y_B-y_A)/(x_B-x_A)=(3-7)/(5-2)=-4/3`

and the equation (obtained using point A, point B would give the same equation):

`y-y_A=m_(AB)(x-x_A)`
`y-7=-4/3(x-2)`
`y=-4/3x+8/3+7=-4/3x+29/3`

In case of height we have only 1 point and slope is still unknown, but it's perpendicular to AB so its slope is an inverse reciprocal of AB's slope.

`m_h=3/4`

And the equation:

`y-y_C=m_h(x-x_C)`
`y-4=3/4(x-9)`
`y=3/4x-27/4+4=3/4x-11/4`

So here we have a system of equations:
`{(y=-4/3x+29/3),(y=3/4x-11/4):}`

We could solve for x:
`-4/3x+29/3=3/4x-11/4`

to get
`x=149/25=5 24/25=5.96`

and substitute to any equation (second one here):
`y=3/4(149/25)-11/4`

to get
`y=43/25=1 18/25=1.72`

Now we have point
`H=(5.96, 1.72)`
which is the other endpoint of height.

We can use pythagorean formula to get distance between H and C:
#CH=sqrt((x_C-x_H)^2+(y_C-y_H)^2) =sqrt((9-5 24/25)^2+(4-1 18/25)^2)=sqrt((3 1/25)^2+(2 7/25)^2) =sqrt((76^2+57^2)/25^2)=sqrt(5776+3249)/25=sqrt(9025)/25=sqrt(25(90*4+1))/25=sqrt(361)/5=19/5#

Or using decimals (calculator may be handy):
#CH=sqrt((9-5.96)^2+(4-1.72)^2)=sqrt((3.04)^2+(2.28)^2) =sqrt(9.2416+5.1984)=sqrt(14.44)=3.8#