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 find the orthocenter of triangle ABC, we need to find the point where the altitudes of the triangle intersect.
Step 1: Sketch the triangle ABC
Using the given coordinates, we can plot the points on a graph:
A(0, 6)
B(4, 6)
C(1, 3)
Step 2: Draw the altitudes
The altitude of a triangle is a perpendicular line from one vertex to the opposite side. We need to find the equations of the three altitudes.
To find the equation of the altitude from A to BC:
The slope of line BC can be found using the coordinates of points B and C. The slope formula is given by:
slope of BC (mBC) = (y2 - y1) / (x2 - x1)
mBC = (6 - 3) / (4 - 1)
mBC = 3/3
mBC = 1
Since the altitude from A is perpendicular to BC, the slope of this altitude will be the negative reciprocal of mBC.
Slope of altitude from A to BC = -1/mBC = -1/1 = -1
To find the equation of the altitude from A(0, 6) with slope -1:
Using the point-slope form of a linear equation:
y - y1 = m(x - x1)
y - 6 = -1(x - 0)
y - 6 = -x
y = -x + 6 (equation of altitude from A to BC)
Similarly, we can find the equations of the altitudes from B and C:
Slope of altitude from B to AC = -1,
Equation: y = 6 (horizontal line passing through B)
Slope of altitude from C to AB = -1/2,
Equation: y = -x + 9 (using point-slope form with point C(1, 3) and slope -1/2)
Step 3: Find the intersection point of the altitudes
To find the orthocenter, we need to find the point where all three altitudes intersect. We can do this mathematically by solving the system of equations formed by the three altitudes.
Solving y = -x + 6 and y = 6, we get:
-x + 6 = 6
-x = 0
x = 0
Substituting x=0 back into any of the equations, we get y = 6.
Therefore, the orthocenter of triangle ABC is H(0, 6).
Step 4: Plot the orthocenter on the graph
Plot the point H(0, 6) on the graph.
Step 5: Connect the orthocenter to the vertices
Draw the lines HA, HB, and HC to complete the triangle HBC.
The final graph should show triangle ABC with its altitudes and orthocenter H(0, 6).
To find the orthocenter of triangle ABC, follow these steps:
Step 1: Sketch the given triangle ABC using the provided vertices.
First, plot the points A(0, 6), B(4, 6), and C(1, 3) on a coordinate plane.
Step 2: Find the slopes of the three sides of triangle ABC.
The slope of a line passing through two points (x1, y1) and (x2, y2) can be calculated using the formula: slope = (y2 - y1) / (x2 - x1).
Let's calculate the slopes of the three sides AB, BC, and AC:
Slope of AB:
(x1, y1) = (0, 6)
(x2, y2) = (4, 6)
slope = (6 - 6) / (4 - 0) = 0 / 4 = 0
Slope of BC:
(x1, y1) = (4, 6)
(x2, y2) = (1, 3)
slope = (3 - 6) / (1 - 4) = -3 / -3 = 1
Slope of AC:
(x1, y1) = (0, 6)
(x2, y2) = (1, 3)
slope = (3 - 6) / (1 - 0) = -3 / 1 = -3
Step 3: Find the negative reciprocal of the slopes.
To find the perpendicular slopes of the sides, invert the slopes and change the sign.
Perpendicular slope of AB: 0
Perpendicular slope of BC: -1/1 = -1
Perpendicular slope of AC: -1/-3 = 1/3
Step 4: Find the equations of the perpendicular bisectors for each side.
The perpendicular bisector of a line segment passes through the midpoint of the line segment and is perpendicular to it.
Find the midpoint and use the perpendicular slope to determine the equation of the perpendicular bisector for each side.
