Posted by Denis Borris on November 11, 2002 at 23:17:01:
In Reply to: distances between points on a graph posted by Treva on November 11, 2002 at 21:49:50:
: I've tried several ways to work this but have not been introduced to this sort of question before and I don't know if I am correct in my work.
: Find "X" so that the point (X,0) is equidistant from the points (6,1) and (-2,5).
: We have been working with the distance formula and midpoint formula. I'm assuming it's a right triangle but have no idea.
: I hope someone can help.
Make it easier by changing (6,1) and (-2,5) to (8,1) and (0,5) : get rid of negative;
now graph those; let (8,1)=A, (0,5)=B, and point on x-axis (X,0)=C.
Join 'em to get triangle BAC (you should have B on y-axis; ok?)
let origin=D; drop perpendicular AE to the x-axis; you now have 2 right triangles:
BCD (hypotenuse BC) and AEC (hypotenuse AC) ; you still with me?
DC=X, so CE = 8-X ; and we have BD=5 and AE=1; ok?
So we want the equality BC=AC, right?
X^2 + 5^2 = (8 - X)^2 + 1^2 ; agree?
X^2 + 25 = 64 - 16X + X^2 + 1
16X = 40
X = 2.5
...now remember what we did at start to the "-2"; so adjust accordingly: OK??
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