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Question 4
The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.

 


Solution

O is the point of intersection of AC and BD. Coordinates of O can be calculated as follows:
x=312=1
y=2+22=2
Now, AC can be calculated as follows:
Diagonals of a square are equal and bisect each other.
AC=(3+1)2+(22)2
=16=4
Hence, sides of square = 2√2 (Using Pythagoras theorem)
Let coordinates of point D be (x1,y1)
AD=22=(x1+1)2+(y12)2
8=(x1+1)2+(y12)2
CD=22=(x13)2+(y12)2
8=(x13)2+(y12)2
From these equations, it is clear that;
(x1+1)2+(y12)2
=(x13)2+(y12)2
(x1+1)2=(x13)2
x12+2x+1=x126x+9
2x+1=6x+9
8x=8
x=1
Value of y1 can be calculated as follows by using the value of x.
CD=22=(x13)2+(y12)2
8=(13)2+(y12)2
4+(y12)2=8
y12=2
y1=4
Hence, D = (1, 4)
Coordinates of B can be calculated using coordinates of O; as follows:
Earlier, we had calculated O = (1, 2)
For BD, using section formula,
1=x+12
x+1=2
x=1
2=y+42
y+4=4
y=0
Hence, B = (1, 0) and D = (1, 4)

Mathematics

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