Question

Find the coordinates of the circumcenter of the triangle with the vertices A(4,12) B(14,6) C(-6,2) This is for question 4 and I need help with graphing it afterward because I think I graphed it wrong.

Answer 1

Lets start by graphing the given points:

Now let's find the circumcenter:

`M_{}AB=((14+4)/(2),(12+6)/(2))`
`M\text{AB}=(9,9)`

And the slope of AB is:

`(12-6)/(4-14)=(6)/(-10)=(3)/(-5)=-(3)/(5)`

So the slope of the perpendicular line must be:

`x\cdot-(3)/(5)=-1`
`x=-(1)/(-(3)/(5))`
`x=(5)/(3)`

The equation of that bisector must be (from the slope-point formula):

`(y-9)=(5)/(3)(x-9)`

`y-9=(5)/(3)x-(45)/(3)`
`y-9=(5)/(3)x-15`
`y=(5)/(3)x-15+9`
`y=(5)/(3)x-6`

For the midpoint of BC we get:

`MBC=((14-6)/(2),(2+6)/(2))`

As shown in the graph.

The slope of BC is:

`(6-2)/(14-(-6))=(4)/(20)=(2)/(10)=(1)/(5)`

So the slope of the perpendicular line must be:

`x\cdot(1)/(5)=-1`
`x=-5`

The equation of that bisector must be:

`y-4=-5(x-4)`
`y-4=-5x+20`
`y=-5x+24`

Now equate the bisectors to find the circumcenter:

`-5x+24=(5)/(3)x-6`
`-5x-(5)/(3)x=-6-24`
`-(20)/(3)x=-30`
`x=(-30\cdot3)/(-20)`
`x=(90)/(20)=(9)/(2)`

And replacing on any of the bisectors:

`y=-5x+24`
`y=-5((9)/(2))+24`
`y=-(45)/(2)+24`
`y=-(45)/(2)+(48)/(2)`
`y=(3)/(2)`

So the circumcenter is:

`((9)/(2),(3)/(2))`