Question

: If \[{z_1} = 1 + 2i\], \[{z_2} = 2 + 3i\], \[{z_3} = 3 + 4i\] then \[{z_1}\], \[{z_2}\], \[{z_3}\] represent the vertices of a/an [Orissa JEE \[2004\]]
A) Equilateral triangle
B) Isosceles triangle
C) Right angled triangle
D) None of these

Answer 1

Hint: in this question we have to find the given point represent whether collinear point or vertices of triangle. In order to find this have to use the formula of area of triangle. If area becomes zero then points are collinear otherwise it represents triangle vertices.



Formula Used:Area of triangle is given by
\[A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&1\\{{a_2}}&{{b_2}}&1\\{{a_3}}&{{b_3}}&1\end{array}} \right|\]
Where
A is area of triangle
\[{a_i}\] is a real part of vertices
\[{b_i}\] is a imaginary part of vertices



Complete step by step solution:Given: \[{z_1} = 1 + 2i\], \[{z_2} = 2 + 3i\]and\[{z_3} = 3 + 4i\] these are collinear
Now area of triangle is given by
\[A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&1\\{{a_2}}&{{b_2}}&1\\{{a_3}}&{{b_3}}&1\end{array}} \right|\]
Where
A is area of triangle
\[{a_i}\] is a real part of vertices
\[{b_i}\] is a imaginary part of vertices
Expand determinant to get area of triangle
\[A = \dfrac{1}{2}({a_1}({b_2} - {b_3}) - {a_2}({b_1} - {b_3}) + {a_3}({b_1} - {b_2}))\]
\[A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&1\\{{a_2}}&{{b_2}}&1\\{{a_3}}&{{b_3}}&1\end{array}} \right|\]
\[A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}1&2&1\\2&3&1\\3&4&1\end{array}} \right|\]
Expand the determinant
\[A = \dfrac{1}{2}(1(3 - 4) - 2(2 - 4) + 3(2 - 3))\]
\[A = \dfrac{1}{2}( - 1 + 4 - 3)\]
\[A = 0\]
We found area of triangle equal to zero so the point represents the collinear points.



Option ‘D’ is correct



Note: We have to remember that collinear point is a point which lies in same line. If all points lie in same line then it never form area so area of triangle is zero for collinear points. Complex number is a number which is a combination of real and imaginary number. So in combination number question we have to represent number as a combination of real and its imaginary part. Imaginary part is known as iota. Square of iota is equal to negative one.