# 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