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?
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
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
If you have questions then ask away