Triangle XYZ has vertices at X(2, 5), Y(8, 5) and Z(8, 1). Find the coordinates of the circumcenter(1 point) Responses
(5,6)
(8,5)
(6,4)
(5,3)
3 answers
The correct answer is (6,4).
Triangle XYZ has vertices at X(2, 5), Y(8, 5) and Z(8, 1). Find the coordinates of the orthocenter of triangle XYZ.(1 point) Responses
(6,4)
(8,5)
(5,6)
(5,3)
(6,4)
(8,5)
(5,6)
(5,3)
To find the orthocenter of triangle XYZ, we need to find the intersection point of the altitudes of the triangle.
The altitude of a triangle is a perpendicular line from a vertex to the opposite side.
Let's find the equation of the altitude from vertex X to side YZ:
1. Find the slope of the line YZ:
slope_YZ = (y2 - y1) / (x2 - x1)
= (5 - 1) / (8 - 8)
= 4 / 0 (undefined)
The slope is undefined because the line YZ is vertical.
2. Since the altitude is perpendicular to YZ, its slope would be the negative reciprocal of the slope_YZ.
Therefore, the slope of the altitude from X is 0.
3. Since the point X(2, 5) lies on the altitude, we can write the equation of the altitude as:
y - y1 = m(x - x1)
y - 5 = 0(x - 2)
y - 5 = 0
y = 5
Therefore, the altitude from X is the horizontal line y = 5.
Now, let's find the equation of the altitude from vertex Y to side XZ:
1. Find the slope of the line XZ:
slope_XZ = (y2 - y1) / (x2 - x1)
= (5 - 1) / (2 - 8)
= 4 / -6
= -2/3
2. Since the altitude is perpendicular to XZ, its slope would be the negative reciprocal of the slope_XZ.
Therefore, the slope of the altitude from Y is 3/2.
3. Since the point Y(8, 5) lies on the altitude, we can write the equation of the altitude as:
y - y1 = m(x - x1)
y - 5 = (3/2)(x - 8)
y - 5 = (3/2)x - 12
y = (3/2)x - 7
Now, let's find the intersection point of the two altitudes:
Since the first altitude is a horizontal line y = 5, and the second altitude is a line with slope (3/2) and y-intercept -7, we can set the two equations equal to each other and solve for x:
5 = (3/2)x - 7
(3/2)x = 12
3x = 24
x = 8
Plugging this value back into either equation, we find that the y-coordinate is 5.
Therefore, the orthocenter of triangle XYZ is the point (8, 5).
The correct answer is (8, 5).
The altitude of a triangle is a perpendicular line from a vertex to the opposite side.
Let's find the equation of the altitude from vertex X to side YZ:
1. Find the slope of the line YZ:
slope_YZ = (y2 - y1) / (x2 - x1)
= (5 - 1) / (8 - 8)
= 4 / 0 (undefined)
The slope is undefined because the line YZ is vertical.
2. Since the altitude is perpendicular to YZ, its slope would be the negative reciprocal of the slope_YZ.
Therefore, the slope of the altitude from X is 0.
3. Since the point X(2, 5) lies on the altitude, we can write the equation of the altitude as:
y - y1 = m(x - x1)
y - 5 = 0(x - 2)
y - 5 = 0
y = 5
Therefore, the altitude from X is the horizontal line y = 5.
Now, let's find the equation of the altitude from vertex Y to side XZ:
1. Find the slope of the line XZ:
slope_XZ = (y2 - y1) / (x2 - x1)
= (5 - 1) / (2 - 8)
= 4 / -6
= -2/3
2. Since the altitude is perpendicular to XZ, its slope would be the negative reciprocal of the slope_XZ.
Therefore, the slope of the altitude from Y is 3/2.
3. Since the point Y(8, 5) lies on the altitude, we can write the equation of the altitude as:
y - y1 = m(x - x1)
y - 5 = (3/2)(x - 8)
y - 5 = (3/2)x - 12
y = (3/2)x - 7
Now, let's find the intersection point of the two altitudes:
Since the first altitude is a horizontal line y = 5, and the second altitude is a line with slope (3/2) and y-intercept -7, we can set the two equations equal to each other and solve for x:
5 = (3/2)x - 7
(3/2)x = 12
3x = 24
x = 8
Plugging this value back into either equation, we find that the y-coordinate is 5.
Therefore, the orthocenter of triangle XYZ is the point (8, 5).
The correct answer is (8, 5).