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 = \(\sqrt{(0 - 8)^{2} + (9 - 3)^{2}}\)

    = \(\sqrt{(-8)^{2} + (6)^{2}}\)

    = \(\sqrt{64 + 36}\)

    = \(\sqrt{100}\)

    = 10 units.

BC = \(\sqrt{(14 - 0)^{2} + (11 - 9)^{2}}\)

    = \(\sqrt{14^{2} + (2)^{2}}\)

    = \(\sqrt{196 + 4}\)

    = \(\sqrt{200}\)

    = 10√2 units.

 

CA = \(\sqrt{(8 - 14)^{2} + (3 - 11)^{2}}\)

     = \(\sqrt{(-6)^{2} + (-8)^{2}}\)

     = \(\sqrt{36 + 64}\)

    = \(\sqrt{100}\)

    = 10 units.

AB\(^{2}\) + CA\(^{2}\) = 100 + 100 = 200 = BC\(^{2}\)

BC\(^{2}\) = AB\(^{2}\) + CA\(^{2}\) ⟹ 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 = \(\sqrt{(2 - (-3))^{2} + (-4 - 2)^{2}}\)

            = \(\sqrt{(5)^{2} + (-6)^{2}}\)

            = \(\sqrt{25 + 36}\)

            = \(\sqrt{61}\) units.

 

 

A’B’ = \(\sqrt{(-2 - (-3))^{2} + (4 - (-2))^{2}}\)

      =  \(\sqrt{1^{2} + 6^{2}}\)

      = \(\sqrt{1 + 36}\)

      = \(\sqrt{37}\) 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 = \(\sqrt{(5 - 1)^{2} + (4 - 2)^{2}}\)

           = \(\sqrt{4^{2} + 2^{2}}\)

           = \(\sqrt{16 + 4}\)

           = \(\sqrt{20}\)

           = \(\sqrt{2 × 2 × 5}\)

           = 2\(\sqrt{5}\) units.

BC = \(\sqrt{(3 - 5)^{2} + (8 - 4)^{2}}\)

     = \(\sqrt{(-2)^{2} + 4^{2}}\)

     = \(\sqrt{4 + 16}\)

     = \(\sqrt{20}\)

     = \(\sqrt{2 × 2 × 5}\)

     = 2\(\sqrt{5}\) units.

 

CD = \(\sqrt{(-1 - 3)^{2} + (6 - 8)^{2}}\)

     = \(\sqrt{(-4)^{2} + (-2)^{2}}\)

     = \(\sqrt{16 + 4}\)

     = \(\sqrt{20}\)

     = \(\sqrt{2 × 2 × 5}\)

     = 2\(\sqrt{5}\) units.

and DA = \(\sqrt{(1 + 1)^{2} + (2 - 6)^{2}}\)

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

           = \(\sqrt{4 + 16}\)

           = \(\sqrt{20}\)

           = \(\sqrt{2 × 2 × 5}\)

           = 2\(\sqrt{5}\) units.

Thus, AB = BC = CD = DA

Diagonal AC = \(\sqrt{(3 - 1)^{2} + (8 - 2)^{2}}\)

                 = \(\sqrt{2^{2} + (-6)^{2}}\)

                 = \(\sqrt{4 + 36}\)

                 = \(\sqrt{40}\)

                 = \(\sqrt{2 × 2 × 2 × 5}\)

                 = 2\(\sqrt{10}\) units.

 Diagonal BD = \(\sqrt{(-1 - 5)^{2} + (6 - 4)^{2}}\)

                 = \(\sqrt{(-6)^{2} + 2^{2}}\)

                 = \(\sqrt{36 + 4}\)

                 = \(\sqrt{40}\)

                 = \(\sqrt{2 × 2 × 2 × 5}\)

                 = 2\(\sqrt{10}\) 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

• Distance Formula
• Distance Properties in some Geometrical Figures
• Conditions of Collinearity of Three Points
• Problems on Distance Formula
• Distance of a Point from the Origin
• Distance Formula in Geometry
• Section Formula
• Midpoint Formula
• Centroid of a Triangle
• Worksheet on Distance Formula
• Worksheet on Collinearity of Three Points
• Worksheet on Finding the Centroid of a Triangle
• Worksheet on Section Formula

10th Grade Math

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