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Question 30882: Given P = (−3, 1) and O the origin, describe the set of all points Q in the plane that satisfy each of the following conditions in turn. If Q is a specific point, give the coordinates. If the points Q form a line or a circle, give an equation of form y = mx+c or ( x−a)2+(y−b)2 = r2.
(a) P~Q is perpendicular to O~P.
(b) O~Q = t O~P, t any real number.
(c) |O~Q| = 4 and the angle between O~P and O~Q is pi/2.
The (~) denotes an arrow above.
THIS IS DRIVING ME CRAZY...
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! 1. Line OP is defined by points (0,0) and (-3,1). We heed to find its gradient.
gradient = (change in y)/(change in x)
gradient = 
gradient = 
gradient = 
The Line PQ is perpendicular to this line, so its gradient is 3. The equation of any straight lines with gradient 3 is y=3x+c where c is a constant.
Now we know that the line PQ goes through point (-3,1), so we have:
y = 3x+c
1 = 3(-3) + c
1 = -9 + c
c = 10
--> PQ is the line y=3x+10
2. Given the line OP is defines as y=-x/3, then line OQ, being a scaler version of it, is given by y=-tx/3.
3. If |OQ|=4, then this means that the point Q makes a circle centred on (0,0), of radius 4. However, we do not want to know every point on the circumference, ie the equation of that circle, since we are also told that line OP makes an angle of pi/2 radians to line OQ. Now pi/2 is 90 degrees. So it is saying that line OQ is perpendicular to line OP.
We already have this info from part1.
OP is y=-x/3 so that OQ here is the line y=3x (also goes through the origin, hence c=0).
So we need to know where the line y=3x crosses the circle made by point Q circling around the origin.
So, equation of the circle is of the form  , where the origin is (0,0) and radius=4, so we get  .
So where does this equal y=3x?
--> 
 or 
Manipulating this, we finally get:
 or 
So, for , 
and for , 
jon.
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