Distance Formula in Geometry

We will discuss here how to use the distance formula in geometry.

1. Show that the points A (8, 3), B (0, 9) and C (14, 11) are the vertices of an isosceles right-angled triangle.

Solution:

AB = √(0βˆ’8)2+(9βˆ’3)2

    = √(βˆ’8)2+(6)2

    = √64+36

    = √100

    = 10 units.

BC = √(14βˆ’0)2+(11βˆ’9)2

    = √142+(2)2

    = √196+4

    = √200

    = 10√2 units.

 

CA = √(8βˆ’14)2+(3βˆ’11)2

     = √(βˆ’6)2+(βˆ’8)2

     = √36+64

    = √100

    = 10 units.

AB2 + CA2 = 100 + 100 = 200 = BC2

BC2 = AB2 + CA2 ⟹ the triangle is right-angled triangle.

and, AB = CA ⟹ the triangle is isosceles.

Here, the triangle ABC is an isosceles right-angled triangle.

 

 

 

2. The point A (2, -4) is reflected in the origin on A’. The point B (-3, 2) is reflected in the x-axis on B’. Compare the distances AB = A’B’.

Solution:

The point A (2, -4) is reflected in the origin on A’.

Therefore, the co-ordinates of A’ = (-2, 4)

The point B (-3, 2) is reflected in the x-axis on B’

Therefore, the co-ordinates of B’ = (-3, -2)

Now, AB = √(2βˆ’(βˆ’3))2+(βˆ’4βˆ’2)2

            = √(5)2+(βˆ’6)2

            = √25+36

            = √61 units.

 

 

A’B’ = √(βˆ’2βˆ’(βˆ’3))2+(4βˆ’(βˆ’2))2

      =  βˆš12+62

      = √1+36

      = √37 units.

 

3. Prove that the points A (1, 2), B (5, 4), C (3, 8) and D (-1, 6) are the vertices of a rectangle.

Solution:

Let A (1, 2), B (5, 4), C (3, 8) and D (-1, 6) be the angular points of the quadrilateral ABCD.

Join AC and BD.

Now AB = √(5βˆ’1)2+(4βˆ’2)2

           = √42+22

           = √16+4

           = √20

           = √2Γ—2Γ—5

           = 2√5 units.

BC = √(3βˆ’5)2+(8βˆ’4)2

     = √(βˆ’2)2+42

     = √4+16

     = √20

     = √2Γ—2Γ—5

     = 2√5 units.

 

CD = √(βˆ’1βˆ’3)2+(6βˆ’8)2

     = √(βˆ’4)2+(βˆ’2)2

     = √16+4

     = √20

     = √2Γ—2Γ—5

     = 2√5 units.

and DA = √(1+1)2+(2βˆ’6)2

           = √22+(βˆ’4)2

           = √4+16

           = √20

           = √2Γ—2Γ—5

           = 2√5 units.

Thus, AB = BC = CD = DA

Diagonal AC = √(3βˆ’1)2+(8βˆ’2)2

                 = √22+(βˆ’6)2

                 = √4+36

                 = √40

                 = √2Γ—2Γ—2Γ—5

                 = 2√10 units.

 Diagonal BD = √(βˆ’1βˆ’5)2+(6βˆ’4)2

                 = √(βˆ’6)2+22

                 = √36+4

                 = √40

                 = √2Γ—2Γ—2Γ—5

                 = 2√10 units.

Therefore, Diagonal AC = Diagonal BD

Thus ABCD is a quadrilateral in which all sides are equal and the diagonals are equal.

Hence required ABCD is a square.

● Distance and Section Formulae


10th Grade Math

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