1
Question
On the Argand plane z1,z2 andz3 are, respectively, the vertices of an isosceles triangle ABC with AC=BC and equal angles are θ. If z4 is the incentre of the triangle, then (z2−z1)(z3−z1)(z4−z1)2=
On the Argand plane z1,z2 andz3 are, respectively, the vertices of an isosceles triangle ABC with AC=BC and equal angles are θ. If z4 is the incentre of the triangle, then (z2−z1)(z3−z1)(z4−z1)2=
Open in App
Solution
The correct option is B 1+secθ
z2−z1|z2−z1|=(z4−z1)|z4−z1|e−iθ/2(clockwise)
z3−z1|z3−z1|=(z4−z1)|z4−z1|eiθ/2(anticlockwise)
Multiplying (1) and (2), we get
(z2−z1)(z3−z1)(z4−z1)2=|z2−z1||z3−z1||z4−z1|2
=(AB)(AC)(AI)2=2(AD)(AC)(AI)2
=2(AD)2(AI)2ACAD=2cos2θ2secθ [∵∠ADC=90°]=(1+cosθ)secθ=1+secθ
z2−z1|z2−z1|=(z4−z1)|z4−z1|e−iθ/2(clockwise)
z3−z1|z3−z1|=(z4−z1)|z4−z1|eiθ/2(anticlockwise)
Multiplying (1) and (2), we get
(z2−z1)(z3−z1)(z4−z1)2=|z2−z1||z3−z1||z4−z1|2
=(AB)(AC)(AI)2=2(AD)(AC)(AI)2
=2(AD)2(AI)2ACAD=2cos2θ2secθ [∵∠ADC=90°]=(1+cosθ)secθ=1+secθ
Suggest Corrections
2
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
