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Test Code 52; Section 1; Question 10

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by djkvakin » Tue Dec 01, 2009 7:44 am
the coordinates are (6,28).
AC=BD=28, hence Y=28
AD=DC=14 (28/2)
X coordinate for D is 20-14 (x coordinate for C - half distance to A) = 6
So we get X(6,28)
B
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by heshamelaziry » Tue Dec 01, 2009 10:30 am
djkvakin wrote:the coordinates are (6,28).
AC=BD=28, hence Y=28
AD=DC=14 (28/2)
X coordinate for D is 20-14 (x coordinate for C - half distance to A) = 6
So we get X(6,28)
B

I understand how the distance between C and A is 28, but when I use the distance formula i get sqrt28 ? could you explain ?
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by djkvakin » Tue Dec 01, 2009 10:48 am
heshamelaziry wrote:
djkvakin wrote:the coordinates are (6,28).
AC=BD=28, hence Y=28
AD=DC=14 (28/2)
X coordinate for D is 20-14 (x coordinate for C - half distance to A) = 6
So we get X(6,28)
B

I understand how the distance between C and A is 28, but when I use the distance formula i get sqrt28 ? could you explain ?
AC=28 - the distance between -8 and +20.
AC=BD. BD is the 'hight' of the point B (and therefore the 'y' coordinate for the point B)
The triangle ABC is a isosceles triangle with 2 equal sides AB=BC. and therefore BD splits AC in half, therefore AD=DC=14. To get a 'x' coordinate of B we need to get 'x' coordinate of D. It could be calculated in 2 ways:
1. 'x' coordinate of A+14
2. 'x' coordinate of C-14

1. -8+14=6
2. 20-14=6

So we get 'x' coordinate of 6 and 'y' coordinate of 28.
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