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
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