Question

Check my answer: The kitchen in Shane’s house is a square. The points P(0, 7) and Q(4, 1) represent the opposite vertices of his kitchen in a drawing on a coordinate grid. Which of these ordered pairs could be a third vertex?
(5, 7)
(-1, 1)
(-1, 2)
(4, 7)
I got (-1,2)

Answer 1

@mathslover

Answer 2

@amistre64

Answer 3

Its (-1,2)

http://openstudy.com/study#/updates/4f47c174e4b00aeb3e3ef2db

Answer 4

?
41
24 (midpoint)
07
?

this looks rather squarish to me; not as , hmmm

Answer 5

Let the required point be : \(x,y\)

then : \(\sqrt{(x-0)^2 + (y-7)^2 }= \sqrt{(x-4)^2 + (y-1)^2 }\)

\(x^2 + y^2 + 49 - 14y = x^2 + 16 - 8x + y^2 + 1 - 2y \)

\(49 - 14y + 8x - 16 -1 + 2y = 0\)

\(32 - 12y + 8x = 0 \)

\(8x - 12y = 32\)

\(2x - 3y = 8 \)



From the options, only (-1,2) satisfies the given relationship.

Answer 6

Oopps, I meant 8x - 12y = -32 and 2x -3y = -8

Answer 7

-1,2 is not going to be it ... in my view

Answer 8

but then i am blind most of the time

Answer 9

center a circle at 2,4

(x-2)^2 + (y-4)^2 = r^2
0 7
---------------------
4 + 9 = 13


(x-2)^2 + (y-4)^2 = 15
-1 2
---------------------
9 + 4

yep

Answer 10

so whats the answer...? @amistre64

Answer 11

i was checking to see if -1 2 worked in my circle idea, and it did ... , so id go with -1,2

Answer 12

@amistre64 oh alrighty thanks guys :)

Answer 13

if we try the other options .... since not all points on a circle are squares ....

(x-2)^2 + (y-4)^2 = 13
5 7
--------------------
9 + 9 not it


(x-2)^2 + (y-4)^2 = 13
-1 1
--------------------
9 + 9 not it


(-1, 2) , we know this is fine


(x-2)^2 + (y-4)^2 = 13
4 7
-------------------
4 + 9 = 13 this one would give me pause to consider

Answer 14

(0, 7) (4, 1)
-7 -7
-----------
4,-6 ; slope is -6/4 = -3/2 ; perp is 2/3

y-4 = 2/3(x-2)

y-4 = 2/3(x-2)
7 4
--------------
3 = 4/3 not it


y-4 = 2/3(x-2)
2 -1
--------------
-2 = -2 , its -1,2 fer sure