If two opposite vertices of a square are (-1, 2) and (3, 2), then the coordinates of the other

Solution:

Let ABCD be a square with A(-1,2) and B(-1,2) (3,2). And Point O is the point where AC and BD intersect.

Now find the coordinates of the points B and D.

Step 1: Find coordinates of point O and distance between A and C.

As we know, the diagonals of a square are equal and bisect each other.

AC= √[(3 + 1)2 + (2 – 2)2] = 4

The coordinates of O can be found using the formula below:

x = (3 – 1)/2 = 1 and y = (2 + 2)/2 = 2

So, O(1,2)

Step 2: Using the Pythagoras theorem find the side of the square

Let a be the square’s side, and AC = 4.

From the right angle triangle, ACD,

a = 2√2

As a result, each square’s side= 2√2

Step 3: Find coordinates of point D

AD and CD length measure are equated.

Assume D’s coordinate are (x1, y1)

AD = √[(x1 + 1)2 + (y1 – 2)2]

When both sides are squared,

AD2 = (x1 + 1)2 + (y1 – 2)2

In the same way, CD2  = (x1 – 3)2 + (y1 – 2)2

Because all of the sides of a square are equal, therefore AD = CD

(x1 + 1)2 + (y1 – 2)2 = (x1 – 3)2 + (y1 – 2)2

x12 + 1 + 2x1 = x12 + 9 – 6x1

8x1 = 8

x1 = 1

By using the value of x, we can calculate the value of y1 as follows.

From step 2: each square’s side = 2√2

CD2 = (x1 – 3)2 + (y1 – 2)2

8 = (1 – 3)2 + (y1 – 2)2

8 = 4 + (y1 – 2)2

y1 – 2 = 2

y1 = 4

As a result, D = (1, 4)

Step 4: Find coordinates of point B

From the BOD line segment

B’s coordinates can be calculated using O’s coordinates, as follows:

We had already calculated as O = (1, 2)

Say B = (x2, y2)

For BD;

1 = (x2 + 1)/2

x2 = 1

And 2 = (y2 + 4)/2

=> y2 = 0

As a result, the coordinates of the required points are as B = (1,0) and D = (1,4)

Last updated at Aug. 16, 2021 by

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Let ABCD be a square having (−1, 2) and (3, 2) as vertices A and C respectively. Let (x, y), (x1, y1) be the coordinate of vertex B and D respectively.

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

∴ AB = BC

`=>sqrt((x+1)^2 + (y-2)^2) = sqrt((x-3)^2 + (y-2)^2)`

=>x2 + 2x + 1 + y2 -4y + 4 = x2 + 9  -6x + y2 + 4 - 4y

⇒ 8x = 8

⇒ x = 1

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

In ΔABC,

AB2 + BC2 = AC2

`=> (sqrt(((1+1)^2)+(y-2)^2))^2 + (sqrt(((1-3)^2)+(y-2)^2))^2 = (sqrt((3+1)^2+(2-2)^2))^2`

⇒ 4 + y2 + 4 − 4y + 4 + y2 − 4y + 4 =16

⇒ 2y2 + 16 − 8 y =16

⇒ 2y2 − 8 y = 0

⇒ y (y − 4) = 0

⇒ y = 0 or 4

We know that in a square, the diagonals are of equal length and bisect each other at 90°. Let O be the mid-point of AC. Therefore, it will also be the mid-point of BD

Coordinate of point O = ((-1+3)/2, (2+2)/2)

`((1+x_1)/2, (y+ y_1)/2) = (1,2)`

`(1+x_1)/2 = 1`

1+x1=2

x1 =1

and

` (y + y_1)/2 = 2`

⇒ y + y1 = 4

If y = 0,

y1 = 4

If y = 4,

y1 = 0

Therefore, the required coordinates are (1, 0) and (1, 4).

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]

x₁2 + 2x₁ + 1 + y₁2 - 4y₁ + 4 = x₁2 + 9 - 6x₁ + y₁2 + 4 - 4y₁ (By Simplifying & Transposing)

8x₁ = 8

x₁ = 1

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

In ΔABC

AB2 + BC2 = AC2 [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 + y₁2 + 4 - 4y₁ + 4 + y₁2 - 4y₁ + 4 = 16

2y₁2 + 16 - 8y₁ = 16

2y₁2 - 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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