Question

What is the orthocenter of a triangle with corners at `(1 ,4 )`, `(5 ,7 )`, and (2 ,3 )#?

Answer 1

Orthocenter is at `(11/7, 25/7)`

Explanation:

There are three vertices given and we need to obtain two altitude linear equations to solve for the Orthocenter.

One negative reciprocal of slope from (1, 4) to (5, 7) and the point (2, 3) gives an altitude equation.

`(y-3)=-1/((7-4)/(5-1))*(x-2)`

`y-3=-4/3(x-2)`

`3y-9=-4x+8`
`4x+3y=17" "` first equation

Another negative reciprocal of slope from (2, 3) to (5, 7) and the point (1, 4) gives another altitude equation.

`y-4=-1/((7-3)/(5-2))*(x-1)`

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

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

`4y-16=-3x+3`

`3x+4y=19" "`second equation

Solve the orthocenter using the first and second equation

`4x+3y=17" "` first equation
`3x+4y=19" "`second equation

Method of elimination using subtraction
`12x+9y=51` first equation after multiplying each term by 3
`underline(12x+16y=76)`second equation after multiplying each term by 4
`0x-7y=-25`

`7y=25`
`y=25/7`

Solve for x using `4x+3y=17" "` first equation and `y=25/7`

`4x+3(25/7)=17" "`

`4x+75/7=17`
`4x=17-75/7`
`x=(119-75)/28`
`x=44/28`
`x=11/7`

Orthocenter is at `(11/7, 25/7)`

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