# CV19

Show that the four points in the Argand plane represented by the complex numbers are the vertices of a square $2+i,4+3i,2+5i,3i\,$

From the given complex numbers,we can derive the points in the plane as $(2,1)(4,3)(2,5)(0,3)\,$

Let the points represent the four sides of a squate respectively $A,B,C,D\,$

Now writing the line equation $AB\,$

${\bar {AB}}=(y-1)=({\frac {3-1}{4-2}})(x-2)\,$

${\bar {AB}}=(y-1)=(x-2)\,$

Simplifying further we get ${\bar {AB}}=x-y-1=0\,$

Now writing the line equation $BC\,$

${\bar {BC}}=(y-3)=({\frac {5-3}{2-4}})(x-4)\,$

${\bar {BC}}=(y-3)=(4-x)\,$

${\bar {BC}}=x+y-7=0\,$

Now writing the line equation $CD\,$

${\bar {CD}}=(y-5)=({\frac {3-5}{0-2}})(x-2)\,$

${\bar {CD}}=(y-5)=(x-2)\,$

${\bar {CD}}=-x+y-3=0\,$

Now writing the line equation $DA\,$

${\bar {DA}}=(y-3)=({\frac {1-3}{2-0}})(x-0)\,$

${\bar {DA}}=(y-3)=(-1)(x)\,$

${\bar {DA}}=x+y-3)=0\,$

The slope of ${\bar {AB}}=1\,$

The slope of ${\bar {BC}}=-1\,$

The slope of ${\bar {CD}}=1\,$

The slope of ${\bar {DA}}=-1\,$

From these slopes,the slopes of opposite sides are equal,which means they are parallel to each other and product of the one to the next is -1,which means they are perpendicular. The lenth of $AB={\sqrt {(2-4)^{2}+(1-3)^{2}}}\,$

$AB={\sqrt {8}}\,$

The lenth of $BC={\sqrt {(4-2)^{2}+(3-5)^{2}}}\,$

$BC={\sqrt {8}}\,$

The lenth of $CD={\sqrt {(2-0)^{2}+(5-3)^{2}}}\,$

$CD={\sqrt {8}}\,$

The lenth of $DA={\sqrt {(2-0)^{2}+(1-3)^{2}}}\,$

$DA={\sqrt {8}}\,$

From the above all the sides are equal. Hence all these conditions prove that the numbers form the vertices of a square.