An equilateral triangle ABC has its centroid at the the origin and the base BC lies along #x+y+1=0#. Gradient of the other two lines are?
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Ans- #2+sqrt3, 2-sqrt3#
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Ans-
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Since the centroid of an equilateral triangle divides each median into segments with a ratio of 2:1, the midpoint of the base BC is also the centroid of the triangle.
Given that the base BC lies along the line (x + y + 1 = 0), its midpoint is the centroid, which is at the origin (0,0).
So, the coordinates of point B and C are (B(2, -2)) and (C(-2, 2)).
Now, we can find the gradients of the lines AB and AC:
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Gradient of AB: [ m_{AB} = \frac{{y_B - y_A}}{{x_B - x_A}} = \frac{{2 - 0}}{{-2 - 2}} = \frac{1}{2} ]
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Gradient of AC: [ m_{AC} = \frac{{y_C - y_A}}{{x_C - x_A}} = \frac{{2 - 0}}{{-2 - 2}} = \frac{1}{2} ]
So, the gradients of the lines AB and AC are both ( \frac{1}{2} ).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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