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)
To find the orthocenter of a triangle, we need to find the intersection of the altitudes.
First, let's find the equation of the line passing through points X(2, 5) and Y(8, 5). This line will be perpendicular to the side YZ.
The slope of the side YZ is given by:
m1 = (y2 - y1) / (x2 - x1) = (1 - 5) / (8 - 8) = -4 / 0 (undefined)
Since the slope is undefined, we know that the side YZ is a vertical line. Therefore, the altitude from vertex X to the side YZ is a horizontal line passing through X(2, 5).
The equation of this line is: y = 5.
Now, let's find the equation of the line passing through points Y(8, 5) and Z(8, 1). This line will be perpendicular to the side XZ.
The slope of the side XZ is given by:
m2 = (y2 - y1) / (x2 - x1) = (1 - 5) / (8 - 2) = -4 / 6 = -2 / 3
Since the slope of the side XZ is -2/3, the slope of the altitude from vertex Y to the side XZ will be the negative reciprocal of -2/3.
The negative reciprocal of -2/3 is 3/2.
Therefore, the equation of the line passing through Y(8, 5) with a slope of 3/2 is given by:
y - 5 = (3/2)(x - 8)
Simplifying, we get: y - 5 = (3/2)x - 12
y = (3/2)x - 12 + 5
y = (3/2)x - 7
Now, let's find the point of intersection of these two lines:
Set y = 5 in the second equation to get:
5 = (3/2)x - 7
(3/2)x = 12
x = 8
Plugging this value of x into the first equation, we get:
y = 5
Therefore, the point of intersection of the altitudes is P(8, 5).
So, the coordinates of the orthocenter of triangle XYZ are (8, 5).