Can someone help me with this problem: Four complex numbers form the vertices of a square in the complex plane. Three of the numbers are -19 + 32i, -5 + 12i, and -22 + 15i. What is the fourth number?

rubikx2910 Feb 24, 2020

#1**+2 **

I just spent ages on this and then my page froze and I lost it all.

Here is the short version. I treated it as a co-ordinate geometry question.

Let A(-19,32) B(-5,12) C(-22,15)

AB is longer than AC so AB is a diagonal and AC is a side.

The midpoint of AB is ((-12,22) That is the centre of the square. I'll call it O

The gradient of AB is -10/7

so

the gradient of the other diagonal is 7/10 =0.7 (the diagonals are perpendicular)

The distance from the centre O to any vertex is sqrt149

So find the point collinear with C and O that is sqrt149 units from O

( this point is the 4th vertex and it is (-2,29)

**So the 4th vertex is -2+29i**

Pictorial representation.

If you have questions then ask away

Melody Feb 24, 2020