I keep getting 0/4. How do you use slope?

Do you use C in point slope form only?

I keep getting 0/4. How do you use slope?

Do you use C in point slope form only?

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Bill F. | Experienced Teacher & Tutor in Round Rock, TXExperienced Teacher & Tutor in Round Roc...

Hi Mara! The orthocenter is where three lines cross: these lines come from each triangle vertex (corner) and go through the triangle and make a perdendicular (at a right angle) intersect with the side opposite the angle each came from. In this case, since this is an acute triangle, they all cross each other inside the triangle. The "trick" is finding the x,y coordinates of that intersection, and so we need to find the equations of those lines to know where they intersect. While doing that, we will also get the slope of each of those lines (3 slopes, not just one).

To find the orthocenter coordinates, we first need to know the slopes of the sides of the triangle. Using those, we can find the slopes of those lines perpendicular to the sides that make-up the orthocenter.

We can use y=mx+b, where m = the slope, which is (change in y, or Δy) / (change in x, or Δx):

Slope of side AB: A = (0,6) and B = (4,6). Δy = 6-6=0, and Δx = 0-4=-4. Δy/Δx = 0/-4 = 0

Slope of side AC: A = (0,6) and C = (1,3). Δy = 6-3=3, and Δx = 0-1=-1. Δy/Δx = 3/-1 = -3

Slope of side BC: B = (4,6) and C = (1,3). Δy = 6-3=3, and Δx = 4-1=3. Δy/Δx = 3/3 = +1

By definition, the slope of a perdendicular line is the negative reciprocal of the original line slope. For example, an original line slope of 4 means the slope of it's perpendicular is -1/4. So for each of our triangle sides:

The slope of line AB = 0, so the **slope of the line from point C perpendicular to AB = 1/0 = undefined (vertical line parallel to the y-axis)**

The slope of line AC = -3, so the **slope of the line from point B perpendicular to AC = +1/3**

The slope of line BC = +1, so the **slope of the line from point A perpendicular to BC = -1/1 = -1**

We know the slope direction (m) of each line, and an x and y coordinate (the corner origin of each line). Let's get the equation for at least two of them and see where they intersect (this will be the orthocenter):

For line B to AC: y - 6 = (1/3)(x - 4); y - 6 = (x/3) - (4/3); 3y - 18 = x - 4, so
**3y - x = 14**

For line A to BC: y - 6 = (-1)(x - 0); y - 6 = -x, so **y + x = 6**

Since these lines intersect at one point (the orthocenter), we can use simultaneous equations to solve for x and/or y:

(3y - x = 14) + (y + x = 6) => 4y = 20, **y = +5**; Substitute this into y + x = 6: 5 + x = 6,
**x = +1**

**So the orthocenter is at coordinates (1,5), and the slopes of all three orthocenter lines are above.**

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