Question

Show that the points A(1, 6), B(4,2), C(1, -2) and D(-2, 2) are the vertices of a square.

Answer 1

Question:

Show that the points A (1,6), B(4,2), C(1,-2), D (-2,2) are the vertices of a rhombus.

Answer:

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Point A (1, 6 )
• Point B (4, 2)
• Point C (1, -2)
• Point D (-2, 2)

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• To show that points are vertices of a square

$\Large{\underline{\underline{\bf{Solution:}}}}$

➞ Here we have to prove that all the sides are equal and diagonals are equal

➞ By distance formula we know that distance between 2 points is given by,

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

➞ First find the distance of AB

    $\sf{AB=\sqrt{(4-1)^{2} +(2-6)^{2} } }$

    $\sf{AB=\sqrt{9+16}}$

    AB = √25

    AB = 5 units-----(1)

➞ Now distance of BC is given by,

    $\sf{BC=\sqrt{(1-4)^{2}+(-2-2)^{2} } }$

    $\sf{BC=\sqrt{9+16}$

    BC = √25

    BC = 5 units----(2)

➞ Distance of CD is given by,

    $\sf{DC=\sqrt{(-2-1)^{2} +(2+2)^{2} } }$

    $\sf{DC=\sqrt{9+16}}$

    DC = 5 units----(3)

➞ Now AD is given by,

    $\sf{AD=\sqrt{(-2-1)^{2} +(2-6)^{2} } }$

    $\sf{AD=\sqrt{9+16}}$

    AD = 5 units----(4)

➞ From equations 1,2, 3, 4,

    AB = BC = CD = DA

➞ Hence ABCD is a rhombus

➞ Distance of AC is given by,

    $\sf{AC=\sqrt{(1-1)^{2}+(-2-6)^{2} } }$

    AC = √64

    AC = 8 units------(5)

➞ Distance of BD is given by,

    $\sf{BD=\sqrt{(-2-4)^{2}+(2-2)^{2} } }$

    BD = √36

    BD = 6 units-----(6)

➞ From equations 5 and 6,

    AC ≠ BD

➞ Hence ABCD is a rhombus.

➞ Hence proved.

$\Large{\underline{\underline{\bf{Notes:}}}}$

➞ The distance formula is given by,

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

Answer 2

Answer :-

★ Question :-

Show that the points A(1, 6), B(4,2), C(1, -2) and D(-2, 2) are the vertices of a rhombus.

★ Concept :-

Here the concept of Distance formula has been used. According to this formula, the distance between two points are given when we have their coordinates as :-

• Distance = √{(x2 - x1)² + (y2 - y1)²}

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★ Solution :-

Given,

» Coordinates of A = (1, 6)

» Coordinates of B = (4, 2)

» Coordinates of C = (1, -2)

» Coordinates of D = (-2, 2)

________________________________

If we need to prove that these give coordinates are the coordinates of a square then first we must prove that they all are equal in length and the diagonals, that is if square is ABCD then AC should be equal to BD.

By using distance formula we get,

• First we will find the distance of AB.

✏ AB = √{(4 - 1)² + (2 - 6)²}

✏ AB = √{9 + 16}

✏ AB = √25

✏ AB = 5 units

• Now we will find the distance of BC.

✏ BC = √{(1 - 4)² + (-2 -2)²}

✏ BC = √{(9 + 16}

✏ AB = √25

✏ AB = 5 units

• Now we will find the distance of CD.

✏ CD = √{(-2 - 1)² + (2 + 2)²}

✏ CD = √{9 + 16}

✏ CD = √25

✏ CD = 5 units

• Now we will be going ahead for AD.

✏ AD = √{(-2 - 1)² + (2 - 6)²}

✏ AD = √{9 + 16}

✏ AD = √25

✏ AD = 5 units

Clearly, AB = BC = CD = AD = 5 units. Hence, ABCD is a square but perfectly a rhombus.

• Now let us check AC.

✏ AC = √{(-1 - 1)² + (-2 - 6)²}

✏ AC = √64

✏ AC = 8 units

• Now let us check BD.

✏ BD = √{(-2 - 4)² + (2 - 2)²}

✏ BD = √36

✏ BD = 6 units

Clearly, AC ≠ BD. Hence, diagonals aren't equal.

Hence, the given coordinates are the coordinates of a parallelogram.

Hence proved.

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★ More to know :-

•Rhombus is a closed 4 sided two dimensional figure which has all the sides equal but diagonals unequal. Rhombus is a square.

• Square is a closed 4 sided two dimensional figure which has all sides equal and diagonals also equal. Every square is a rectangle.

• Rectangle is also a closed 4 sided two dimensional figure which has opposite sides equal and diagonals bisect each other.

• Parallelogram is a closed 4 sided two dimensional figure which had opposite sides parallel. Rhombus,square and rectangle are all parallelogram.