Find the coordinates of the orthocenter of a triangle with vertices at each set of points on a coordinate plane. a. (0,0), (7,3), (3,51)
b. (5,3), (8,6), (4,10)
.
a. The orthocenter is .
(Type an ordered pair. Simplify your answer.)

b. The orthocenter is.
(Type an ordered pair. Simplify your answer.)

9514 1404 393

  a. (24, 2)

  b. (8, 6)

Step-by-step explanation:

The orthocenter is the point where the altitudes of the triangle coincide. For an obtuse triangle, it will be outside the triangle. For a right triangle, it will be the right angle vertex. Of course, each altitude is a line through a vertex and perpendicular to the opposite side.

To make finding the orthocenter as easy as possible, we want to find a simple way to write an equation for a line through a point and perpendicular to the segment between two other points. For segment end points (x1, y1) and (x2, y2), the line between them will have slope (y2 -y1)/(x2 -x1). The perpendicular line has a slope that is the opposite reciprocal of this: -(x2 -x1)/(y2 -y1). Then the point-slope equation of the perpendicular through point (x3, y3) is ...

  y -y3 = -(x2 -x1)/(y2 -y1)(x -x3)

This can be rearranged to something resembling standard form:

  (x2 -1x)(x -x3) +(y2 -y1)(y -y3) = 0

  (x2 -x1)x +(y2 -y1)y = (x2 -x1)x3 +(y2 -y1)y3 . . . . line through P3, ⊥ to P1P2

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For the first altitude, we choose P1 = (0, 0), P2 = (7, 3), and P3 = (3, 51). This gives ...

For the second altitude, we choose P1 = (0, 0), P2 = (3, 51), and P3 = (7, 3). This gives ...

Finding the point of intersection of these two equations can be done an of a number of ways. Cramer's Rule provides a quick solution:

  x = ((3)(-174) -(51)(-174))/(7·51 -3·3) = 8352/348 = 24

  y = ((-174)(3) -(-174)(7))/348 = 696/348 = 2

The orthocenter for the triangle is (24, 2).

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We can use the method just described, or we can verify the triangle is a right triangle, as it appears to be when the points are plotted.

The slope of P1P2 is (6 -3)/(8 -5) = 3/3/ = 1.

The slope of P2P3 is (10 -6)/(4 -8) = 4/-4 = -1

These slopes have a product of -1, so the segments are perpendicular. The vertex P2 has the right angle, hence is the orthocenter.

The orthocenter for the triangle is (8, 6).