Triangle ABC has vertices A(0,6), B(4,6) and C(1,3) Find the orthocenter of triangle ABC.
Please show your work so that I can understand how to do this type of problem. Thank You I really appreciate your help ^_^
To find the orthocenter of triangle ABC, we need to first find the equations of the three altitudes and then find their point of intersection.
Step 1: Find the slopes of the sides of the triangle.
The slope of AB can be found using the formula:
slope_AB = (y2 - y1) / (x2 - x1)
= (6 - 6) / (4 - 0)
= 0 / 4
= 0
Similarly, the slope of BC can be found using the formula:
slope_BC = (y2 - y1) / (x2 - x1)
= (6 - 3) / (4 - 1)
= 3 / 3
= 1
The slope of AC can be found using the formula:
slope_AC = (y2 - y1) / (x2 - x1)
= (3 - 6) / (1 - 0)
= -3 / 1
= -3
Step 2: Find the slopes of the altitudes.
The slope of the altitude from A can be found by taking the negative reciprocal of the slope of BC:
slope_altitude_A = -1 / slope_BC
= -1 / 1
= -1
The slope of the altitude from B can be found by taking the negative reciprocal of the slope of AC:
slope_altitude_B = -1 / slope_AC
= -1 / (-3)
= 1/3
The slope of the altitude from C can be found by taking the negative reciprocal of the slope of AB:
slope_altitude_C = -1 / slope_AB
= -1 / 0 (Note that the slope of AB is 0, which means it's a vertical line)
= undefined
Step 3: Find the equations of the altitudes.
The equation of a line can be written in the point-slope form: y - y1 = m(x - x1), where m is the slope of the line.
Using this formula, the equation of the altitude from A can be written as:
y - 6 = (-1)(x - 0)
y - 6 = -x
Similarly, the equation of the altitude from B can be written as:
y - 6 = (1/3)(x - 4)
3y - 18 = x - 4
x - 3y + 14 = 0
And the equation of the altitude from C can be written as:
x - 1 = 0 (since the slope is undefined, the line is a vertical line passing through the point C(1,3))
Step 4: Find the intersection point of the altitudes.
To find the orthocenter, we need to find the point where all three altitudes intersect.
The intersection can be found by solving the simultaneous equations of the altitudes.
Solving the equations of the altitudes:
y - 6 = -x ........(1)
x - 3y + 14 = 0 ........(2)
x - 1 = 0 ........(3)
From equation (3), we have x = 1. Substituting this into equations (1) and (2) gives:
y - 6 = -1
y = 5
x - 3(5) + 14 = 0
x - 15 + 14 = 0
x - 1 = 0
So, the intersection point of the altitudes (and hence the orthocenter) is (1, 5).
Therefore, the orthocenter of triangle ABC is H(1, 5).
To find the orthocenter of triangle ABC, we need to find the intersection point of the altitudes of the triangle. An altitude is a line segment that passes through a vertex of the triangle and is perpendicular to the opposite side.
Step 1: Find the slopes of the sides of the triangle.
The slope of a line passing through two points (x1, y1) and (x2, y2) is given by (y2 - y1) / (x2 - x1).
Slope of AB:
(x1, y1) = (0, 6)
(x2, y2) = (4, 6)
slope_AB = (6 - 6) / (4 - 0) = 0 / 4 = 0
Slope of BC:
(x1, y1) = (4, 6)
(x2, y2) = (1, 3)
slope_BC = (3 - 6) / (1 - 4) = -3 / -3 = 1
Slope of AC:
(x1, y1) = (0, 6)
(x2, y2) = (1, 3)
slope_AC = (3 - 6) / (1 - 0) = -3 / 1 = -3
Step 2: Find the equations of the perpendicular lines to the sides passing through the vertices.
For AB, the perpendicular line passes through point C(1, 3). The slope of a line perpendicular to another line with slope m is -1/m.
slope_perpendicular_AB = -1/0 (since slope_AB = 0, the perpendicular slope is undefined)
So, the equation of the line perpendicular to AB passing through C(1, 3) is x = 1.
For BC, the perpendicular line passes through point A(0, 6). The slope of the line is -1.
So, the equation of the line perpendicular to BC passing through A(0, 6) is y - 6 = -1(x - 0).
y = -x + 6
For AC, the perpendicular line passes through point B(4, 6). The slope of the line is -1/1 = -1.
So, the equation of the line perpendicular to AC passing through B(4, 6) is y - 6 = -1(x - 4).
y = -x + 10
Step 3: Find the point of intersection of the altitudes.
Now that we have the equations of the three lines, we can find the point of intersection by solving the system of equations.
Since the line AB is vertical (x = 1), it intersects with the line y = -x + 6 at x = 1 and y = 5. So, the point of intersection is (1, 5).
Similarly, the line BC (y = -x + 6) and AC (y = -x + 10) intersect at x = 2 and y = 4. So, the point of intersection is (2, 4).
Therefore, the orthocenter of triangle ABC is at point (2, 4).
Note: In this example, the orthocenter is inside the triangle. However, the orthocenter can also be outside the triangle if the triangle is obtuse.
Ok Thank You ^_^!!!!
the orthocentre is the intersection of the altutides.
Method:
Find the equation of two of those altitudes.
Solve the two equations to find their intersection point.
How?
Make a rough sketch
Pick any point and find the slope of the opposite side.
The slope of the altitude to that side is the negative reciprocal of the slope of that side.
Now you have the slope and a point on that line, find the equation for the line.
Repeat the above for a second altitude, solve the two equations.
PS, just noticed how nice your points are. One of the lines is a horizontal line, so the altitude from (1,3) to that line is x = 1