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secondary 4 | E Maths
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Date Posted: 4 years ago
Views: 232
A (-2, 5)
B (4, 8)
C (0, y)
Since AC = BC,
sqrt [(0 - (-2))^2 + (y - 5)^2] = sqrt [(0 - 4)^2 + (y - 8)^2]
sqrt [4 + (y - 5)^2] = sqrt [16 + (y - 8)^2]
Squaring both sides,
4 + (y - 5)^2 = 16 + (y - 8)^2
4 + y2 - 10y + 25 = 16 + y2 - 16y + 64
29 - 10y = 80 - 16y
6y = 51
y = 17/2 = 8.5
So C is (0, 8.5).
B (4, 8)
C (0, y)
Since AC = BC,
sqrt [(0 - (-2))^2 + (y - 5)^2] = sqrt [(0 - 4)^2 + (y - 8)^2]
sqrt [4 + (y - 5)^2] = sqrt [16 + (y - 8)^2]
Squaring both sides,
4 + (y - 5)^2 = 16 + (y - 8)^2
4 + y2 - 10y + 25 = 16 + y2 - 16y + 64
29 - 10y = 80 - 16y
6y = 51
y = 17/2 = 8.5
So C is (0, 8.5).
Up to the square both sides part you are correct, but we cannot just cancel the squares after that. It is mathematically incorrect. We must expand and solve the equation.
Tysm ! Was thinking of expanding , but I didn’t know that I’ll have to cancel the y^2
In such questions the y^2 will almost always cancel out (an exception is only when A and B are at the same height y = k and equally far from the x-axis, where C can have two values).
Note that had angle ACB been 90 degrees, there is an easier approach than forming such equations.
Note that had angle ACB been 90 degrees, there is an easier approach than forming such equations.
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Let C be (k, 0). If you need to remove the square to find the value of k, u have to use the special products of (a + b)² = a² + 2ab + b² to expand the square.
Date Posted:
4 years ago
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For (b)
Date Posted:
4 years ago
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for (b)
Date Posted:
4 years ago
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