# Thread: Circumcenter and orthrocenter of a triangle

1. ## Circumcenter and orthrocenter of a triangle

The points are X(70,15), Y(59,95), Z(52,28)
So far, I have slope of XY=-7.273, YZ=9.571, ZX=-0.722
Perp slope of XY= 0.138, YZ=-0.104, ZX=1.385
I need to find the circumcenter and the Orthrocenter of this triangle.. PLEASE HELP i dont know what im doing..
THANKS~

2. $You\ have\ the\ slopes\ of\ lines\ perpendicular\ to\ all\ 3\ sides.$

$The\ orthocentre\ is\ the\ point\ of\ intersection\ of\ the\ 3\ lines\ that\ are\ perpendicular$
$to\ a\ side\ and\ pass\ through\ the\ opposite\ vertex.$

$Hence\ the\ perpendicular\ line\ to\ the\ line\ XY\ contains\ the\ point\ Z,$
$the\ perpendicular\ to\ YZ\ contains\ X,$
$the\ perpendicular\ to\ XZ\ contains\ Y.$

$Therefore,\ write\ line\ equations\ with\ the\ slopes\ you\ calculated\$
$that\ contain\ these\ points.$

$Admittedly,\ you\ only\ need\ 2\ of\ these\ 3\ lines,$
$and\ the\ point\ of\ intersection\ found\ by\ solving\ the\ 2\ simultaneous$
$equations\ gives\ you\ the\ orthocentre.$

$The\ circumcentre\ is\ the\ centre\ of\ a\ circle\ that\ contains$
$X,\ Y,\ and\ Z\ on\ the\ circumference.$
$This\ means\ X,\ Y,\ and\ Z\ are\ equidistant\ from\ this\ centre.$

$You\ must\ bisect\ 2\ of\ the\ 3\ sides\ to\ find\ this.$
$Find\ the\ midpoints\ of\ 2\ sides.$
$Write\ the\ equations\ of\ the\ perpendicular\ lines\ containing\$
$these\ midpoints.$
$Solve\ the\ pair\ of\ simultaneous\ equations\ to\ find\ the\ circumcentre.$

$Draw\ 2\ arbitrary\ triangles\ with\ the\ lines\ mentioned\ to\ guide\ you.$

3. $Also,\ it\ would\ be\ a\ good\ idea\ to\ write\ 3\ sets\ of$
$simultaneous\ equations\ for\ both\ orthocentre\ and\ circumcentre$
$and\ verify\ that\ in\ both\ cases\ all\ 3\ sets\ give\ the\ same\ solution.$

4. Sorry, i couldn't thoroughly understand... Could you put it in simpler terms or a different way? Sorry i got lost in the description
I'm a freshmen in highschool

5. That's ok, Jacobban.

For the orthocentre, your first step is to imagine one side as the base of the triangle.
Just take any triangle to begin with.
Let the base be horizontal.
Now, draw the perpendicular height.
For instance, say the base is the line AB in a triangle with vertices A, B and C.
C is the top vertex of the triangle.

The question you must be able to answer is...

How do we write the equation of the perpendicular line from the base AB to the point C ?

You do it by first writing the slope of AB, using $slope=\frac{y_2-y_1}{x_2-x_1}$

Then you write the slope of the perpendicular line that contains C.
To do that, you turn the first slope upside-down and change the sign.

If the slope of AB is 2, then the perpendicular line's slope is $-\frac{1}{2}$

To write the equation of the perpendicular line containing C, you use the C co-ordinates in $y-y_1=m_p(x-x_1)$

where $m_p$ is the perpendicular slope.

Study the diagrams in the .pdf file.

You have to do that procedure for 2 lines, then solve the simultaneous equations to find the orthocentre.
After this, you will have no bother with solving simultaneous equations and working with perpendicular lines.

I'll get you started....

The slope of XY is $\frac{95-15}{59-70}=-\frac{80}{11}$
Therefore, any line perpendicular to XY has slope = $\frac{11}{80}$

The one containing Z is the line $y-28=\frac{11}{80}(x-52)$

80y-2240=11x-572
80y=11x+1668

Repeat for another side, to get a 2nd equation.

Solving the 2 equations gives the co-ordinates of the orthocentre.

6. THANKS SO MUCH! and how do you find the circumcenter????
thats just what i needed!

7. The circumcentre is the centre of the circle for which the given points are points on the circle's circumference.

All points on a circle circumference are equidistant from it's centre.

If you take any 2 points on the circumference and draw lines from them to the circle centre, and between the 2 points themselves, you make an isosceles triangle.

Remember an isosceles triangle has 2 sides equal.
On a circle circumference, these sides are the radius of the circle.
So, circles and isosceles triangles are well related.

An isosceles triangle is made of 2 back to back right-angled triangles.

This means, if we have an isosceles triangle, bisect it's base and draw the perpendicular bisector, that bisector must pass through the opposite vertex.

In this case (circumcentre), that perpendicular must go through the circle centre.

So, the procedure for locating the circumcentre is.....

Bisect 2 sides of the triangle by
1. writing the side midpoints
2. writing the side slopes
3. writing the perpendicular slopes
4. writting the line equations that have the perpendicular slopes and contain the midpoint.

Do this for 2 sides.
Then solve the simultaneous equations for these 2 perpendicular bisectors.
The solution is the x and y co-ordinates of the circumcentre.

If you get stuck, just ask again..

8. i did up to number 3
so i got
18/13 and -7/67
and now i dont know what to do for number 4

9. You've got the slopes of the perpendicular lines to YZ and XZ.

There are an infinite amount of perpendiculars, however.

You need the ones going through the midpoints of those lines.

The midpoint of YZ is $\frac{59+52}{2},\frac{95+28}{2}=(55.5,61.5)$

The midpoint of XZ is (61,21.5)

You now write the lines equations with these slopes and points.
These 2 lines will intersect at the circle centre.

$y-61.5=-\frac{7}{67}(x-55.5)$

Write the equation of the 2nd bisector and solve the resulting simultaneous equations.