Question

find the distance between the circumcentre and orthocentre of the triangle formed by the vertices (3,4)(3,-2), and (5,4)

Answer 1

Answer:

√10

Step-by-step explanation:

Let's put A = (3, 4),  B = (3, -2),  C = (5, 4).

There's a little observation here that saves a lot of work.  Namely, the x coordinates of A and B are the same, and the y coordinates of x and z are the same.  This means that AB is vertical and AC is horizontal, so the triangle ABC is a right angled triangle, with the right angle being at A.

It follows that A is the orthocentre of the triangle (the altitude from C to AB meets AB at A, and the altitude from B to AC meets AC at A).

Also, being a right angled triangle, it is circumscribed by a circle with diameter BC.  This means the circumcentre is the midpoint of BC and so is at

D = (B + C) / 2 = (4, 1).

The distance between the orthocentre and the circumcentre is then just the distance between A and D, and this is

√( (3-4)² + (4-1)² ) = √(1+9) = √10.

Answer 2

Step-by-step explanation