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Question

A(3,4) and C(1,−1) are the two opposite angular points of a square ABCD. Find the coordinates of the other two vertices.

A
(92,12)
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B
(12,52)
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C
(12,12)
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D
(92,52)
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Solution

The correct options are
A (92,12)
B (12,52)
Given-
ABCD is a square with A(3,4) & C(1,1).
To find out-
The coordinates of B & C
Solution-
Let the coordinates of B=(x,y).
Since ABCD is a square,
AB=BCAB2=BC2.
Using distance formula,
(x3)2+(y4)2=(x1)2+(y+1)2
4x+10y23=0
x=(2310y4) .........(i).
Now, in ΔABC,
AB2+BC2=AC2
(x3)2+(y4)2+(x1)2+(y+1)2=(31)2+(4+1)2
x2+y24x3y=0.
Substituting for x from (i),
(2310y4)2+y24(2310y4)3y=0
4y212y+5=0
(2y1)(2y5)=0
y=(12,52).
Substituting for y in (i),
(x,y)=(92,12) & (12,52)
Now abscissae of A & C are 3 & 1, respectively.
Abscissa of B will be in between 1 & 3.
So here, 92 will be the abscissa of B.
Coordinates of B & D are (92,12) & (12,52), respectively.

389577_240707_ans.png

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