Question

What is a point equal distance from all four points a(10,45) (25,20) (42,37) and (33,52) use the midpoint ,slope ,distance and equation of a line to solve

Answer 1

If you plot these points accurately on graph paper, you should get a quadrilateral. Have you plotted yet? If not please do so. Tell me if you get a rectangle because you should.

Answer 2

analytical method
call the point (x ,y )
write eqns using distance formula

Answer 3

Yes you get a quadrilateral almost like a trapezoid @calculusfunctions

Answer 4

@edsgirl17, you should get a rectangle.

Answer 5

However technically you are correct because a rectangle is a trapezoid with two pairs of parallel sides.

Answer 6

Now to find a point which is equidistant from all four points, we need to locate the center of this rectangle. Do you know how to do that? Do you want to try, instead of me just giving you the solution like last time? If you need help, I'll be right here. But I have to go soon so please don't take too long.

Answer 7

Is it point (27.5, 38.5)

Answer 8

@calculusfunctions ??

Answer 9

lol I haven't actually worked it out myself. I thought you would show me your steps, and then I could tell you whether you were doing it correctly or not. But now I will work it out and let tou know if you're correct. Did you check the answer key already?

Answer 10

Yes the answer should be 25,37 but I can't figure out how , I have worked on this question for four hours now and keep getting different answers

Answer 11

The answer key says (25, 37)? Because according to my calculations it should be (30, 40). Which means that either I or the answer key is wrong or perhaps you wrote the vertices above incorrectly. Could you double check and get back to me please. I will be logging out in a few minutes so if you don't catch me before I log out then I hope someone else will help or I will put the solution up, if you still require it, when I return.

Answer 12

How did you get 30,40 ?? No there are right I checked

Answer 13

You are right, I had made an error in my calculations. I will write the complete solution for you right now. It'll take me about 10 minutes to type. so please be patient. Thank you and once again I apologize for my error.

Answer 14

The vertices given are (10, 45), (25, 20), (42, 37) and (33, 52).

To find a point which is equidistant equidistant from the vertices, we need to find the circumcenter. The circumcenter is the point of intersection of the perpendicular bisectors. Thus find the equations of any two perpendicular bisectors, and determine their point of intersection.

To find the equation of a perpendicular bisector of a side
i). Find the midpoint of the side,
ii). find the slope of the side, and the negative reciprocal of this slope is the slope of the perpendicular bisector.
iii) using the slope of the perpendicular bisector and the midpoint, find the equation of the perpendicular bisector.

In our example, let A(10, 45), B(25, 20), C(42, 37) and D(33, 52). The four sides of the rectangle are AB, BC, CD and AD. Find the perpendicular bisectors of any two of these sides by following the above steps and then find their point of intersection by either the substitution or elimination method.

First let's choose side AB.

Midpoint of side AB = (35/2, 65/2)

Slope of side AB = -5/3

the slope of the perpendicular bisector of side AB is m = 3/5

Using m = 3/5 and the midpoint (35/2, 65/2) find the equation y = mx + b. Hence
\[\frac{ 65 }{ 2 }=(\frac{ 3 }{ 5 })(\frac{ 35 }{ 2 })+b\]
\[\frac{ 65 }{ 2 }-\frac{ 21 }{ 2 }=b\]
\[b =22\]

\[y =\frac{ 3 }{ 5 }x +22\] is the equation of the perpendicular bisector of side AB.


Now let's choose side BC.

Midpoint of side BC = (67/2, 57/2)

Slope of side BC = 17/17 = 1

the slope of the perpendicular bisector of side BC is m = -1.

Using m = -1 and the midpoint (67/2, 57/2) find y = mx + b. Thus
\[\frac{ 57 }{ 2 }=(-1)(\frac{ 67 }{ 2 })+b\]
\[\frac{ 57 }{ 2 }+\frac{ 67 }{ 2 }=b\]
\[b =62\]
\[y =-x +62\] is the equation of the perpendicular bisector of side BC.

Now let the right sides of the two equations equal to each other. Thus

\[\frac{ 3 }{ 5 }x +22=-x +62\]
Multiply both sides by 5 to obtain
\[3x +110=-5x +310\]
\[8x =200\]
\[x =25\]

Substitute x = 25 into either equation. Let's choose the easier one. Hence let's substitute x = 25 int y = -x + 62. Hence

y = -25 + 62

y = 37

Therefore the point (25, 37) is equidistant from all the given vertices.


I hope this helps. Once again I apologize for earlier. I was in a rush and wasn't really thinking, although that is no excuse! Please let me know if this helps and if you need anything else from me, I'll be back later.

Answer 15

Thanks you sooo much!!!!