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
To sketch the graph of ABC, plot the given points A(0, 6), B(4, 6), and C(1, 3) on a coordinate plane.
Next, draw the triangle ABC by connecting the three points A, B, and C with line segments. Label the vertices A, B, and C accordingly.
To find the orthocenter of triangle ABC, we need to determine the point where the altitudes of the triangle intersect. The altitude of a triangle is a perpendicular line segment drawn from a vertex to the opposite side.
We can find the equations of the line segments AB, AC, and BC using the slope formula.
The slope of AB is:
m_AB = (y_B - y_A) / (x_B - x_A) = (6 - 6) / (4 - 0) = 0/4 = 0
The slope of AC is:
m_AC = (y_C - y_A) / (x_C - x_A) = (3 - 6) / (1 - 0) = -3/1 = -3
The slope of BC is:
m_BC = (y_C - y_B) / (x_C - x_B) = (3 - 6) / (1 - 4) = -3/-3 = 1
Since the slopes of AB, AC, and BC are 0, -3, and 1, respectively, the lines AB, AC, and BC are horizontal, downward-sloping, and upward-sloping, respectively.
To find the equations of the lines that are perpendicular to AB, AC, and BC and pass through the opposite vertices, we need to find the negative reciprocal of the slopes.
The negative reciprocal of m_AB is 1/0, which is undefined for a vertical line. Since AB is a horizontal line passing through point (0, 6), the equation for the line perpendicular to AB and passing through point B(4, 6) is x = 4.
The negative reciprocal of m_AC is 1/(-3) = -1/3. The line perpendicular to AC and passing through point C(1, 3) has a slope of -1/3. We can find the equation in point-slope form:
y - y_C = (-1/3)(x - x_C)
y - 3 = (-1/3)(x - 1)
3y - 9 = -x + 1
x + 3y = 10
The negative reciprocal of m_BC is -1. The line perpendicular to BC and passing through point A(0, 6) has a slope of -1. we can find the equation in point-slope form:
y - y_A = (-1)(x - x_A)
y - 6 = -1(x - 0)
y - 6 = -x
x + y = 6
Now, we have three equations: x = 4, x + 3y = 10, and x + y = 6. We can find the intersection point of these lines by solving the system of equations.
Using the equation x = 4, substitute x = 4 into the other two equations:
4 + 3y = 10
3y = 6
y = 2
4 + 2 = 6
The intersection point of x = 4 and x + 3y = 10 is (4, 2).
Using the equation x + 3y = 10, substitute x = 4 into the equation:
4 + y = 6
y = 2
The intersection point of x = 4 and x + y = 6 is (4, 2).
Thus, the orthocenter of triangle ABC is the point (4, 2).