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a)integral co-ordinates b)co-ordinates which are rational c)at least one co-ordinate irrational d)co-ordinates which are irrational
If a is an odd integer, b will not be an integer. |dw:1356846803078:dw|
Height of triangle c, is also the y-coordinate of the upper vertex. It's irrational.
how did u get a/2?
The triangle is equilateral, and all sides measure "a" long. The vertical height is the perpendicular bisector of the side that lies on the x-axis. Each half, therefore, measure a/2 long.
can u possibly explain this?i got confused.. http://gyazo.com/077e2faa197346bf8450280343875ee2
Yes, I'll explain the whole thing. The problem states "two vertices of an equilateral triangle have integral coordinates" That means those coordinates are integers.
To simplify the calculations, I set one vertex at (0, 0), and the orther one at (a, 0), with the condition that a is an integer.
Because the triangle is equilateral, the third vertex lies in the perpendicular bisector of the first side which has coordinates (0, 0) and (a, 0).
The x-coordinate of the third vertex is the same as the x-coordinate of the midpoint of the first side. As such it's (x1 + x2)/2, but in our case, x1 is 0. So if x2, which is a, is odd, then the x-coordinate of the third vertex will not be an integer.
So at least one coordinate may be not an integer. That rules out the first choice.
is it C?
That is b, the x-coordinate of the third vertex. b = a/2, and if a is odd, a/2 is not an integer.
No, not yet. I was referring to small letter b in my drawing. The answer is C, you're right.