If (x + 2) is a factor of x3 − 6x2 + kx + 10, k =
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The coordinates of point F are (15/4 , -5/2)
- The coordinates of point G = (-3/4 , -3)
From the image attached:
AB divided into 4 similar parts, hence
- E is the mid-point of AB
- D is the mid-point of AE
- F is the mid-point of EB
Note that the rule of the mid-point states that:
When M (x , y) is the mid-point of the segment of AB, where A (x1 , y1) and B (x2 , y2), Then x = (x1 + x2)/2 and that of y = (y1 + y2)/2
Then lets solve for points E, F, D
Since A (-3 , -1) and B (6 , -3),
E is the mid-point of AB
Then E =[(-3 + 6)/2, (-1 + -3)/2]
= (3/2 , -2)
Since F is the mid-point of EB,
E (3/2 , -2) , B (6 , -3)
Then F = [(3/2 + 6)/2 , (-2 + -3)/2]
= (15/4 , -5/2)
Since D is the mid-point of AE, A (-3 , -1), E (3/2 , -2)
Then D = [(-3 + 3/2)/2 , (-1 + -2)/2]
= (-3/4 , -3/2)
Note also that:
BC is said to be an horizontal segment due to the fact that B and C have similar y - coordinate and G can be found on BC.
Since the y-coordinate of G is similar to y-coordinate of B and C
Then y-coordinate of G is = -3
Since DG ⊥ BC
Then DG = vertical segment
So G has similar x-coordinate of D
Then The x-coordinate of G = -3/4
The G = (-3/4 , -3)
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