Question 4
The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.
O is the point of intersection of AC and BD. Coordinates of O can be calculated as follows using mid point formula:
x=x1+x22,y=y1+y22(mid point between (x1,y1) and (x2,y2))
x=3−12=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+(2−2)2
=√16=4
Hence, sides of square = 2√2 (Using Pythagoras theorem)
Let coordinates of point D be (x1,y1)
AD=2√2=√(x1+1)2+(y1−2)2
⇒8=(x1+1)2+(y1−2)2
CD=2√2=√(x1−3)2+(y1−2)2
⇒8=(x1−3)2+(y1−2)2
From these equations, it is clear that AD=CD;
(x1+1)2+(y1−2)2
=(x1−3)2+(y1−2)2
⇒(x1+1)2=(x1−3)2
⇒x12+2x1+1=x12−6x1+9
⇒2x1+1=−6x1+9
⇒8x1=8
⇒x1=1
Value of y1 can be calculated as follows by using the value of x1.
CD=2√2=√(x1−3)2+(y1−2)2
⇒8=(1−3)2+(y1−2)2
⇒4+(y1−2)2=8
⇒y1−2=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)