The midpoints of the vertices of a triangle are (1,2), (3,4) and (5,0). The sum of the x coordinates of the three vertices is then
(a) 4
(b) 6
(c) 7
(d) 9
(e) 17/2
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Triangle and Midpoint Meet
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drawing the midpoints and using the properties of similar triangles, we can derive the vertices:
(-1,6), (3,-2), (7,2)
Answer is D
Note that by joining the midpoints of the sides of a triangle, you'll be effectively creating four congruent triangles.
(-1,6), (3,-2), (7,2)
Answer is D
Note that by joining the midpoints of the sides of a triangle, you'll be effectively creating four congruent triangles.
Last edited by avenus on Tue May 12, 2009 1:50 am, edited 1 time in total.
Another way to solve the problem,
let the vertices be (x1,y1), (x2,y2), (x3,y3):
let the point (5,0) be the midpoint of (x1,y1) and (x2,y2)
|||y, (3,4) (x2, y2) and (x3,y3)
(1,2) (x1,y1) and (x3,y3)
Then applying the midpoint formula to all the midpoints and adding only the x-cordinates of will give the sum as 9.
Thus, D.
Hope it helps.
let the vertices be (x1,y1), (x2,y2), (x3,y3):
let the point (5,0) be the midpoint of (x1,y1) and (x2,y2)
|||y, (3,4) (x2, y2) and (x3,y3)
(1,2) (x1,y1) and (x3,y3)
Then applying the midpoint formula to all the midpoints and adding only the x-cordinates of will give the sum as 9.
Thus, D.
Hope it helps.
Try again guys. This is a question that could really come on the GMAT. Be careful in analysis.Pranay wrote:Another way to solve the problem,
let the vertices be (x1,y1), (x2,y2), (x3,y3):
let the point (5,0) be the midpoint of (x1,y1) and (x2,y2)
|||y, (3,4) (x2, y2) and (x3,y3)
(1,2) (x1,y1) and (x3,y3)
Then applying the midpoint formula to all the midpoints and adding only the x-cordinates of will give the sum as 9.
Thus, D.
Hope it helps.
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scoobydooby
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let the vertices be (a,b); (c, d) and (e, f)
the mid points are (1,2), (3,4) and (5,0).
a+c=2 (applying midpoint formula)
c+e=6
a+e=10
adding them all up, 2(a+c+e)=18
or a+c+e=9 (a, c and e are the x co-ordinates of the vertices)
hence, D
dtweah, where are we all getting mistaken? what are we missing?
the mid points are (1,2), (3,4) and (5,0).
a+c=2 (applying midpoint formula)
c+e=6
a+e=10
adding them all up, 2(a+c+e)=18
or a+c+e=9 (a, c and e are the x co-ordinates of the vertices)
hence, D
dtweah, where are we all getting mistaken? what are we missing?
scoobydooby wrote:let the vertices be (a,b); (c, d) and (e, f)
the mid points are (1,2), (3,4) and (5,0).
a+c=2 (applying midpoint formula)
c+e=6
a+e=10
adding them all up, 2(a+c+e)=18
or a+c+e=9 (a, c and e are the x co-ordinates of the vertices)
hence, D
quote]
What is the solution to the system:
x+y=2
y+x=6
I bet you wouldn't go about it by adding 2 and 6.
Why not find x and y?
You guys are correct. D is the answer. I was thinking of another problem. My bad. I owe you all one. (laugh)scoobydooby wrote:let the vertices be (a,b); (c, d) and (e, f)
the mid points are (1,2), (3,4) and (5,0).
a+c=2 (applying midpoint formula)
c+e=6
a+e=10
adding them all up, 2(a+c+e)=18
or a+c+e=9 (a, c and e are the x co-ordinates of the vertices)
hence, D
dtweah, where are we all getting mistaken? what are we missing?
