Question

If `A(x_1,y_1),B(x_2,y_2) and C(x_3,y_3)` are the vertices of traingle ABC and `x_(1)^(2)+y_(1)^(2)=x_(2)^(2)+y_(2)^(2)=x_3^(2)+y_(3)^(2)`, then show that `x_1 sin2A+x_2sin2B+x_3sin2C=y_1sin2A+y_2sin2B+y_3sin 2C=0`.

Answer 1

Since `x_(1)^(2)+y_(1)^2=x_(2_^2=x_(3)^2`, circumcentre of `DeltaABC` is `(0,0)`.
Now, coordinates of circumcentre are
`((x_1sin2A+x_2sin2B+x_3sin2C)/(sin2A+sin2B+sin2C),(y_1sin2A+y_2sin2b+y_3sin2C)/(sin2A+sin2B+sin2C))=(0,0)`
Therefore,`x_1sin2A+x-2sin2B+x_3sin2C=y_1sin2A+y_2sin2B+y_3sin2C=0`