Question : The two opposite vertices of a square are (-1,2) and (3,2). Find the coordinates of the other two vertices.
Explanation by Vanshika
Solution :
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Let
A (-1,2)
B (3,2)
By Distance formula
AB = √[(x2-x1)2+(y2-y1)2]
AB = √[(3+1)2+(2-2)2]
AB = √[16+0]
AB = 4 units
We know,
Diagonal of a square, d = a√2
so, a =d/√2 = d√2/2
a = 4√2/2 = 2√2 units.
so, a =d/√2 = d√2/2
a = 4√2/2 = 2√2 units.
AB = BC = a = 2√2 units
Again by Distance formula
AB = √[(x+1)2+(y-2)2]
2√2= √[(x+1)2 + (y-2)2]
SBS
(2√2)2 = (x+1)2 + (y-2)2
SBS
(2√2)2 = (x+1)2 + (y-2)2
8 = x2+1+2x+y2+4-4y
8-4-1=x2+y2+2x-4y
3 = x2+y2+2x-4y ---------- (1)
8-4-1=x2+y2+2x-4y
3 = x2+y2+2x-4y ---------- (1)
Similarly
BC = √[(x-3)2+(y-2)2]
BC = √[(x-3)2+(y-2)2]
2√2= √[(x-3)2+(y-2)2]
SBS
8 = (x-3)2+(y-2)2
8 = x2+9-6x+y2+4-4y
8 = x2+9-6x+y2+4-4y
8-9-4 = x2+y2-6x-4y
-5 = x2+y2-6x-4y ------------- (2)
Subtracting equation (2) from (1)
3 = x2+y2+2x-4y
-5 = x2+y2-6x-4y
+ - - + +
--------------------------
8= 0 + 0 + 8x + 0
--------------------------
x=8/8
x = 1
+ - - + +
--------------------------
8= 0 + 0 + 8x + 0
--------------------------
x=8/8
x = 1
Now putting x=1 in equation (1)
3 = (1)2+y2+2(1)-4y
3 = 1 +y2+2-4y
3-1-2=y2-4y
0=y2-4y
0=y2-4y
0=y(y-4)
y= 0 or y-4=0
3 = 1 +y2+2-4y
3-1-2=y2-4y
0=y2-4y
0=y2-4y
0=y(y-4)
y= 0 or y-4=0
y=0 or y=4
Hence, the coordinates of other two opposite vertices of the square are (1,0) and (1,4).