steven_ghoos wrote:Thank you again! It's very helpfull to me.
For the coordinate plane question, I hope they will not ask many like this. I also still have a small question (in orange):
Point P and Q lies on the same circle with center at (0, 0).
Thus, (s² + t²) = (-√3)² + 1² = 3 + 1 = 4
Is this always true for points on a circle with center (0,0)? X1² + y1² = x2² + y2²
Again line segments OP and OQ are perpendicular.
Thus (slope of OP)*(slope of OQ) = -1
Slope of OP = 1/(-√3) = -(1/√3)
=> Slope of OQ = (t - 0)/(s - 0) = t/s = (-1)/(-1/√3) = √3
=> t = √3s
Thus, (s² + (√3s)²) = 4
=> (s² + 3s²) = 4
=> s² = 1
=> s = ±1
As point Q lies in the first quadrant s = 1.
Answer to the question in orange :
Since those two points are on the circumference of the circle, the line joining the center of the circle with the points (x1,y1) and (x2,y2), is the radius. Hence they are same.
If instead of center being at (0,0), it is at some arbitrary point (a,b), then you'll have :
(x1-a)^2+(y1-b)^2 = (x2-a)^2+(y2-b)^2
