triangle ABC has vertices (0,0), (10,5), (5,65) what is the orthocenter?

Answer 1

The orthocenter is the point of intersection of the three altitudes of the triangle. Since all three altitudes meet at the same point the intersection of two of the altitudes is sufficient to determine the coordinates of the orthocenter.

Let m₁ represent the slope of the line passing through (0,0), and (10,5).

Let m₂ represent the slope of the line passing through (0,0) and (5,65).

m₁ = 5/10 = 1/2

m₂ = 13

The altitude from (5,65) is perpendicular to the side with endpoints (0,0) and (10,5) and that side has a slope of 1/2. The slope of that altitude is -2 (since perpendicular lines have slopes that are negative reciprocals). Let's reference that altitude as l₁. The equation of l₁: y - 65 = -2(x - 5) or y = -2x +75.

The altitude from (10,5) is perpendicular to the side with slope 13. So its equation (l₂) is y - 5 = (-1/13)(x -10) or

y = (-1/13)x +10/13 + 65/13

y = (-1/13)x + 75/13

Use the substitution method to determine the point of intersection of l₁ and l₂.

(-1/13)x + 75/13 = -2x +75 (now multiply every term 13 to clear fractions)

-x + 75 = -26x + 75(13)

25x = 75(13) - 75

x = [75(13 - 1) / 25] = 3(12) = 36

When x = 36:

y = -2(36) + 75 = -72 + 75 = 3

The coordinates of the orthocenter: (36,3). This point is in the exterior of the triangle.

Try finding the equation of the altitude from (0,0) perpendicular to its opposite side (that has a slope of -12). Then find its point of intersection with y = -2x + 75. The result should be (36,3).

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