Match the following COLUMN - I COLUMN - IIA)If z1, z2, z3 are verticesof an equilateral triangle thenp)z1−z22+3z3−z22=0B)If z1, z2, z3 are vertices of A, B, C of an isosceles right angled triangle respectively with right angle at C, then q)z2−z32=3z1−z2z3−z1C)If z1, z2, z3 are vertices of A, B, C of an isosceles triangle respectively and angles B & C areeach equal to π6then r)z1−z22=2z1−z3z3−z2D)If z1, z2, z3 are vertices of A, B, C respectivelyof a triangle ABC in which∠A=π6,∠B=π2, ∠C=π3, then s)z1−z22+z2−z32+z3−z12=0 (A)(B)(C)(D)1)srqp2)rqsq3)psrr4)qpps

a) BA¯=BC¯eiπ3;z1−z2=z3−z2eiπ3………(1)CA¯=CB¯e−iπ3⇒z1−z3=z2−z3e−iπ3…..(2)From (1) & (2) z2−z1z3−z2=z3−z2z3−z1⇒∑z12=∑z1z2⇒a→sb) CA¯=CB¯eiπ2⇒z1−z3=z2−z3(−i)…..(1)BA¯=2BC¯eiπ4z1−z22=2Z3−Z22(i)⇒z1−z22=2z3−z22−z3+z1−z2+z3=2z3−z2z1−z3⇒b→rc) BA¯=13BC¯eiπ6,CA¯=−13BC¯e−iπ6⇒BC¯2=−3BA¯⋅CA¯⇒z2−z32=−3z3−z1z2−z13z1−z2z3−z1 d) BA¯=3BC¯eiπ2Squaring we get z1−z22=3z3−z22(−1)⇒d→p

Match the following COLUMN - I COLUMN - IIA)If z1, z2, z3 are verticesof an equilateral triangle thenp)z1−z22+3z3−z22=0B)If z1, z2, z3 are vertices of A, B, C of an isosceles right angled triangle respectively with right angle at C, then q)z2−z32=3z1−z2z3−z1C)If z1, z2, z3 are vertices of A, B, C of an isosceles triangle respectively and angles B & C areeach equal to π6then r)z1−z22=2z1−z3z3−z2D)If z1, z2, z3 are vertices of A, B, C respectivelyof a triangle ABC in which∠A=π6,∠B=π2, ∠C=π3, then s)z1−z22+z2−z32+z3−z12=0 (A)(B)(C)(D)1)srqp2)rqsq3)psrr4)qpps

a) BA¯=BC¯eiπ3;z1−z2=z3−z2eiπ3………(1)CA¯=CB¯e−iπ3⇒z1−z3=z2−z3e−iπ3…..(2)From (1) & (2) z2−z1z3−z2=z3−z2z3−z1⇒∑z12=∑z1z2⇒a→sb) CA¯=CB¯eiπ2⇒z1−z3=z2−z3(−i)…..(1)BA¯=2BC¯eiπ4z1−z22=2Z3−Z22(i)⇒z1−z22=2z3−z22−z3+z1−z2+z3=2z3−z2z1−z3⇒b→rc) BA¯=13BC¯eiπ6,CA¯=−13BC¯e−iπ6⇒BC¯2=−3BA¯⋅CA¯⇒z2−z32=−3z3−z1z2−z13z1−z2z3−z1 d) BA¯=3BC¯eiπ2Squaring we get z1−z22=3z3−z22(−1)⇒d→p

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Match the following
COLUMN - I COLUMN - II
A) If z₁, z₂, z₃ are vertices
of an equilateral triangle then p) ${(z_{1} - z_{2})}^{2} + 3 {(z_{3} - z_{2})}^{2} = 0$
B) If z₁, z₂, z₃ are vertices of A, B, C of an isosceles right angled triangle
respectively with right angle at C,
then q) ${(z_{2} - z_{3})}^{2} = 3 (z_{1} - z_{2}) (z_{3} - z_{1})$
C) If z₁, z₂, z₃ are vertices of A, B, C of an isosceles triangle respectively and angles B & C are
each equal to $\frac{π}{6}$ then
r) ${(z_{1} - z_{2})}^{2} = 2 (z_{1} - z_{3}) (z_{3} - z_{2})$
D) If z₁, z₂, z₃ are vertices of A, B, C respectively
of a triangle ABC in which $∠ A = \frac{π}{6}, ∠ B = \frac{π}{2}, ∠ C = \frac{π}{3}, then$ s) ${(z_{1} - z_{2})}^{2} + {(z_{2} - z_{3})}^{2} + {(z_{3} - z_{1})}^{2} = 0$
(A) (B) (C) (D)
1) s r q p
2) r q s q
3) p s r r
4) q p p s

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Detailed Solution

$\bar{BA} = \bar{BC} e^{\frac{iπ}{3}}; z_{1} - z_{2} = (z_{3} - z_{2}) e^{i \frac{π}{3}}$ ………(1)
$\bar{CA} = \bar{CB} e^{\frac{- iπ}{3}} \Rightarrow (z_{1} - z_{3}) = (z_{2} - z_{3}) e^{\frac{- iπ}{3}}$ …..(2)
From (1) & (2) $\frac{z_{2} - z_{1}}{z_{3} - z_{2}} = \frac{z_{3} - z_{2}}{z_{3} - z_{1}}$
$\Rightarrow \sum z_{1}^{2} = \sum z_{1} z_{2} \Rightarrow a \to s$
b) $\bar{CA} = \bar{CB} e^{\frac{iπ}{2}} \Rightarrow z_{1} - z_{3} = (z_{2} - z_{3}) (- i)$ …..(1)
$\bar{BA} = \sqrt{2} \bar{BC} e^{\frac{iπ}{4}} {(z_{1} - z_{2})}^{2} = 2 {(Z_{3} - Z_{2})}^{2} (i)$ $\begin{array}{l} \Rightarrow {(z_{1} - z_{2})}^{2} = 2 {(z_{3} - z_{2})}^{2} \frac{(- z_{3} + z_{1})}{- z_{2} + z_{3}} \\ = 2 (z_{3} - z_{2}) (z_{1} - z_{3}) \Rightarrow b \to r \end{array}$

c) $\bar{BA} = \frac{1}{\sqrt{3}} \bar{BC} e^{\frac{iπ}{6}}, \bar{CA} = \frac{- 1}{\sqrt{3}} \bar{BC} e^{\frac{- iπ}{6}}$
$\begin{array}{l} \Rightarrow {\bar{BC}}^{2} = - 3 \bar{BA} \cdot \bar{CA} \\ \Rightarrow {(z_{2} - z_{3})}^{2} = - 3 (z_{3} - z_{1}) (z_{2} - z_{1}) \\ 3 (z_{1} - z_{2}) (z_{3} - z_{1}) \end{array}$

d) $\bar{BA} = \sqrt{3} \bar{BC} e^{\frac{iπ}{2}}$
Squaring we get
${(z_{1} - z_{2})}^{2} = 3 {(z_{3} - z_{2})}^{2} (- 1) \Rightarrow d \to p$

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