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____From Worksheet on Distance Formula____ to HOME PAGE__

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