CHECK GIVEN FOUR POINTS FORM SQUARE

(i)  Plot the given points in the graph and draw the square.

(ii)  Now we need to find the length of all sides.

(iii)  In a square length of all sides are equal.

(iv)  Length of diagonals are also equal

If we prove the above two conditions(iii, iv) are true, we can decide that the given points form a square.

Question 1 :

Examine whether the given points 

A (-9, -7), B (-6, -7), C (-6, -4) and D (-9, -4)

form a square.

Solution :

Step 1 :

Step 2 :

Finding the length of all sides.

Distance Between Two Points (x₁, y₁) and (x₂, y₂)

√(x₂ - x₁)² + (y₂ - y₁)²

Length of side AB :

Here x₁ = -9, y₁ = -7, x₂ = -6  and  y₂ = -7

=  √(-6-(-9))² + (-7-(-7))²

=  √(-6+9)² + (-7+7)²

=  √3² + 0²

=  √9

=  3 units ---(1)

Length of side BC :

Here x₁ = -6, y₁ = -7, x₂ = -6  and  y₂ = -4

=  √(-6-(-6))² + (-4-(-7))²

=  √(-6+6)² + (-4+7)²

=  √0² + 3²

=  √9

=  3 units ---(2)

Length of side CD :

Here x₁ = -6, y₁ = -4, x₂ = -9  and  y₂ = -4

=  √(-9-(-6))² + (-4-(-4))²

=  √(-9+6)² + (-4+4)²

=  √(-3)² + 0²

=    √9

=  3 units ---(3)

Length of side DA :

Here x₁ = -9, y₁ = -4, x₂ = -9  and  y₂ = -7

=  √(-9-(-9))² + (-7-(-4))²

=  √(-9+9)² + (-7+4)²

=  √0² + 3²

=  √9

=  3 units ---(4)

(1) = (2) = (3) = (4)

Length of all sides are equal.

Step 3 :

Length of diagonal AC :

Here x₁ = -9, y₁ = -7, x₂ = -6  and  y₂ = -4

=  √(-6-(-9))² + (-4-(-7))²

=  √(-6+9)² + (-4+7)²

=  √3² + 3²

=  √9 + 9

=  √18 units

Length of diagonal BD :

Here x₁ = -6, y₁ = -7, x₂ = -9  and  y₂ = -4

=  √(-9-(-6))² + (-4-(-7))²

=  √(-9+6)² + (-4+7)²

=  √(-3)² + 3²

=  √9 + 9

=  √18 units

Length of diagonal AC  =  Length of diagonal BD

So, the given points are vertices of square.

Question 2 :

Examine whether the given points

A (1, 2), B (2, 2), C (2, 3) and D (1, 3)

forms a square.

Solution :

Step 1 :

Step 2 :

Let the given points are A (1, 2) and B (2, 2) and C (2, 3) and D (1, 3).

Length of side AB :

Here x₁ = 1, y₁ = 2, x₂ = 2  and  y₂ = 2

=  √(2-1)² + (2-2)²

=  √(1)² + (0)²

=  √1 + 0²

=  1 unit ----(1)

Length of side BC :

Here x₁ = 2, y₁ = 2, x₂ = 2  and  y₂ = 3

=  √(2-2)² + (3-2)²

=  √(0)² + (1)²

=  √0² + 1

=  1 unit ----(2)

Length of side CD :

Here x₁ = 2, y₁ = 3, x₂ = 1  and  y₂ = 3

=  √(1-2)² + (3-3)²

=  √(-1)² + (0)²

=  √1 + 0²

=  1 unit ----(3)

Length of side CD :

Here x₁ = 1, y₁ = 3, x₂ = 1  and  y₂ = 2

=  √(1-1)² + (2-3)²

=  √(0)² + (-1)²

=  √0 + 1

=  1 unit  ----(4)

(1) = (2) = (3) = (4)

Step 3 :

Length of diagonal AC :

Here x₁ = 1, y₁ = 2, x₂ = 2  and  y₂ = 3

=  √(2-1)² + (3-2)²

=  √(1)² + 1²

=  √1 + 1

=  √2 units

Length of diagonal BD :

Here x₁ = 2, y₁ = 2, x₂ = 1  and  y₂ = 3

=  √(1-2)² + (3-2)²

=  √(-1)² + (1)

=  √1 + 1

=  √2 units

Length of diagonal AC  =  Length of diagonal BD

So, the given points are vertices of square.

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