If two opposite vertices of a square are (5, 4) and (1, −6), find the coordinates of its remaining two vertices.
Solution
The distance d between two points `(x_1,y_1)` and `(x_2,y_2)` is given by `the formula
`d= sqrt((x_1- x_2)^2+(y_1 -y_2)^2)`
In a square, all the sides are of equal length. The diagonals are also equal to each other. Also in a square, the diagonal is equal to `sqrt2` times the side of the square.
Here let the two points which are said to be the opposite vertices of a diagonal of a square be A(5,4) and C(1,−6).
Let us find the distance between them which is the length of the diagonal of the square.
`AC = sqrt((5 - 1)^2 + (4 = 6)^2)`
`=sqrt((4)^2 + (10)^2)`
`= sqrt(16 + 100)`
`AC = 2sqrt29`
Now we know that in a square,
Side of the square = `"Diagonal of the square"/sqrt2`
Substituting the value of the diagonal we found out earlier in this equation we have,
Side of the square = `(2sqrt29)/sqrt2`
side of the square = `sqrt58`
Now, a vertex of a square has to be at equal distances from each of its adjacent vertices.
Let P(x, y) represent another vertex of the same square adjacent to both ‘A’ and ‘C
`AP = sqrt((5 - x)^2 + (4 - y)^2)`
`CP = sqrt((1 - x)^2 + (-6-y)^2)`
But these two are nothing but the sides of the square and need to be equal to each other.
AP = CP
`sqrt((5 - x)^2 + (4 - y)) = sqrt((1 - x)^2 + (-6 - y)^2)`
Squaring on both sides we have,
`(5 -x)^2 + (4 - y)^2 = (1 - x)^2 + (-6 - y)^2`
`25 + x^2 - 10x + 16 + y^2 - 8y = 1 + x^2 - 2x + 36 + y^2 + 12y`
8x + 20y = 4
2x + 5y = 1
From this we have, x = `(1- 5y)/2`
Substituting this value of ‘x’ and the length of the side in the equation for ‘AP’ we have,
`AP = sqrt((5 - x)^2 + (4 - y)^2)`
`sqrt(58) = sqrt((5 - x)^2 + (4 - y)^2)`
Squaring on both sides,
`58 = (5 - x)^2 + (4 - y)^2`
`58 = (5 - ((1 - 5y)/2))^2 + (4 - y)^2`
`58 = ((9 + 5y)/2)^2 + (4 - y)^2`
`58 = (81 + 25y^2 + +90y)/4 + 16 + y^2 - 8y`
`232 = 81 + 25y^2 + 90y + 64 + 4y^2 - 32y`
`87 = 29y^2 + 58y`
We have a quadratic equation. Solving for the roots of the equation we have,
`29y^2 + 58y - 87 = 0`
`29y^2 + 87y - 29y - 87 = 0`
29y(y + 3) - 29(y + 3) = 0
(y + 3)(29y - 29) = 0
(y + 3)(y - 1) = 0
The roots of this equation are −3 and 1.
Now we can find the respective values of ‘x’ by substituting the two values of ‘y’
When y = -3
`x = (1 - 5(-3))/2`
`= (1 + 15)/2`
x= 8
when y = 1
`x = (1- 5(1))/2`
`= (1-5)/2`
x = -2
Therefore the other two vertices of the square are (8, -3) and (-2, 1).
