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 rightangled 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 rightangled triangle.
and, AB = CA ⟹ the triangle is isosceles.
Here, the triangle ABC is an isosceles rightangled triangle.
2. The point A (2, 4) is reflected in the origin on A’. The point B (3, 2) is reflected in the xaxis on B’. Compare the distances AB = A’B’.
Solution:
The point A (2, 4) is reflected in the origin on A’.
Therefore, the coordinates of A’ = (2, 4)
The point B (3, 2) is reflected in the xaxis on B’
Therefore, the coordinates 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
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