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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?

 Feb 24, 2020
 #1
avatar+108679 
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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  laugh

 Feb 24, 2020
 #2
avatar+82 
+1

Thank you so much! That's a really smart way of doing it, wow!

rubikx2910  Mar 1, 2020

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