Find the coordinates of the orthocenter of Triangle A B C with vertices ALeft-parenthesis 2 comma 6 right-parenthesis, BLeft-parenthesis 8 comma 6 right-parenthesis, and CLeft-parenthesis 6 comma 2 right-parenthesis.
A. Left-parenthesis 5 comma 4 right-parenthesis
B. Left-parenthesis 1 comma 2 right-parenthesis
C. Left-parenthesis 6 comma 4 right-parenthesis
D. Left-parenthesis 6 comma 8 right-parenthesis
Why not just say A(2,6) , B(8,6) and C(6,2)
we need the intersection of altitudes.
since AB is horizontal, x = 6 is the altitude from C to AB
Slope of AC = -1, so slope of altitude from B to AC = +1
equaltion of that altitude: y = x + b
(8,6) lies on it
6 = 8 + b , b = -2
altitude from B to AC is y = x - 2
but x = 6
then y = 6-2 = 4
the orthocentre is (6,4)
To find the coordinates of the orthocenter of triangle ABC, the following steps can be followed:
Step 1: Calculate the slopes of the lines AB, BC, and AC.
The slope of a line passing through two points (x1, y1) and (x2, y2) is given by:
slope = (y2 - y1) / (x2 - x1)
Using the given coordinates:
Slope of AB = (6 - 6) / (8 - 2) = 0/6 = 0
Slope of BC = (6 - 2) / (8 - 6) = 4/2 = 2
Slope of AC = (6 - 2) / (8 - 2) = 4/6 = 2/3
Step 2: Calculate the negative reciprocals of the slopes.
The negative reciprocal of a slope 'm' is given by:
negative reciprocal = -1/m
Negative reciprocal of AB = -1/0 (undefined)
Negative reciprocal of BC = -1/2
Negative reciprocal of AC = -3/2
Step 3: Find the equations of the perpendicular bisectors for each side.
The equation of a perpendicular bisector passing through the midpoint (x, y) of two points (x1, y1) and (x2, y2) is given by:
y = negative reciprocal * (x - (x1 + x2)/2) + (y1 + y2)/2
For the three sides AB, BC, and AC:
Equation of the perpendicular bisector of AB: y = (undefined) * (x - (2 + 8)/2) + (6 + 6)/2
As the slope of AB is undefined, the line is vertical and has an equation of the form x = constant, passing through the midpoint of (2 + 8)/2 = 5
So, the equation of the perpendicular bisector of AB is x = 5.
Equation of the perpendicular bisector of BC: y = -1/2 * (x - (8 + 6)/2) + (6 + 2)/2
y = -1/2 * (x - 7) + 4
y = -1/2 * x + 7/2 + 4
y = -1/2 * x + 15/2
Equation of the perpendicular bisector of AC: y = -3/2 * (x - (2 + 6)/2) + (6 + 2)/2
y = -3/2 * (x - 4) + 4
y = -3/2 * x + 6 + 4
y = -3/2 * x + 10
Step 4: Calculate the point of intersection of any two perpendicular bisectors.
To find the orthocenter, we need to find the intersection of any two perpendicular bisectors. Let's find the intersection point of the lines x = 5 and y = -1/2 * x + 15/2.
Substituting the equation x = 5 into y = -1/2 * x + 15/2:
y = -1/2 * 5 + 15/2
y = -5/2 + 15/2
y = 10/2 = 5
Therefore, the intersection point of x = 5 and y = -1/2 * x + 15/2 is (5, 5).
Step 5: Verify the intersection point lies within the triangle.
To verify if the intersection point (5, 5) lies within the triangle, we can check if the coordinates satisfy the inequalities of the slopes.
For AB: 0 < 2 (PASSED)
For BC: 2 < 2 (NOT PASSED)
For AC: 2/3 < 2 (PASSED)
Since the point (5, 5) does not satisfy all the inequalities, it is not the orthocenter of the triangle.
Therefore, the correct answer is none of the given options.
To find the coordinates of the orthocenter of Triangle ABC, we can use the property that the orthocenter is the point of intersection of the altitudes of the triangle.
An altitude of a triangle is a line segment drawn from a vertex perpendicular to the opposite side. To find the equations of the altitudes, we need to determine the slopes of the lines containing the sides of the triangle.
Let's start by finding the slopes of the sides AB, BC, and AC.
The slope of AB is given by (y2 - y1)/(x2 - x1) = (6 - 6)/(8 - 2) = 0/6 = 0.
The slope of BC is given by (y2 - y1)/(x2 - x1) = (6 - 2)/(8 - 6) = 4/2 = 2.
The slope of AC is given by (y2 - y1)/(x2 - x1) = (2 - 6)/(6 - 2) = -4/4 = -1.
Now, let's find the equations of the lines containing the altitudes.
The altitude from vertex A to side BC has a slope of -1/2 (negative reciprocal of the slope of BC). Using point-slope form, the equation of this line is y - 6 = (-1/2)(x - 8), which simplifies to y = (-1/2)x + 10.
The altitude from vertex B to side AC has a slope of 2 (reciprocal of the slope of AC). Using point-slope form, the equation of this line is y - 6 = 2(x - 8), which simplifies to y = 2x - 10.
The altitude from vertex C to side AB has a slope of 0 (horizontal line). The equation of this line is simply y = 2.
Now, we need to find the intersection point of these three lines. We have the following three equations:
1) y = (-1/2)x + 10
2) y = 2x - 10
3) y = 2
To find the x-coordinate, we can equate equations 1 and 2:
(-1/2)x + 10 = 2x - 10
Multiplying both sides by 2 to clear the fraction, we get:
-x + 20 = 4x - 20
Rearranging terms, we have:
5x = 40
Dividing both sides by 5, we get:
x = 8
Substituting this value of x into equation 1, we can find the y-coordinate:
y = (-1/2)(8) + 10
y = -4 + 10
y = 6
Therefore, the coordinates of the orthocenter are (8, 6).
None of the given options (A, B, C, D) match the coordinates of the orthocenter.
Therefore, the answer is none of the above options.