Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

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

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

Answered by mr W last updated on 05/May/21

Commented by EnterUsername last updated on 05/May/21

Thank you, Sir

$$\mathrm{Thank}\:\mathrm{you},\:\mathrm{Sir} \\ $$

Commented by EnterUsername last updated on 05/May/21

Sir, where is the solution ? ��

Commented by mr W last updated on 05/May/21

DM^(→) =(2−1,−1−1)=(1,−2) ∣DM∣=(√((2−1)^2 +(−1−1)^2 ))=(√5) ∣MA∣=((∣DM∣)/2)=((√5)/2) say A(x_A ,y_A ) MA^(→) =(x_A −2,y_A +1) MA^(→) ⊥DM^(→) : (x_A −2)×1+(y_A +1)×(−2)=0 ⇒x_A −2y_A −4=0 ∣DM∣^2 =(x_A −2)^2 +(y_A +1)^2 =(((√5)/2))^2 =(5/4) (2y_A +4−2)^2 +(y_A +1)^2 =(5/4) 4y_A ^2 +8y_A +4+y_A ^2 +2y_A +1=(5/4) y_A ^2 +2y_A +1=(1/4) (y_A +1)^2 =(1/4) y_A =±(1/2)−1=−(1/2) or −(3/2) x_A =3 or 1 ⇒A=3−(i/2) or 1−((3i)/2) ⇒answer (A) is correct.

$$\overset{\rightarrow} {{DM}}=\left(\mathrm{2}−\mathrm{1},−\mathrm{1}−\mathrm{1}\right)=\left(\mathrm{1},−\mathrm{2}\right) \\ $$$$\mid{DM}\mid=\sqrt{\left(\mathrm{2}−\mathrm{1}\right)^{\mathrm{2}} +\left(−\mathrm{1}−\mathrm{1}\right)^{\mathrm{2}} }=\sqrt{\mathrm{5}} \\ $$$$\mid{MA}\mid=\frac{\mid{DM}\mid}{\mathrm{2}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${say}\:{A}\left({x}_{{A}} ,{y}_{{A}} \right) \\ $$$$\overset{\rightarrow} {{MA}}=\left({x}_{{A}} −\mathrm{2},{y}_{{A}} +\mathrm{1}\right) \\ $$$$\overset{\rightarrow} {{MA}}\bot\overset{\rightarrow} {{DM}}: \\ $$$$\left({x}_{{A}} −\mathrm{2}\right)×\mathrm{1}+\left({y}_{{A}} +\mathrm{1}\right)×\left(−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}_{{A}} −\mathrm{2}{y}_{{A}} −\mathrm{4}=\mathrm{0} \\ $$$$\mid{DM}\mid^{\mathrm{2}} =\left({x}_{{A}} −\mathrm{2}\right)^{\mathrm{2}} +\left({y}_{{A}} +\mathrm{1}\right)^{\mathrm{2}} =\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\left(\mathrm{2}{y}_{{A}} +\mathrm{4}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}_{{A}} +\mathrm{1}\right)^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\mathrm{4}{y}_{{A}} ^{\mathrm{2}} +\mathrm{8}{y}_{{A}} +\mathrm{4}+{y}_{{A}} ^{\mathrm{2}} +\mathrm{2}{y}_{{A}} +\mathrm{1}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${y}_{{A}} ^{\mathrm{2}} +\mathrm{2}{y}_{{A}} +\mathrm{1}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left({y}_{{A}} +\mathrm{1}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${y}_{{A}} =\pm\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}=−\frac{\mathrm{1}}{\mathrm{2}}\:{or}\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${x}_{{A}} =\mathrm{3}\:{or}\:\mathrm{1} \\ $$$$\Rightarrow{A}=\mathrm{3}−\frac{{i}}{\mathrm{2}}\:{or}\:\mathrm{1}−\frac{\mathrm{3}{i}}{\mathrm{2}} \\ $$$$\Rightarrow{answer}\:\left({A}\right)\:{is}\:{correct}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com