# If Two Opposite Vertices of a Square Are (5, 4) and (1, −6), Find the Coordinates of Its Remaining Two Vertices. - Mathematics

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).

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.2 | Q 55 | Page 17