If a,b,c and u,v,w are complex numbers representing the vertices of two triangles such that c=(1−r)a+rb and w=(1−r)u+rv where r is a complex number, then the two triangles
A
have the same area
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B
are similar
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C
are congruent
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D
none of these
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Solution
The correct option is B are similar Let A(a),B(b),C(c) and P(u),Q(v),R(w)
Now c=(1−r)a+rb and w=(1−r)u+rv ⇒c−ab−a=w−uv−u=r
⇒∣∣c−ab−a∣∣=∣∣w−uv−u∣∣=|r|
⇒ACAB=PRPQ=|r| and ∠CAB=∠RPQ Hence both the triangles are similar