Triangle AOB has vertices A(4,4), O(0,0), and B(8,0). EF right bisects AB at P. GH right bisects OA at Q. Determine the coordinates of the circumcentre of triangle DEF.
I really need help, please. Today, my teacher just gave us all this work for over the weekend saying that the book would explain it all, but I've read the chapter a million time and I can't even begin to understand the question.
(4,0) would be the correct centre for the circumcentre of the original triangle ABO
in general, the circumcentre is the intersection of any two of the rightbisectors of two of the sides of the original triangle.
The right bisector of a side goes through the midpoint of that side and is perpendicular to it. I think that is why they mentioned P and Q.
e.g. to find the equation of the right bisector of OA first find Q which would be (2,2). The slope of OA is 1 so the slope of the perpendicular is -1, (the negative reciprocal)
The equation of the right bisector is then y = -x + b
sub in (2,2)
2 = -2 + b, ..... b = 4
equation is y = -x + 4
Find the equation of the right bisector of AB in the same way.
Solving, you should find that (4,0) will be the intersection of those two lines
Your question makes no sense to me.
How long is EF, do we know the endpoints E and F?
You want the circumcentre of triangle DEF, but D is not mentioned in your data.
Please restate correctly.
that's all the question says. I've tried to interpret it different ways, but I just don't know... The back of the book says the answer is (4,0), but there's no more info.
thank you so very much!
You've been so helpful!
Your suppose to use AOB, not DEF since the book make an error.
I can definitely help you understand and solve the problem. Let's break it down step by step.
Step 1: Find the equation of the line EF by finding the midpoint of AB and the negative reciprocal of AB's slope.
- The midpoint of AB can be found by using the midpoint formula:
Midpoint formula: ((x1 + x2)/2, (y1 + y2)/2)
In this case, the coordinates of A are (4, 4) and the coordinates of B are (8, 0). So the midpoint of AB is:
Midpoint of AB: ((4 + 8)/2, (4 + 0)/2) = (6, 2)
- The slope of AB can be found using the slope formula:
Slope formula: (y2 - y1)/(x2 - x1)
In this case, the coordinates of A are (4, 4) and the coordinates of B are (8, 0). So the slope of AB is:
Slope of AB: (0 - 4)/(8 - 4) = -1
- The negative reciprocal of the slope of AB is the slope of the line EF. So the slope of EF is:
Slope of EF: -(1/(-1)) = 1
- We now have the midpoint (6, 2) and the slope 1. Using the point-slope form of a line (y - y1) = m(x - x1), we can plug in the values:
Point-slope form of a line: y - y1 = m(x - x1)
In this case, the point (x1, y1) is (6, 2) and the slope m is 1. So the equation of EF is:
Equation of EF: y - 2 = 1(x - 6)
'Step 2: Repeat the same process to find the equation of GH by finding the midpoint of OA and the negative reciprocal of OA's slope.
' The coordinates of O are (0, 0) and the coordinates of A are (4, 4). So the midpoint of OA is:
Midpoint of OA: ((0 + 4)/2, (0 + 4)/2) = (2, 2)
The slope of OA can be found using the slope formula:
Slope formula: (y2 - y1)/(x2 - x1)
In this case, the coordinates of O are (0, 0) and the coordinates of A are (4, 4). So the slope of OA is:
Slope of OA: (4 - 0)/(4 - 0) = 1
The negative reciprocal of the slope of OA is the slope of the line GH. So the slope of GH is:
Slope of GH: -(1/1) = -1
We now have the midpoint (2, 2) and the slope -1. Using the point-slope form of a line (y - y1) = m(x - x1), we can plug in the values:
Point-slope form of a line: y - y1 = m(x - x1)
In this case, the point (x1, y1) is (2, 2) and the slope m is -1. So the equation of GH is:
Equation of GH: y - 2 = -1(x - 2)
Step 3: To find the circumcentre of triangle DEF, we need to find the intersection point of lines EF and GH. You can find this point by solving the simultaneous equations formed by the equations of EF and GH.
The simultaneous equations are:
Equation of EF: y - 2 = 1(x - 6)
Equation of GH: y - 2 = -1(x - 2)
We can solve these equations to find the values of x and y.
Using substitution or elimination method, let's solve the equations:
Equating the values of y - 2:
1(x - 6) = -1(x - 2)
x - 6 = -x + 2
2x = 8
x = 4
Plugging the value of x in one of the equations, let's find y:
y - 2 = 1(4 - 6)
y - 2 = 1(-2)
y - 2 = -2
y = 0
Therefore, the intersecting point of EF and GH is (4, 0).
This point is the circumcentre of triangle DEF.
So, the coordinates of the circumcentre of triangle DEF are (4, 0).
I hope this explanation helps! If you have any further questions, feel free to ask.