Question

Q.Choose the correct answer.
Q. The points (1/√3 , 1), (2/√3, 2),(1/√3,3) are the vertices of
a) Isosceles triangle
b) Equilateral
c) Right Triangle
d) None of above

Answer 1

Answer to Question #91483 - Math - Analytic Geometry

**Question**: Choose the correct answer.

The points $(1/\sqrt{3}, 1)$, $(2/\sqrt{3}, 2)$, $(1/\sqrt{3}, 3)$ are the vertices of

a) Isosceles triangle

b) Equilateral

c) Right Triangle

d) None of above

**Solution**:

Let the three points are named as:

$$A: \left(\frac {1}{\sqrt {3}}, 1\right) \quad , \qquad B: \left(\frac {2}{\sqrt {3}}, 2\right) \quad , \qquad C: \left(\frac {1}{\sqrt {3}}, 3\right)$$

If they represent the three vertices of a triangle ABC, then the length of the three sides are calculated as:

$$\left| A B \right| = \sqrt {\left(2 - 1\right) ^ {2} + \left(\frac {2}{\sqrt {3}} - \frac {1}{\sqrt {3}}\right) ^ {2}} = \sqrt {1 ^ {2} + \left(\frac {1}{\sqrt {3}}\right) ^ {2}} = \sqrt {1 + \frac {1}{3}} = \sqrt {\frac {4}{3}} = \frac {2}{\sqrt {3}}$$ $$\left| B C \right| = \sqrt {\left(3 - 2\right) ^ {2} + \left(\frac {1}{\sqrt {3}} - \frac {2}{\sqrt {3}}\right) ^ {2}} = \sqrt {1 ^ {2} + \left(\frac {- 1}{\sqrt {3}}\right) ^ {2}} = \sqrt {1 + \frac {1}{3}} = \sqrt {\frac {4}{3}} = \frac {2}{\sqrt {3}}$$ $$\left| A C \right| = \sqrt {\left(3 - 1\right) ^ {2} + \left(\frac {1}{\sqrt {3}} - \frac {1}{\sqrt {3}}\right) ^ {2}} = \sqrt {2 ^ {2} + (0) ^ {2}} = \sqrt {4} = 2$$

As we can see that, the two sides are equal. That is

$$\left| A B \right| = \left| B C \right| = \frac {2}{\sqrt {3}}$$

Therefore, the given points are vertices of an Isosceles triangle

The triangle ABC has been plotted, it is obvious that the two sides,

$$\left| A B \right| = \left| B C \right|$$

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