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Coordinates of Point B??

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by adi_800 » Wed Apr 28, 2010 10:16 am
What is the easiest way to solve this problem??? I used distance formula twice and then got the answer..

10. What are the coordinates of point B in the xy-plane
above?
(A) (6, 12)
(B) (6, 28)
(C) (8, 20)
(D) (12, 20)
(E) (14, 28)
[/img]

OA is B
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by Stuart@KaplanGMAT » Wed Apr 28, 2010 11:06 am
adi_800 wrote:What is the easiest way to solve this problem??? I used distance formula twice and then got the answer..

10. What are the coordinates of point B in the xy-plane
above?
(A) (6, 12)
(B) (6, 28)
(C) (8, 20)
(D) (12, 20)
(E) (14, 28)
Hi,

I'm not sure what you mean by the distance formula, but because of the information shown we know that:

1) since we have an isoscles triangle, the x-coord of B is halfway between A and C. Since A to C is 28 units, B is 14 units from either end (so -8 + 14 = 6 or +20 - 14 = 6).

2) since AC = BD, we know that BD is 28 units. The base of the triangle is at y-coordinate 0, so the apex of the triangle (i.e. point B) is at y-coord 0 + 28 = 28.

Therefore, B is at (6, 28).
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by adi_800 » Wed Apr 28, 2010 11:13 am
How did you deduce the x-coord of B is halfway between A and C... I did not get how u used isosceles triangle for that...
N distance formula states...
distance between two points A(x1,y1) and B(x2,y2) is given by Rt((y2-y1)^2 + (x2-x1)^2)
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by Stuart@KaplanGMAT » Wed Apr 28, 2010 1:10 pm
adi_800 wrote:How did you deduce the x-coord of B is halfway between A and C... I did not get how u used isosceles triangle for that...
N distance formula states...
distance between two points A(x1,y1) and B(x2,y2) is given by Rt((y2-y1)^2 + (x2-x1)^2)
From the diagram, we see that BD is perpendicular to the base of the triangle.

For isosceles triangles, if you drop a line from the vertex between the two equal sides that's perpendicular to the base (i.e. draw the height of the triangle), that line will bisect the base of the triangle.
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