Question

show that the points A (0,0) B(3,0), C (4,1) and D(1,1) from a parallelogram.

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Answer 1

Answer:

Step-by-step explanation:

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

• Point A (0,0)
• Point B (3,0)
• Point C (4,1)
• Point D(1,1)

$\Large{\underline{\underline{\sf{To\:Prove:}}}}$

• The points form a parallelogram

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

→ We need to find the distance between the points AB, BC, CD , DA by using distance formula

→ Distance between two points is given by the formula,

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

→ Distance of  AB is given by

  $AB=\sqrt{(3-0)^{2}+(0-0)^{2} }$

  $AB=\sqrt{9}$

  AB = 3 units------equation 1

→ Distance of BC is given by

  $BC=\sqrt{(4-3)^{2} +(1-0)^{2} }$

  $BC=\sqrt{1+1}$

  BC = √2 units------equation 2

→ Distance of CD is given by

  $CD=\sqrt{(1-4)^{2}+(1-1)^{2} }$

  $CD=\sqrt{(-3)^{2} }$

  $CD=\sqrt{9}$

  CD = 3 units------equation 3

→ Distance of DA is given by

   $DA=\sqrt{(0-1)^{2} +(0-1)^{2} }$

   $DA=\sqrt{1+1}$

   DA = √2 units------equation 4

→ From equation 1, 2 , 3, 4 we can see that

   AB = CD

   BC = DA

→ The opposite sides of the quadrilateral are equal.

→ Hence the points form a parallelogram

→ Hence proved.

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

→ Distance formula is given by

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

→ In a parallelogram,

• Opposite sides are equal
• Opposite sides are parallel
• Opposite angles are equal
• Adjacent angles are supplementary
• The diagonals of a parallelogram bisect each other.