IMO B

I've recently been given this solution:

t/s*(-1/root(3)) = -1 ---> since lines OP and OQ are perpendicular

also let
s^2+t^2=r^2=3+1=4 ---> (since P lies on the same circle and O is origin)

solve for s

s^2+3s^2=4 => s=+/-1

LevelOne wrote:I've recently been given this solution:

t/s*(-1/root(3)) = -1 ---> since lines OP and OQ are perpendicular

also let
s^2+t^2=r^2=3+1=4 ---> (since P lies on the same circle and O is origin)

solve for s

s^2+3s^2=4 => s=+/-1

how did u replace t^2 with 3s^2 ?

Can u expand more on your first solution, if u can please.

abhinav85 wrote:IMO B

ok, how ?

IMO you don't need any math here to solve the problem.

you can notice in the diagram, to be a 90 degree angle, the two points must be in reverse order with respect to the quadrant they are in. think about (1,0 and 0,1) the angle b/w them is 90 degrees. The same with (-1,0) and (0,1).

here the semi-circle is in two quads. the point (s) is in the first quad which means it is positive. since the point given is [-sqrt(3),1],

(s,t) would be the reverse with respect to the first quadrant.

[1, sqrt(3)] would be (s,t) in this case, which would make s = 1.

hope that made sense.

if you are looking for a more mathematical approach, you can find it by making two 30-60-90 triangles with the x axis. Check out the link below.

https://www.beatthegmat.com/what-is-the- ... 27181.html

ssmiles08 wrote:IMO you don't need any math here to solve the problem.

you can notice in the diagram, to be a 90 degree angle, the two points must be in reverse order with respect to the quadrant they are in. think about (1,0 and 0,1) the angle b/w them is 90 degrees. The same with (-1,0) and (0,1).

here the semi-circle is in two quads. the point (s) is in the first quad which means it is positive. since the point given is [-sqrt(3),1],

(s,t) would be the reverse with respect to the first quadrant.

[1, sqrt(3)] would be (s,t) in this case, which would make s = 1.

hope that made sense.

if you are looking for a more mathematical approach, you can find it by making two 30-60-90 triangles with the x axis. Check out the link below.

https://www.beatthegmat.com/what-is-the- ... 27181.html

Thanks for pointing this link !