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Prove that the points representing the complex numbers $\left ( 7+5\mathit{i} \right )$, $\left ( 5+2\mathit{i} \right )$, $\left ( 4+7\mathit{i} \right )$ and $\left ( 2+4\mathit{i} \right )$ form a parallelogram. (Plot the points and use midpoint formula).

1 Answer

  • If $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ are represented by the points $A(x_2,y_2),B(x_2,y_2)$ on the complex Argand plane,then their sum $(x_1+x_2)+i(y_1+y_2)$ is represented by the point $C(x_1+x_2,y_1+y_2)$ which completes the parallelogram $OACB$.
Step 1:
Let $z_1=(7+5i),z_2=(5+2i),z_3=(4+7i),z_4=(2+4i)$ be represented by the points $A,B,C,D$ in the Argand plane.
Plotting the points $A(7,5),B(5,2),C(4,7),D(2,4)$ we have:
Step 2:
The diagonals of the quadrilateral $DBAC$ are $DA$ and $BC$.
Th midpoint of the diagonal $DA$ is $\big(\large\frac{7+2}{2},\frac{4+5}{2}\big)$ i.e $(\large\frac{9}{2},\frac{9}{2}\big)$ and the midpoint of the diagonal $BC$ is $\big(\large\frac{5+4}{2},\frac{2+7}{2}\big)$ i.e $\big(\large\frac{9}{2},\frac{9}{2}\big)$.
The diagonals bisect each other.
Therefore $DBAC$ is a parallelogram.
answered Jun 10, 2013 by sreemathi.v
edited Jun 10, 2013 by sreemathi.v

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