Two opposite vertices of a square are - 1 2 and 3 to find the coordinates of other two vertices

Given:

Two opposite vertices of a square are $(-1, 2)$ and $(3, 2)$.

To do:

We have to find the coordinates of other two vertices.

Solution:

Let ABCD be the given square and $A (-1, 2)$ and $C (3, 2)$ are the opposite vertices.
Let the coordinates of $B$ be $(x, y)$. Join AC.

This implies,

$AB=BC=CD=DA$

We know that,

The distance between two points $ \mathrm{A}\left(x_{1}, y_{1}\right) $ and $ \mathrm{B}\left(x_{2}, y_{2}\right) $ is $ \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} $.

Therefore,

$ \mathrm{AB}=\mathrm{BC} $

Squaring on both sides, we get,

$ \mathrm{AB}^{2}=\mathrm{BC}^{2} $

$ \Rightarrow (x+1)^{2}+(y-2)^{2}=(x-3)^{2}+(y-2)^{2} $

$ \Rightarrow x^{2}+2 x+1+y^{2}-4 y+4 =x^{2}-6 x+9+y^{2}-4 y+4 $

$ \Rightarrow 2 x+6 x-4 y+4 y=13-5 $

$ \Rightarrow 8 x=8 $

$ \Rightarrow x=1 $.........(i)

$ \mathrm{ABC} $ is a right angled triangle.

$ \Rightarrow \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2} $

$ \Rightarrow (3+1)^{2}+(2-2)^{2}=x^{2}+2 x+1+y^{2}-4 y +4+x^{2}-6 x+9+y^{2}-4 y+4 $

$ (4)^{2}=2 x^{2}+2 y^{2}-4 x-8 y+18 $

$ 16=2 x^{2}+2 y^{2}-4 x-8 y+18 $

$ \Rightarrow 2 x^{2}+2 y^{2}-4 x-8 y+18-16=0 $

$ \Rightarrow 2 x^{2}+2 y^{2}-4 x-8 y+2=0 $

$ \Rightarrow 2 (x^{2}+y^{2}-2 x-4 y+1)=0 $

$ \Rightarrow x^{2}+y^{2}-2x-4y+1=0 $

Substituting $ x=1 $, we get,

$ (1)^{2}+y^{2}-2(1)-4 y+1=0 $

$ \Rightarrow 1+y^2-2-4y+1=0 $

$ \Rightarrow y^2-4y=0 $

$ \Rightarrow y(y-4)=0 $

$ \Rightarrow(y)(y-4)=0 $

$ y=0 $ or $ y-4=0 $

$ y=0 $ or $ y=4 $

Therefore, the other points of the square are $(1,0)$ and $(1,4)$.

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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).

Last updated at Aug. 16, 2021 by

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Ex 7.4, 4 The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices. The two given vertices of the square ABCD are A(−1, 2) and C(3, 2). Let B(x, y) be the unknown vertex. We know, all the sides of square are equal. AB = CB Using Distance Formula, AB = √((𝑥+1)^2+(𝑦−2)^2 ) CB = √((𝑥−3)^2+(𝑦−2)^2 ) Now, AB = CB √((𝑥+1)^2+(𝑦−2)^2 )=√((𝑥−3)^2+(𝑦−2)^2 ) Cancelling the square roots (𝑥+1)^2+(𝑦−2)^2= (𝑥−3)^2+(𝑦−2)^2 (𝑥+1)^2= (𝑥−3)^2+(𝒚−𝟐)^𝟐−(𝒚−𝟐)^𝟐 (𝑥+1)^2=(𝑥−3)^2 (𝑥+1)^2−(𝑥−3)^2=0 𝑥^2+1+2𝑥−(𝑥^2+9−6𝑥)=0 𝑥^2+1+2𝑥−𝑥^2−9+6𝑥=0 8𝑥−8=0 8𝑥=8 𝑥=8/8 𝑥=1 Join AC. Since all angles of square is 90° Δ ABC is a right angled triangle Applying Pythagoras theorem, 〖𝐴𝐶〗^2=𝐴𝐵^2+𝐵𝐶^2 [√((3+1)^2+(2−2)^2 )]^2 =[√((𝑥+1)^2+(𝑦−2)^2 )]^2+ [√((𝑥−3)^2+(𝑦−2)^2 )]^2 〖(4)〗^2+(0)^2=(𝑥+1)^2+(𝑦−2)^2+(𝑥−3)^2+(𝑦−2)^2 16 = 𝑥^2+1+2𝑥+𝑦^2+4−4𝑦+𝑥^2+9−6𝑥+𝑦^2+4−4𝑦 16 = 〖2𝑥〗^2−4𝑥+2𝑦^2−8𝑦+18 Putting x = 1 16 = 〖2(1)〗^2−4(1)+2𝑦^2−8𝑦+18 16 = 2 − 4 + 〖2𝑦〗^2−8𝑦+18 16 = 16 + 〖2𝑦〗^2−8𝑦 16 − 16 = 〖2𝑦〗^2−8𝑦 0 = 〖2𝑦〗^2−8𝑦 0 = 2𝑦 (𝑦−4) 𝑦 (𝑦−4) = 0 So, 𝑦=0 and 𝑦 −4=0 So, 𝑦=0 and 𝑦=4 Thus, the other two vertices are (1, 0) and (1, 4). The Square will look like this.

The two opposite vertices of a square are 1,2 and 3,2. Find the co ordinates of the other two vertices.

B (1, 0) or B ( 1,4) and D(2, 4) or D(2,0)

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B (2,3) or B ( 2,4) and D(3, 4) or D(3,0)

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B (2,0) or B ( 1,4) and D(2, 4) or D(1,0)

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B (0,1) or B (0,2) and D(3, 1) or D(3,-2)

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