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Isosceles triangle: What could be coord's of its vertices?
- Thread starter JayJay06
- Start date
Hello, Jay Jay!
Your question is not supplied with adequate information.
You can place the triangle any where in the co ordinate plane, if nothing specific is given with the question. You can place any point in the coordinate plane as one of the vertices (call it X). Then you draw a straight line. From the line, you cut the length of the side (the length of the sides are equal, this being an issosceles triangle). And thus we'll get another vertex (call it Y). Now take the length of the side as the radius and X and Y respectively as the centres, draw two arcs. Let the point of intersection be Z, another vertex.
This is a common procedure of drawing a triangle . But in case of problem of coordinate we must be given some clue regarding the vertex.
Thank you .
Your question is not supplied with adequate information.
You can place the triangle any where in the co ordinate plane, if nothing specific is given with the question. You can place any point in the coordinate plane as one of the vertices (call it X). Then you draw a straight line. From the line, you cut the length of the side (the length of the sides are equal, this being an issosceles triangle). And thus we'll get another vertex (call it Y). Now take the length of the side as the radius and X and Y respectively as the centres, draw two arcs. Let the point of intersection be Z, another vertex.
This is a common procedure of drawing a triangle . But in case of problem of coordinate we must be given some clue regarding the vertex.
Thank you .
As has been previously noted, you could place the isosceles triangle anywhere on the coordinate plane. But, you might want to consider placing it in a position where it would be easy to determine the coordinates, and where the coordinates of the vertices involve the smallest possible number of different variables.
Here's one possibility. Put the vertex of the isosceles triangle on the y-axis, and the base of the isosceles triangle along the x-axis. Then, possible coordinates might be (0, a) for the vertex, and (b, 0) and (-b, 0) for the endpoints of the base. You can easily verify that the distances from (0, a) to (b, 0) and from (0, a) to (-b, 0) are equal.
Here's one possibility. Put the vertex of the isosceles triangle on the y-axis, and the base of the isosceles triangle along the x-axis. Then, possible coordinates might be (0, a) for the vertex, and (b, 0) and (-b, 0) for the endpoints of the base. You can easily verify that the distances from (0, a) to (b, 0) and from (0, a) to (-b, 0) are equal.