Co-ordinate geometry

abhasjha wrote:slope of PO = a/b
slope of QO = t/s
PO and QO is perpendicular so [a/b]*[t/s]=-1, so t/s = -b/a
In this case, b = √3, a = -1 , so t/s = [-√3]/[-1] = √3/1, so s =1

Alternatively

from the figure we can see that OP=OQ = radius of the semicircle. Also Angle POQ = 90degree

thus PQ^2 = OP^2 + OQ^2-------eqn1

using distance formula (O is origin)
OP = sqrt [(1-0)^2 + (sqrt -3 -0)^2] = sqrt [4] = 2 = OQ

also OQ = sqrt [ (s-0)^2 + (t-0)^2] = sqrt[s^2 + t^2]
OQ^2 = s^2+t^2 =4 ----eqn2

now PQ = 2sqrt 2 when we put values of OP and OQ in eqn 1

using distance formula length of PQ = sqrt [ (s+ sqrt3)^2 + (t-1)^2]
PQ^2 = (s+sqrt3)^2 + (t-1)^2 = 8--------eqn3

soving eqn 2 and 3 we get sqrt 3 s = t.Equating this in eqn 2 we get s=1

Gurpreet singh wrote:Please explain this part

In this case, b = √3, a = -1

Here's one approach:

Zoser wrote:

Here's one approach:

H Brent,

Can you explain why you assumed in your solution that the second triangle is 30:60:90? like the traingle on the left?

I know the r=2 but this does not mean that the second triangle is the same as the triangle to the left.

Wham am I missing here?

The three angles along the x-axis must sum to 180.
In Brent's solution, the leftmost angle along the x-axis = 30.
The figure indicates that the middle angle along the x-axis = 90.
Thus, the rightmost angle along the x-axis = 180-30-90 = 60.
Result:
The triangle on the right is also a 30-60-90 triangle.