Look at the picture.
![\vec{BE}=[-2-(-6);\ 4-10]=[4;\ -6]](https://tex.z-dn.net/?f=%5Cvec%7BBE%7D%3D%5B-2-%28-6%29%3B%5C%204-10%5D%3D%5B4%3B%5C%20-6%5D)
![D(x;\ y)\\\\\vec{ED}=[x-(-2);\ y-4]=[x+2;\ y-4]\\\\\vec{ED}=\vec{BE}\ therefore\\\\ \ [x+2;\ y-4]=[4;\ -6]\to x+2=4\ and\ y-4=-6\\\\x=2\ and\ y=-2\\\\Answer:\ \boxed{A.\ (2;\ -2)}](https://tex.z-dn.net/?f=D%28x%3B%5C%20y%29%5C%5C%5C%5C%5Cvec%7BED%7D%3D%5Bx-%28-2%29%3B%5C%20y-4%5D%3D%5Bx%2B2%3B%5C%20y-4%5D%5C%5C%5C%5C%5Cvec%7BED%7D%3D%5Cvec%7BBE%7D%5C%20therefore%5C%5C%5C%5C%20%5C%20%5Bx%2B2%3B%5C%20y-4%5D%3D%5B4%3B%5C%20-6%5D%5Cto%20x%2B2%3D4%5C%20and%5C%20y-4%3D-6%5C%5C%5C%5Cx%3D2%5C%20and%5C%20y%3D-2%5C%5C%5C%5CAnswer%3A%5C%20%5Cboxed%7BA.%5C%20%282%3B%5C%20-2%29%7D)
Please help i am very confused
1 answer:
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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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