ABCD is a rhombus. Its diagonals AC and BD inter- sect at M and satisfy BD=2AC. If the points D and M are represented by the complex numbers 1+i and 2−i, respectively, then A is represented by (A) 3−i/2 (B) 3+i/2 (C) 1+3i/2 (D) 1−3i/2


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Question Number 140205 by EnterUsername last updated on 05/May/21

$A B C D is a rhombus . Its diagonals A C and B D inter -$ $sect at M and satisfy B D = 2 A C . If the points D and$ $M are represented by the complex numbers 1 + i and$ $2 - i, respectively, then A is represented by$ $(A) 3 - i / 2 (B) 3 + i / 2 (C) 1 + 3 i / 2 (D) 1 - 3 i / 2$
	Answered by mr W last updated on 05/May/21

	Commented by EnterUsername last updated on 05/May/21

	$Thank you, Sir$
	Commented by EnterUsername last updated on 05/May/21
	Sir, where is the solution ? ��
	Commented by mr W last updated on 05/May/21

	$\vec{D M} = (2 - 1, - 1 - 1) = (1, - 2)$ $∣ D M ∣= \sqrt{{(2 - 1)}^{2} + {(- 1 - 1)}^{2}} = \sqrt{5}$ $∣ M A ∣= \frac{∣ D M ∣}{2} = \frac{\sqrt{5}}{2}$ $s a y A (x_{A}, y_{A})$ $\vec{M A} = (x_{A} - 2, y_{A} + 1)$ $\vec{M A} ⊥ \vec{D M} :$ $(x_{A} - 2) \times 1 + (y_{A} + 1) \times (- 2) = 0$ $\Rightarrow x_{A} - 2 y_{A} - 4 = 0$ $∣ D M ∣^{2} = {(x_{A} - 2)}^{2} + {(y_{A} + 1)}^{2} = {(\frac{\sqrt{5}}{2})}^{2} = \frac{5}{4}$ ${(2 y_{A} + 4 - 2)}^{2} + {(y_{A} + 1)}^{2} = \frac{5}{4}$ $4 y_{A}^{2} + 8 y_{A} + 4 + y_{A}^{2} + 2 y_{A} + 1 = \frac{5}{4}$ $y_{A}^{2} + 2 y_{A} + 1 = \frac{1}{4}$ ${(y_{A} + 1)}^{2} = \frac{1}{4}$ $y_{A} = \pm \frac{1}{2} - 1 = - \frac{1}{2} o r - \frac{3}{2}$ $x_{A} = 3 o r 1$ $\Rightarrow A = 3 - \frac{i}{2} o r 1 - \frac{3 i}{2}$ $\Rightarrow a n s w e r (A) i s c o r r e c t .$
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