Two opposite vertices of a square are (5, 4) and (1, -6) then the remaining vertices are

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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Answer : Given :If two opposite vertices of a square are ( 5,4 ) and ( 1,6)

To find : the coordinates of its remaining vertices

Let B (x, y).

We have AB = BC

Squaring both sides

=> AB2 = BC2

=> (x-5)2 +(y-4)2 = (x-1)2 +(y-6)2

=> x2 +25 -10x +y2 +16 - 8y = x2 +1 -2x + y2 +36 -12y

=> 4 - 8x +4y =0

=>2x - y =1 .........(1)

also AC2 = (5-1)2 + (4-6)2 = 16 + 4 = 20

In right triangle ABC, by Pythagoras theorem,

AC2 = AB2+BC2

=> AC2 = 2AB2

=> 20 = 2 ( (x-5)2 +(y-4)2 )

=> 20 = 2 ( x2 + 25 -10x+ y2 +16 -8y )

=> 10= x2 + 25 -10x+ y2 +16 -8y

=> x2 -10x+ y2 -8y + 31 =0

=> x2 -10x+ (2x+1)2 - 8(2x+1) + 31 =0 {using eq 1}

=> x2 -10x+ 4x2 + 1 + 4x -16x-8 + 31 =0

=>5 x2 -22x +24 =0

D = b2 -4ac

=> D = (-22)2 - 4(5)(24)

=> D = 484 - 480

=> D =4

=> D1/2 = 2

x = (-b+D1/2) / 2a , (-b - D1/2) / 2a

=> x = (22 +2) / 10 , (22-2) / 10

=> x = 2.4 , 2

=> y = 2x+1 , => y = 5.8 , 5

therefore the points are ( 2.4 , 5.8) and ( 2,5) Answer

Given:

Two opposite vertices of a square are $(5, 4)$ and $(1, -6)$.

To do:

We have to find the coordinates of its remaining two vertices.

Solution:

Let ABCD be the given square and $A (5, 4)$ and $C (1, -6)$ 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-5)^{2}+(y-4)^{2}=(x-1)^{2}+(y+6)^{2} $

$ \Rightarrow x^{2}-10 x+25+y^{2}-8 y+16 =x^{2}-2 x+1+y^{2}+12 y+36 $

$ \Rightarrow -10 x+2 x-8 y-12 y=37-41 $

$ \Rightarrow -8 x-20 y=-4 $

$ \Rightarrow -4(2 x+5 y)=-4 $

$ \Rightarrow 2 x+5 y=1 $

$ \Rightarrow 2 x=1-5 y $$ \Rightarrow x=\frac{1-5 y}{2} $

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

$ \Rightarrow \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2} $$ \Rightarrow (5-1)^{2}+(4+6)^{2}=x^{2}-10 x+25+y^{2}-8 y +16+x^{2}-2 x+1+y^{2}+12 y+36 $$ (4)^{2}+(10)^{2}=2 x^{2}+2 y^{2}-12 x+4 y+78 $

$ 16+100=2 x^{2}+2 y^{2}-12 x+4 y+78 $

$ \Rightarrow 2 x^{2}+2 y^{2}-12 x+4 y+78-16-100=0 $

$ \Rightarrow 2 x^{2}+2 y^{2}-12 x+4 y-38=0 $

$ \Rightarrow x^{2}+y^{2}-6 x+2 y-19=0 $

Substituting $ x=\frac{1-5 y}{2} $

$ (\frac{1-5 y}{2})^{2}+y^{2}-6(\frac{1-5 y}{2})+2 y-19=0 $

$ \Rightarrow \frac{1+25 y^{2}-10 y}{4}+y^{2}-3(1-5 y)+2 y-19=0 $

$ \Rightarrow 1+25 y^{2}-10 y+4 y^{2}-12(1-15 y)+8 y-76=0 $

$ \Rightarrow 1+25 y^{2}-10 y+4 y^{2}-12+60 y+8 y-76=0 $

$ \Rightarrow 29 y^{2}+58 y-87=0 $$ \Rightarrow y^{2}+2 y-3=0 $$ \Rightarrow y^{2}+3 y-y-3=0 $$ \Rightarrow y(y+3)-1(y+3)=0 $

$ \Rightarrow(y+3)(y-1)=0 $

$ y+3=0 $ or $ y-1=0 $