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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 = β(0β8)2+(9β3)2
= β(β8)2+(6)2
= β64+36
= β100
= 10 units.
BC = β(14β0)2+(11β9)2
= β142+(2)2
= β196+4
= β200
= 10β2 units.
CA = β(8β14)2+(3β11)2
= β(β6)2+(β8)2
= β36+64
= β100
= 10 units.
AB2 + CA2 = 100 + 100 = 200 = BC2
BC2 = AB2 + CA2 βΉ the triangle is right-angled triangle.
and, AB = CA βΉ 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 Aβ. The point B (-3, 2) is reflected in the x-axis on Bβ. Compare the distances AB = AβBβ.
Solution:
The point A (2, -4) is reflected in the origin on Aβ.
Therefore, the co-ordinates of Aβ = (-2, 4)
The point B (-3, 2) is reflected in the x-axis on Bβ
Therefore, the co-ordinates of Bβ = (-3, -2)
Now, AB = β(2β(β3))2+(β4β2)2
= β(5)2+(β6)2
= β25+36
= β61 units.
AβBβ = β(β2β(β3))2+(4β(β2))2
= β12+62
= β1+36
= β37 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 = β(5β1)2+(4β2)2
= β42+22
= β16+4
= β20
= β2Γ2Γ5
= 2β5 units.
BC = β(3β5)2+(8β4)2
= β(β2)2+42
= β4+16
= β20
= β2Γ2Γ5
= 2β5 units.
CD = β(β1β3)2+(6β8)2
= β(β4)2+(β2)2
= β16+4
= β20
= β2Γ2Γ5
= 2β5 units.
and DA = β(1+1)2+(2β6)2
= β22+(β4)2
= β4+16
= β20
= β2Γ2Γ5
= 2β5 units.
Thus, AB = BC = CD = DA
Diagonal AC = β(3β1)2+(8β2)2
= β22+(β6)2
= β4+36
= β40
= β2Γ2Γ2Γ5
= 2β10 units.
Diagonal BD = β(β1β5)2+(6β4)2
= β(β6)2+22
= β36+4
= β40
= β2Γ2Γ2Γ5
= 2β10 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.
β Distance and Section Formulae
10th Grade Math
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