Use coordinates to prove that the nine-point center is the midpoint of the segment from the orth

Use coordinates to prove that the nine-point center is the...

Key Concepts

Midpoint Formula

The midpoint formula provides an algebraic method to find the midpoint of a line segment by averaging the coordinates of its endpoints. This formula is critical in proving that the nine-point center is exactly the midpoint of the segment joining the circumcenter and the orthocenter, thereby establishing its position on the Euler line.

Orthocenter

The orthocenter is the point where the three altitudes of a triangle intersect. In coordinate settings, its coordinates can be computed by solving the equations of the altitudes. The relationship between the orthocenter and other centers like the circumcenter and nine-point center is a fundamental concept in triangle geometry.

Circumcenter

The circumcenter is the center of the circle that passes through all three vertices of the triangle (circumcircle) and is equidistant from these vertices. In coordinate geometry, it is determined by finding the intersection point of the perpendicular bisectors of the triangle’s sides, serving as an essential triangle center in many geometric proofs.

Nine-Point Circle

The nine-point circle of a triangle is a circle that passes through nine significant points: the three midpoints of the sides, the three feet of the altitudes, and the three midpoints of the segments joining the vertices to the orthocenter. The center of this circle, called the nine-point center, has a special position relative to other triangle centers.

Coordinate Geometry

Coordinate geometry involves using a coordinate system to represent geometric figures and employ algebraic methods to solve geometric problems. Applying coordinates to a triangle allows for the precise computation of important points such as the circumcenter, orthocenter, and nine-point center, as well as the use of formulas like the midpoint formula.

Euler Line

The Euler line is a straight line that, in any non-equilateral triangle, contains the circumcenter, centroid, and orthocenter. Additionally, the nine-point center lies on this line. In the context of coordinate proofs, showing that the nine-point center is the midpoint of the segment joining the circumcenter and orthocenter is a key step in demonstrating its alignment on the Euler line.

Transcript

00:01 So for this question we're asked to show that the midpoint of the orthocenter and the circumcenter of a triangle is the nine point center for the triangle.

00:10 So the orthocenter we're given in the text, the formula for the orthocenter is 0, negative ac over b if our triangle looks like this.

00:48 And so a is at a comma 0, c is at c comma 0, and b is at 0 comma b.

01:08 We're also told that the nine point center, which is what this midpoint should equal, is a plus c over 4, comma b squared minus a c over 4b.

01:31 We're not, however, told what the circumcenter is.

01:34 So first we're going to have to find the circumcenter.

01:37 Now the circumcenter is the intersection of perpendicular bisectors.

01:43 So to find the circumcenter, we're going to have to first find the equation of two perpendicular bisectors to sides of this triangle.

01:49 And then from there, we can see where they intersect.

01:52 That'll be the circumcenter.

01:53 And then we'll use that on the orthocenter to find the midpoint and compare that to the nine point center.

01:59 So if we're finding the circumcenter, first we'll have the perpendicular bisector of ac, because since ac is a horizontal line, this is going to be a vertical line, so it's relatively easy to find.

02:23 It's just a vertical line through the midpoint of that side, and that midpoint is going to be halfway between a and c, so it's a plus c over 2, comma, 0.

02:34 This is just going to be x equals a plus c over 2.

02:42 Next we'll find the perpendicular bisector of another point, say ab.

02:47 It's going to take a little bit more work because this is not a vertical or horizontal line.

02:58 It has to go through the midpoint of ab, which is just a over 2, comma, b over 2, because we have a plus 0 over 2 for the x coordinate and 0 plus b over 2 for the y coordinate.

03:25 And the slope will be b minus 0 over 0 minus a.

03:43 This is the slope of ab.

03:47 So that's negative b over a, meaning the slope of the perpendicular bisector will be positive a over b.

04:11 So we'll use point slope form here to get the equation of the line.

04:15 Y minus y1 equals the slope times x minus x 1.

04:26 So we have y minus b over 2 equals a over b times x minus a over 2.

04:34 We'll distribute that a over b.

04:37 We get y minus b over 2 equals a over b x minus a squared over 2 b.

04:47 And i will add b over 2 to both sides.

04:54 We get y equals a over bx minus a squared over 2b plus b over 2...