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

**Solution:**

Let's draw a figure of a square with the two opposite vertices (-1, 2) and (3, 2),

Let ABCD be a square having known vertices A (- 1, 2) and C (3, 2) respectively.

Let B(x₁, y₁) and D(x₂, y₂) be the two unknown vertex

We know that the sides of a square are equal to each other.

Therefore, AB = BC

By Using Distance formula on AB = AC with A (- 1, 2), B(x₁, y₁) and C (3, 2)

√ [(x₁ - (-1))^{2} + (y₁ - 2)^{2}] = √ [(x₁ - 3)^{2} + (y₁ - 2)^{2}]

x₁^{2} + 2x₁ + 1 + y₁^{2} - 4y₁ + 4 = x₁^{2} + 9 - 6x₁ + y₁^{2} + 4 - 4y₁ (By Simplifying & Transposing)

8x₁ = 8

x₁ = 1

We know that in a square, all interior angles are 90 degrees.

In ΔABC

AB^{2} + BC^{2} = AC^{2} [By Pythagoras theorem]

The distance formula is used to find the distance between AB, BC, and AC

(x₁ - (-1))^{2} + (y₁ - 2)^{2 }+ (x₁ - 3)^{2} + (y₁ - 2)^{2} = [3 - (-1)]^{2} + [ 2 - 2 ]^{2}

By using x₁ = 1,

(1 + 1)^{2} + (y₁ - 2)^{2} + (1 - 3)^{2} + (y₁ - 2)^{2} = 16

4 + y₁^{2} + 4 - 4y₁ + 4 + y₁^{2} - 4y₁ + 4 = 16

2y₁^{2} + 16 - 8y₁ = 16

2y₁^{2} - 8y₁ = 0

y₁ (y₁ - 4) = 0

y₁ = 0 or 4

Now, we have got the coordinates of point B(1, 0)

Let's plot the square on a graph as shown below:

We see that the vertex opposite to (1, 0) is (1, 4)

Hence, for point D we have the coordinates x₂ = 1, y₂ = 4

Hence the required vertices are B (1, 0) and D (1, 4).

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 7

**Video Solution:**

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

Maths NCERT Solutions Class 10 Chapter 7 Exercise 7.4 Question 4

**Summary:**

The two opposite vertices of a square are (- 1, 2) and (3, 2). Then the coordinates of the other two vertices are B (1, 0) and D (1, 4).

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