Q.41. Complex numbers.

Question 41.
If A(Z1) ,B(Z2) and C(Z3) are collinear points then prove that
(Z3-Z1)*(conjugate of Z2-conjugate of Z1)=(conjugate of Z3-conjugate of Z1)*(Z2-Z1).
 
Answer 41.
Shifting origin to A and rotating the coordinate axes such that B and C are on x-axis,
A(Z1) = (0, 0), B(Z2) = (a, 0) and C(Z3) = (ka, 0)
=> their conjugates are
A(Z1') = (0, 0), B(Z2') = (a, 0) and C(Z3') = (ka, 0)

LHS
= (Z3 - Z1) * (Z2' - Z1')
= (ka, 0) * (a, 0)
= (ka^2, 0) and

RHS
= (Z3' - Z1') * (Z2 - Z1)
= (ka, 0) * (a, 0)
= (ka^2, 0)

=> LHS = RHS
=> (Z3-Z1)*(conjugate of Z2-conjugate of Z1)=(conjugate of Z3-conjugate of Z1)*(Z2-Z1).

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