Probability
Full Member
- Joined
- Jan 26, 2012
- Messages
- 425
OK I have some points. A(5, -3), B(-1, 1) and C(0.2)
I have calculated the slope that passes through AB
y2 - y1 = 1 - (-3) = - 4/6 = - 2/3
x2 - x1 - 1 - 5
Slope AB = - 2/3
Coordinates of the midpoints AB
(1/2 (x1+x2), 1/2(y1 + y2))
(1/2(5 + -1), 1/2(-3 + 1)
MAB = {4/2, -2/2}
MBC = {-1/2, 3/2}
Finding the equation of perpendicular bisector AB
MAB = 2/3
Equation is;
bAB : y - (-2/2) = 2/3(x - 4/2) this implies y = 2/3x - 3
At this point I am unsure if an error is present, the reason being is that in my previous thread Parametric Equations, I concluded that y = (3x - 8) / (2)
so if then (3(4/2) - 8)) / (2) = - 2/2 = - 1, then using the equation above just found;
y = 2/3x - 3 implies 2/3 (4/2) - 3 = - 1.66 or - 5/3
Looking at this I am suspecting that the centre of the circle lies at points (2, -1.66)
How am I doing?
I have calculated the slope that passes through AB
y2 - y1 = 1 - (-3) = - 4/6 = - 2/3
x2 - x1 - 1 - 5
Slope AB = - 2/3
Coordinates of the midpoints AB
(1/2 (x1+x2), 1/2(y1 + y2))
(1/2(5 + -1), 1/2(-3 + 1)
MAB = {4/2, -2/2}
MBC = {-1/2, 3/2}
Finding the equation of perpendicular bisector AB
MAB = 2/3
Equation is;
bAB : y - (-2/2) = 2/3(x - 4/2) this implies y = 2/3x - 3
At this point I am unsure if an error is present, the reason being is that in my previous thread Parametric Equations, I concluded that y = (3x - 8) / (2)
so if then (3(4/2) - 8)) / (2) = - 2/2 = - 1, then using the equation above just found;
y = 2/3x - 3 implies 2/3 (4/2) - 3 = - 1.66 or - 5/3
Looking at this I am suspecting that the centre of the circle lies at points (2, -1.66)
How am I doing?