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The triangle formed by the points 1, (1 + i)/√2 and i as vertices in the Argand diagram is
- a)scalene
- b)equilateral
- c)isosceles
- d)right angled
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The triangle formed by the points 1, (1 + i)/√2 and i as vertice...
2, and (-i)/2 in the complex plane is an equilateral triangle.
To see why, first note that the distance between any two of these points is the same:
- The distance between 1 and (1+i)/2 is |(1+i)/2 - 1| = |i/2| = 1/2.
- The distance between (1+i)/2 and (-i)/2 is |(-i)/2 - (1+i)/2| = |(-1-i)/2| = 1/2.
- The distance between (-i)/2 and 1 is |1 - (-i)/2| = |(2+i)/2| = 1/2.
Therefore, the triangle has three sides of length 1/2, which is enough to prove that it is equilateral.
Alternatively, we can use complex numbers to compute the distances and angles between the points. Let a = 1, b = (1+i)/2, and c = (-i)/2. Then the complex numbers corresponding to these points are:
- a = 1 + 0i
- b = 1/2 + i/2
- c = 0 - i/2
The distance between a and b is given by |b-a| = |(1/2 + i/2) - 1| = |(-1/2 + i/2)| = 1/2. Similarly, the distances between b and c, and between c and a are also 1/2.
The angle between the vectors ab and ac can be computed using the dot product:
ab · ac = Re[(b-a)(c-a)*] = Re[(1/2 + i/2)(-i/2 - 1)*] = Re[-1/4 + i/4] = -1/4
where * denotes complex conjugation. Therefore,
cos(∠BAC) = (ab · ac) / (|ab| |ac|) = (-1/4) / (1/2 * 1/2) = -1/2
which means that ∠BAC = 120 degrees (or 2π/3 radians). Similarly, we can compute the angles at B and C to be 120 degrees as well. Therefore, the triangle is equilateral.
To see why, first note that the distance between any two of these points is the same:
- The distance between 1 and (1+i)/2 is |(1+i)/2 - 1| = |i/2| = 1/2.
- The distance between (1+i)/2 and (-i)/2 is |(-i)/2 - (1+i)/2| = |(-1-i)/2| = 1/2.
- The distance between (-i)/2 and 1 is |1 - (-i)/2| = |(2+i)/2| = 1/2.
Therefore, the triangle has three sides of length 1/2, which is enough to prove that it is equilateral.
Alternatively, we can use complex numbers to compute the distances and angles between the points. Let a = 1, b = (1+i)/2, and c = (-i)/2. Then the complex numbers corresponding to these points are:
- a = 1 + 0i
- b = 1/2 + i/2
- c = 0 - i/2
The distance between a and b is given by |b-a| = |(1/2 + i/2) - 1| = |(-1/2 + i/2)| = 1/2. Similarly, the distances between b and c, and between c and a are also 1/2.
The angle between the vectors ab and ac can be computed using the dot product:
ab · ac = Re[(b-a)(c-a)*] = Re[(1/2 + i/2)(-i/2 - 1)*] = Re[-1/4 + i/4] = -1/4
where * denotes complex conjugation. Therefore,
cos(∠BAC) = (ab · ac) / (|ab| |ac|) = (-1/4) / (1/2 * 1/2) = -1/2
which means that ∠BAC = 120 degrees (or 2π/3 radians). Similarly, we can compute the angles at B and C to be 120 degrees as well. Therefore, the triangle is equilateral.
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