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


10th Grade Math

From Worksheet on Distance Formula to HOME PAGE


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.



Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



Share this page: What’s this?

Recent Articles

  1. Adding 5-digit Numbers with Regrouping | 5-digit Addition |Addition

    Mar 18, 24 02:31 PM

    Adding 5-digit Numbers with Regrouping
    We will learn adding 5-digit numbers with regrouping. We have learnt the addition of 4-digit numbers with regrouping and now in the same way we will do addition of 5-digit numbers with regrouping. We…

    Read More

  2. Adding 4-digit Numbers with Regrouping | 4-digit Addition |Addition

    Mar 18, 24 12:19 PM

    Adding 4-digit Numbers with Regrouping
    We will learn adding 4-digit numbers with regrouping. Addition of 4-digit numbers can be done in the same way as we do addition of smaller numbers. We first arrange the numbers one below the other in…

    Read More

  3. Worksheet on Adding 4-digit Numbers without Regrouping | Answers |Math

    Mar 16, 24 05:02 PM

    Missing Digits in Addition
    In worksheet on adding 4-digit numbers without regrouping we will solve the addition of 4-digit numbers without regrouping or without carrying, 4-digit vertical addition, arrange in columns and add an…

    Read More

  4. Adding 4-digit Numbers without Regrouping | 4-digit Addition |Addition

    Mar 15, 24 04:52 PM

    Adding 4-digit Numbers without Regrouping
    We will learn adding 4-digit numbers without regrouping. We first arrange the numbers one below the other in place value columns and then add the digits under each column as shown in the following exa…

    Read More

  5. Addition of Three 3-Digit Numbers | With and With out Regrouping |Math

    Mar 15, 24 04:33 PM

    Addition of Three 3-Digit Numbers Without Regrouping
    Without regrouping: Adding three 3-digit numbers is same as adding two 3-digit numbers.

    Read More