pentagon coordinates

[browncutie]

There are 4 coordinates given: (5,0), ( 4,-3), (0,5), (-4,-3). The points form a circle and the origin on the coordinate grid is the centre of the circle. Where should the last point of the pentagon be placed? What coordinates would it be?

 

[Deleted member 4993]

There are 4 coordinates given: (5,0), ( 4,-3), (0,5), (-4,-3). The points form a circle and the origin on the coordinate grid is the centre of the circle. Where should the last point of the pentagon be placed? What coordinates would it be?

Since the shape of pentagon does not have any other constraint (like it is not regular pentagon with equal sides) - you can put the fifth point anywhere - as long as it is not co-linear with any other two vertices.

 

[browncutie]

DO u think u can show me how to get the last point cause i need to use a specific formula to figure out the answer of the coordinates.

 

[Mrspi]

DO u think u can show me how to get the last point cause i need to use a specific formula to figure out the answer of the coordinates.

You have not given us any specifics about the pentagon. As you were told in a previous response, if the pentagon is NOT a regular pentagon, then there are no limitations on where the final vertex might be located.

Please re-read your problem; if the pentagon is to be regular, please tell us that, and we'll look at your problem again.

 

[soroban]

Hello, browncutie!

There are 4 points given: (5,0), ( 4,-3), (0,5), (-4,-3).
The points form a circle and the origin is the centre of the circle.
Where should the last point of the pentagon be placed?
What coordinates would it be?

$\displaystyle \text{The four points are exactly 5 units from the origin.}$
. . $\displaystyle \text{Hence, the circle has radius 5.}$

$\displaystyle \text{The fifth vertex can be any other point }(x,\,y)\text{ where: }\:x^2+y^2\:=\:25$

 

[galactus]

If you plot your points you can see the pentagon has dimensions shown in the diagram.

$\displaystyle AB=5\sqrt{2}, \;\ BC=\sqrt{10}, \;\ CD=6, \;\ DE=\sqrt{10}, \;\ EA=5\sqrt{2}$

It would appear the fifth point belongs at (-5,0).

Pardon my 'crooked' pentagon.