Question

Let: ${z_1} = a + ib$, ${z_2} = c + id$. If the points represented by complex numbers \[{z_1},{z_2}\]and ${z_1} - {z_2}$are collinear, then
$A)ad + bc = 0$
$B)ad - bc = 0$
$C)ab + cd = 0$
$D)ab - cd = 0$

Answer 1

Hint: First, the colinear is the points three or more than three that are lying on the straight line.
Below three points, like take two points and determinant we get a line so which is trivially proved if the points are below three lines.
But in the question, they don’t give points, they are given as the elements represent a complex plane like real and imaginary combined numbers.
Formula used: \[\left| {a + ib} \right| = \sqrt {{a^2} + {b^2}} \]

Complete step by step answer:
From the given that the points in the complex numbers, \[{z_1},{z_2}\]and ${z_1} - {z_2}$ are collinear which means taking modulus before the common separation equal to taking common numbers first and then modulus, which is $\left| {{z_1} - {z_2}} \right| = \left| {{z_1}} \right| - \left| {{z_2}} \right|$
Now substitute the values from the given that ${z_1} = a + ib$ and ${z_2} = c + id$ in the collinear form $\left| {{z_1} - {z_2}} \right| = \left| {{z_1}} \right| - \left| {{z_2}} \right|$.
Thus, we get, \[\left| {{z_1} - {z_2}} \right| = \left| {{z_1}} \right| - \left| {{z_2}} \right| \Rightarrow \left| {a + ib - (c + id)} \right| = \left| {a + ib} \right| - \left| {c + id} \right|\]
Now separating the real and imaginary values on the left-hand side we get, \[ \Rightarrow \left| {(a - c) + i(b - d)} \right| = \left| {a + ib} \right| - \left| {c + id} \right|\]
By applying the formula, we get, $\sqrt{{(a - c)^2} + {(b - d)^2}} = \sqrt {{a^2} + {b^2}} + \sqrt {{c^2} + {d^2}} $
Since${(a - b)^2} = {a^2} + {b^2} - 2ab,{(a + b)^2} = {a^2} + {b^2} + 2ab$,
now applying the general formula, we get,
${a^2} + {c^2} + {b^2} + {d^2} - 2(ac + bd) = {a^2} + {c^2} + {b^2} + {d^2} + 2\sqrt {{{(a + b)}^2}{{(c + d)}^2}} $
Thus, by canceling the common terms, we get $ - 2(ac + bd) = 2\sqrt {{{(a + b)}^2}{{(c + d)}^2}} $
Now exchange the root into the square root to the left side we get, ${a^2}{c^2} + {b^2}{d^2} + (2acbd) = {a^2}{c^2} + {a^2}{d^2} + {b^2}{c^2} + {b^2}{d^2}$
Again, canceling the common terms, we get, \[{a^2}{d^2} + {b^2}{c^2} - 2abcd = 0\]
As per the formula${(a - b)^2} = {a^2} + {b^2} - 2ab$, thus we get ${(ad - bc)^2} = 0$.
Equating the root to the right side we get, $ad - bc = 0$

So, the correct answer is “Option B”.

Note: We need to calculate all the terms as per the formula a plus b and a minus by the whole square.
Or we are able to mistake the calculation part and may get the other options but which is wrong.
Also, in the collinear forms, we are able to use the slope formula for the real points.
But in this question the points are real and imaginary; the slope formula is not used.