ABC has vertices A(0, 6), B(4, 6), and 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 slopes you had to derive. (3 points)
To find the orthocenter of triangle ABC, we need to first sketch the graph of triangle ABC, and then find the intersection of the altitudes.
Plotting the given points A(0, 6), B(4, 6), and C(1, 3) in a coordinate plane, we have:
The coordinates of vertices A, B, and C are as follows:
A(0, 6), B(4, 6), C(1, 3).
Now, let's find the equation of the line passing through A(0, 6) and perpendicular to the line containing B(4, 6) and C(1, 3).
The slope of the line containing B(4, 6) and C(1, 3) can be calculated as:
m = (y2 - y1) / (x2 - x1)
= (3 - 6) / (1 - 4)
= (-3) / (-3)
= 1
The slope of the line perpendicular to this line will be -1/m = -1/1 = -1.
Using the point-slope form of a line, where the point is A(0, 6) and the slope is -1:
y - y1 = m(x - x1)
y - 6 = -1(x - 0)
y - 6 = -x
Converting the equation to slope-intercept form gives:
y = -x + 6
This is the equation of one of the altitudes of triangle ABC.
Similarly, let's find the equation of the line passing through B(4, 6) and perpendicular to the line containing A(0, 6) and C(1, 3).
The slope of the line containing A(0, 6) and C(1, 3) can be calculated as:
m = (y2 - y1) / (x2 - x1)
= (3 - 6) / (1 - 0)
= (-3) / (1)
= -3
The slope of the line perpendicular to this line will be -1/m = -1/-3 = 1/3.
Using the point-slope form of a line, where the point is B(4, 6) and the slope is 1/3:
y - y1 = m(x - x1)
y - 6 = (1/3)(x - 4)
y - 6 = (1/3)x - 4/3
Converting the equation to slope-intercept form gives:
y = (1/3)x - 4/3 + 6
y = (1/3)x + 14/3
This is the equation of another altitude of triangle ABC.
Finally, let's find the equation of the line passing through C(1, 3) and perpendicular to the line containing A(0, 6) and B(4, 6).
The slope of the line containing A(0, 6) and B(4, 6) is 1.
The slope of the line perpendicular to this line will be -1/m = -1/1 = -1.
Using the point-slope form of a line, where the point is C(1, 3) and the slope is -1:
y - y1 = m(x - x1)
y - 3 = -1(x - 1)
y - 3 = -x + 1
Converting the equation to slope-intercept form gives:
y = -x + 4
This is the equation of the third altitude of triangle ABC.
Now, we have three equations representing the altitudes of triangle ABC:
1) y = -x + 6
2) y = (1/3)x + 14/3
3) y = -x + 4
Next, we need to find the intersection points of these lines. By solving the system of equations, we can find the coordinates of the orthocenter.
Solving equations 1) and 2) yields:
-x + 6 = (1/3)x + 14/3
3(-x + 6) = 3((1/3)x + 14/3)
-3x + 18 = x + 14
-4x = -4
x = 1
Substituting x = 1 into equation 1) gives:
y = -1 + 6
y = 5
Therefore, the first intersection point is (1, 5).
Solving equations 2) and 3) yields:
(1/3)x + 14/3 = -x + 4
(1/3)x + x = 4 - 14/3
(3/3)x + (3/3)x = 12/3 - 14/3
6/3x = -2/3
x = -1/3
Substituting x = -1/3 into equation 2) gives:
y = (1/3)(-1/3) + 14/3
y = -1/9 + 14/3
y = -1/9 + 42/9
y = 41/9
Therefore, the second intersection point is (-1/3, 41/9).
Finally, solving equations 1) and 3) yields:
-y + 6 = -x + 4
-y = -x - 2
y = x + 2
Substituting y = x + 2 into equation 1) gives:
x + 2 = -x + 6
2x = 4
x = 2
Substituting x = 2 into equation 3) gives:
y = -2 + 4
y = 2
Therefore, the third intersection point is (2, 2).
Hence, the orthocenter of triangle ABC is the intersection point of the altitudes, which is (2, 2).