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secondary 3 | A Maths
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Date Posted: 2 years ago
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2 portions:
(1) Right angle
(2) isosceles
(1) To show an angle is perpendicular,
Multiplication of Gradient of 2 lines = -1
In other words:
Gradient of AB * Gradient of BC = -1
Grad of AB = (5 + 1) / (3-1)
= 3
Grad of BC = (-1-1) / (9-3)
= -2/6
= -1/3
:. Grad of AB * Grad of BC = 3 * -1/3
= -1
Hence AB perpendicular to BC,
And ABC is a right angle
(2) showing isosceles:
Length of AB = Length of BC
AB = square root ( (1-3)^2 + (5+1)^2)
= square root (40)
BC = square root ( (3-9)^2 + (-1-1)^2)
= square root (40)
= AB
Hence AB = BC
From (1) and (2), :. ABC is a right angled
Isosceles triangle (shown) //
(1) Right angle
(2) isosceles
(1) To show an angle is perpendicular,
Multiplication of Gradient of 2 lines = -1
In other words:
Gradient of AB * Gradient of BC = -1
Grad of AB = (5 + 1) / (3-1)
= 3
Grad of BC = (-1-1) / (9-3)
= -2/6
= -1/3
:. Grad of AB * Grad of BC = 3 * -1/3
= -1
Hence AB perpendicular to BC,
And ABC is a right angle
(2) showing isosceles:
Length of AB = Length of BC
AB = square root ( (1-3)^2 + (5+1)^2)
= square root (40)
BC = square root ( (3-9)^2 + (-1-1)^2)
= square root (40)
= AB
Hence AB = BC
From (1) and (2), :. ABC is a right angled
Isosceles triangle (shown) //
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