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An equilateral triangle is one in which all three sides are of equal length. If two vertices of an equilateral triangle are (0, 4) and (0, 0), find the third vertex. How many of these triangles are possible?
Seeking the needed steps here.
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If we take (0,4) and (0,0) as the two points defining the base then the midpoint is (0,2) and the horizontal line y=2 is the line of symmetry that goes through the vertex. So it must have its y coordinate 2.
If you use Pythag. on the half triangle made by splitting the equilateral triangle in two along that line (y=2) you can determine the required x coordinate(s). There will be two answers as the triangle may 'sit' left of the y axis or right of it.
Bob
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This interactive tool of MIF's may help visualise what's going on: Interactive Cartesian Coordinates
It has several options you can choose from (eg, regular or irregular shapes, choice of the number of sides, entry of custom coordinates).
However, it will only display one shape at a time, and, unlike Geogebra, it lacks mathematical precision...but it's accurate enough to give a pretty good idea.
Last edited by phrontister (2023-10-14 10:24:18)
"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson
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If we take (0,4) and (0,0) as the two points defining the base then the midpoint is (0,2) and the horizontal line y=2 is the line of symmetry that goes through the vertex. So it must have its y coordinate 2.
If you use Pythag. on the half triangle made by splitting the equilateral triangle in two along that line (y=2) you can determine the required x coordinate(s). There will be two answers as the triangle may 'sit' left of the y axis or right of it.
Bob
Thank you. I will work this out on paper.
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This interactive tool of MIF's may help visualise what's going on: Interactive Cartesian Coordinates
It has several options you can choose from (eg, regular or irregular shapes, choice of the number of sides, entry of custom coordinates).
However, it will only display one shape at a time, and, unlike Geogebra, it lacks mathematical precision...but it's accurate enough to give a pretty good idea.
Thank you for the link. I truly appreciate your reply.
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