1) Triangle PQR has vertices P(-2, -1) Q(-7, 4) R(-1, 1)
a) Show PR is perpendicular to QR.
Now for this I thought if i found the gradients of both lines using y2-y1/x2-x1 then by showing they both multipy to make -1.
But that gave me 8/5 x 3 which doesnt equal -1 so that doesnt show that they are perpendicular...so i think ive gone wrong here somewhere.
2) PQ is a diameter of a circle. Find an equation of the circle when:
a) P is (2, 0) and Q is (6,0)
With this i assumed that if i found the midpoint of the line then that would give me the centre of the circle which I could then place into the equation :
(x-a)^2 + (y-b)^2 = r^2
Then to find the radius i thought by finding the magnitude of the line then i could find the diameter and half it.
I got the answer to be (x-1)^2 + (y-3)^2 = 4 but the book tells me its
(x-4)^2 + y^2 = 4
Could someone attempt to explain where ive gone wrong on these two questions.
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Circle Geometry watch
- Thread Starter
- 09-11-2004 20:58
- 09-11-2004 21:22
If two lines are perpendicular ((m1)(m2))=-1
Let m1=mPR and m2=mQR. Therefore:
((m1)(m2))=((2)(-1/2))=-1 Hence PR is perpendicular to QR.
Q. E. D.
2a) P(2,0) Q(6,0) and PQ is the diameter of the circle. Find the mid-point to find the centre of the cirlce.
=> = (((8)/(2)),((0)/(2)))=(4,0)
Cartesian equation of a circle:
(((x-a)^2)+((y-b)^2))=(r^2) where a and b are the x and y co-ordinates, respectively, of the centre of the circle and r is the radius of the circle.
r: The radius is the magnitude of the distance between one side of the diameter to the midpoint of the circle. Let us consider P(2,0). Since the co-ordinates of the centre are (4,0). We can tell that the radius is 4-2=2units.
Therefore the equation of the respective circle is:
=>(((x-4)^2)+(y^2))=(4) Q. E. D.
- Thread Starter
- 09-11-2004 22:31