Equation for the perpendicular bisector of AB:
Midpoint of AB = [(0 + 4) / 2, (6 + 6) / 2] = (2, 6)
Using the slope-intercept form (y = mx + b), where m is the perpendicular slope and (x, y) is a point on the line:
y = (0)(x - 2) + 6
y = 6
Equation for the perpendicular bisector of BC:
Midpoint of BC = [(4 + 1) / 2, (6 + 3) / 2] = (5/2, 9/2)
Using the slope-intercept form (y = mx + b), where m is the perpendicular slope and (x, y) is a point on the line:
y = (-1)(x - 5/2) + 9/2
y = -x + 8
Equation for the perpendicular bisector of AC:
Midpoint of AC = [(0 + 1) / 2, (6 + 3) / 2] = (1/2, 9/2)
Using the slope-intercept form (y = mx + b), where m is the perpendicular slope and (x, y) is a point on the line:
y = (1/3)(x - 1/2) + 9/2
y = (1/3)x + 7/6
Step 5: Find the point of intersection of any two perpendicular bisectors.
Select any two perpendicular bisectors and solve their equations simultaneously to find their point of intersection, which is the orthocenter.
Let's find the point of intersection for the perpendicular bisectors of AB and BC:
Equation of the perpendicular bisector of AB: y = 6
Equation of the perpendicular bisector of BC: y = -x + 8
Solving these equations simultaneously, we get:
6 = -x + 8
-x = 2
x = -2
Substituting x = -2 back into one of the equations, we get:
y = 6
Therefore, the point of intersection is (-2, 6), which is the orthocenter of triangle ABC.
Step 6: Plot the orthocenter on the triangle's graph.
Finally, plot the orthocenter (-2, 6) on the sketched graph of triangle ABC.
We have now found the orthocenter of triangle ABC.
To sketch a graph of triangle ABC, we plot the three given vertices: A(0, 6), B(4, 6), and C(1, 3) on a coordinate plane.
Now, let's find the orthocenter of triangle ABC. The orthocenter is the point of intersection of the altitudes of a triangle. To find it, we need to determine the equations of the three altitudes.
Step 1: Find the slopes of the sides of the triangle
- The slope of side AB = (change in y / change in x) = (6 - 6) / (4 - 0) = 0
- The slope of side BC = (6 - 3) / (4 - 1) = 1
- The slope of side AC = (3 - 6) / (1 - 0) = -3
Step 2: Find the perpendicular slopes (negative reciprocals) of the sides
- The perpendicular slope to AB is undefined (as the slope of AB is 0)
- The perpendicular slope to BC = -1 (negative reciprocal of 1)
- The perpendicular slope to AC = 1/3 (negative reciprocal of -3)
Step 3: Find the equation of the altitude passing through A
- Since side BC is opposite to vertex A, the altitude from A is perpendicular to side BC.
- Using the point-slope form of a line (y - y1 = m(x - x1)), where (x1, y1) = (0, 6) and m = 1/3, we have y - 6 = (1/3)(x - 0) = (1/3)x
Simplifying, we get y = (1/3)x + 6
Step 4: Find the equation of the altitude passing through B
- Similarly, the altitude from B is perpendicular to side AC.
- Using the point-slope form with (x1, y1) = (4, 6) and m = -3 (negative reciprocal of AC's slope), we have y - 6 = -3(x - 4) = -3x + 12
Simplifying, we get y = -3x + 18
Step 5: Find the equation of the altitude passing through C
- For the altitude from C, we need to find the equation of a line perpendicular to side AB passing through the point (1, 3).
- Since the slope of AB is 0, the perpendicular slope is undefined (vertical line).
- Therefore, the equation of the altitude passing through C is x = 1
Step 6: Find the point of intersection of the altitudes (orthocenter)
- We can find the orthocenter by finding the point of intersection of any two of the altitudes.
- From Step 3, we have the equation y = (1/3)x + 6
- From Step 4, we have the equation y = -3x + 18
To find the point of intersection, we set these equations equal to each other and solve for x:
(1/3)x + 6 = -3x + 18
Multiply both sides by 3 to eliminate fractions:
x + 18 = -9x + 54
10x = 36
x = 3.6
Substitute this value back into either of the equations to find the y-coordinate:
y = (1/3)(3.6) + 6
y = 1.2 + 6
y = 7.2
Hence, the orthocenter of triangle ABC is approximately (3.6, 7.2).
Finally, we can plot the orthocenter on the graph of triangle ABC.