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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 = (08)2+(93)2

    = (8)2+(6)2

    = 64+36

    = 100

    = 10 units.

BC = (140)2+(119)2

    = 142+(2)2

    = 196+4

    = 200

    = 10√2 units.

 

CA = (814)2+(311)2

     = (6)2+(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 = (2(3))2+(42)2

            = (5)2+(6)2

            = 25+36

            = 61 units.

 

 

A’B’ = (2(3))2+(4(2))2

      =  12+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 = (51)2+(42)2

           = 42+22

           = 16+4

           = 20

           = 2×2×5

           = 25 units.

BC = (35)2+(84)2

     = (2)2+42

     = 4+16

     = 20

     = 2×2×5

     = 25 units.

 

CD = (13)2+(68)2

     = (4)2+(2)2

     = 16+4

     = 20

     = 2×2×5

     = 25 units.

and DA = (1+1)2+(26)2

           = 22+(4)2

           = 4+16

           = 20

           = 2×2×5

           = 25 units.

Thus, AB = BC = CD = DA

Diagonal AC = (31)2+(82)2

                 = 22+(6)2

                 = 4+36

                 = 40

                 = 2×2×2×5

                 = 210 units.

 Diagonal BD = (15)2+(64)2

                 = (6)2+22

                 = 36+4

                 = 40

                 = 2×2×2×5

                 = 210 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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