Question 184480: help please:don't know where to start
3 points A(3,1, B(-1,-11) and C (-5,9) nd circle that passes through them.
slope of line AB
cordinate of mid point that passes through AB
equation of perpidicular bisector of AB(1)
show that the line corresponding to parametiic equations x=t-14,y=-1/3t is same as 1 above
perpendicular bisector of line segment AC
centre of circle through A,B and c
equation of the circle
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! help please:don't know where to start
3 points A(3,1, B(-1,-11) and C (-5,9) nd circle that passes through them.
slope of line AB
cordinate of mid point that passes through AB
equation of perpidicular bisector of AB(1)
show that the line corresponding to parametiic equations x=t-14,y=-1/3t is same as 1 above
perpendicular bisector of line segment AC
centre of circle through A,B and c
equation of the circle
---------------------
Slope of the line thru A and B:
Slope, m, = diffy/diffx
m = (1+11)/(3+1)
m = 3
------------------
Midpoint of AB: (it's a point, it doesn't pass thru anything)
Find the average of x and y separately.
x1 = (3-1)/2 = 1
y1 = (1-11)/2 = -5
--> the midpoint is (1,-5)
---------------------------
equation of perpidicular bisector of AB(1):
Lines perpendicular to AB will have a slope that's the negative inverse of AB's slope, -1/3.
Then use
y-y1 = m*(x-x1) where (x1,y1) is the midpoint.
y+5 = (-1/3)*(x-1)
y = (-1/3)x -14/3 slope-intercept form
x+3y = -14 standard form
-----------------------
show that the line corresponding to parametiic equations x=t-14,y=-1/3t is same as 1 above
x=t-14
y=-1/3t
Eliminate t by substitution
x=t-14
t = x+14
y = (-1/3)*(x+14)
y = (-1/3)x - 14/3 Same as above.
-----------------------------
perpendicular bisector of line segment AC
Do this one as I did the one above.
--------------------
The rest later.
|
|
|
