Question

Show that the following points are vertices of a
square:
(i) (5, 6), (1,5), (2, 1) and (6,2)
(3, 2), (0, 5), (-3, 2) and (0, -1).​

Answer 1

Answer:

AB=

$\sqrt{17}$

BC=

$\sqrt{17}$

CD=

$\sqrt{17}$

AD=

$\sqrt{17}$

AC=

$\sqrt{34}$

BD=

$\sqrt{34}$

ii)Given:

A(3, 2) = (x₁, y₁)

B(0,5) = (x₂, y₂)

C(-3, 2) = (x₃, y₃)

D(0, -1) = (x₄, y₄)

To Prove:

Given points (ABCD) are the vertices of square.

Solution:

By applying the distance formula,

Distance of AB = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

(x

2

−x

1

)

2

+(y

2

−y

1

)

2

On substituting the values, we get

= \sqrt{(0-3)^2+(5-2)^2}

(0−3)

2

+(5−2)

2

= \sqrt{9+9}

9+9

= \sqrt{18}

18

Distance of BC = \sqrt{(x_3-x_2)^2+(y_3-y_2)^2}

(x

3

−x

2

)

2

+(y

3

−y

2

)

2

On substituting the value, we get

= \sqrt{(-3-0)^2+(2-5)^2}

(−3−0)

2

+(2−5)

2

= \sqrt{9+9}

9+9

= \sqrt{18}

18

Distance of CD = \sqrt{(x_4-x_3)^2+(y_4-y_3)^2}

(x

4

−x

3

)

2

+(y

4

−y

3

)

2

On substituting the values, we get

= \begin{gathered}\sqrt{(0-(-3))^2\\+(-1-2)^2}\end{gathered}

= \sqrt{9+9}

9+9

= \sqrt{18}

18

Distance of DA = \sqrt{(x_4-x_1)^2+(y_4-y_1)^2}

(x

4

−x

1

)

2

+(y

4

−y

1

)

2

On substituting the values, we get

= \sqrt{(0-3)^2+(-1-2)^2}

(0−3)

2

+(−1−2)

2

= \sqrt{9+9}

9+9

= \sqrt{18}

18

All the four sides are equal. Thus, the given points are the vertices of a square.

Step-by-step explanation:

Since AB=BC=CD=AD and diagonals AC= BD

Therefore, ABCD is a square.