Question

Triangle ABC has vertices A(0,6) B(4,6) C(1,3). Sketch a graph of ABC and use it to find the orthocenter of ABC. Then list the steps you took to find the orthocenter, including any necessary points or slope you had to derive.

Answer 1

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.

Answer 2

The orthocenter of a triangle in the point of concurrency of the altitudes of

a triangle.

• The coordinates of the orthocenter of triangle ΔABC is $\underline{(1, 5)}$

Steps to find the orthocenter:

The given vertices are; A(0, 6), B(4, 6), and C(1, 3)

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Step 1; Find the slopes of at least two altitudes of the triangle

The slope of an altitude is equal to $-\dfrac{1}{m}$, where m is the slope of the

opposite side of the triangle.

$Slope, \, \overline{AB} =\dfrac{6-6}{4-0} = 0$, the altitude line from C to $\overline{AB}$ is line x = 1

$Slope, \, \overline{BC} =\dfrac{6-3}{4-1} = 1$, slope of altitude from A to $\overline{BC}$ = -1

Step 2: Find the equation of altitude from A to $\overline{BC}$ in point and slope form:

The equation of the altitude from A to $\overline{BC}$ in point and slope form is

therefore;

y - 6 = (-1)×(x - 0) = -x

y = 6 - x

Step 3: Find the point of intersection of two or the three altitudes

The orthocenter is the point the altitudes intersect, therefore, at the

orthocenter, we have;

y = 6 - x

x = 6 - y

Given that an altitude is the line x = 1, at the point of intersection, (at the

orthocenter) we have;

x = 1 and x = 6 - y by transitive property of equality, we have;

1 = 6 - y

y = 6 - 1 = 5

y = 5

Therefore, at the orthocenter, x = 1, y = 5, the coordinates of the orthocenter is $\underline{(1, 5)}$

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