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Prove that the three distinct points $z_1,z_2$, and $z_3$ lie on the same straight line iff $z_3 - z_2 = c(z_2 - z_1)$ for some real number $c$ and $z$ is complex.

I know that two vectors are parallel iff one is a scalar multiple of the other, thus $z$ is parallel to $w$ iff $z = cw$. So, from that, does that mean $z_3 - z_2 = c(z_2 - z_1)$ are parallel thus making it lie on the same line?

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Perhaps it will be easier to reason with: -z_2 + z_3 = c(z_2 - z_1). – Patrick McLaren Jan 19 '13 at 4:33
@PatrickMcLaren Nope it doesn't. Can you explain further why it will be easier to reason with that? – Q.matin Jan 19 '13 at 4:39
up vote 3 down vote accepted

If in a complex plane $z_1$ is represented by point $A, z_2$ by $B$ and $z_3$ by $C$, then $z_3-z_2$ represents the vector $\vec {BC}$, and $z_2-z_1$ represents vector $\vec {AB}$. Thus, these points $A,B,C$ are collinear iff $\vec{AB}$ and $\vec{BC}$ are parallel as they already have one point $B$ in common $\implies \vec {BC}=c\vec{AB}\implies z_3-z_2=c(z_2-z_1)$.

Here, $c$ is taken real because when we multiply by a complex no. , it's real part gives the scaling and imaginary part rotates it. Here, we need the vectors to be parallel ,so we omit the rotation and $c$ has to be real only.

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Great explaination! I understand now, it helped out a lot. Thank you a lot, Avatar! – Q.matin Jan 19 '13 at 4:51
i am glad that the answer served its purpose. – Aang Jan 19 '13 at 5:12

If $z_k=x_k+iy_k$ for $k=1,2,3$

As $z_3-z_2=c(z_2-z_1),$

If $c=0, z_3=z_2$ and if $z=\infty, z_2=z_1$ so $c$ non-zero finite number.

$\implies x_3-x_2+i(y_3-y_2)=c\{x_2-x_1+i(y_2-y_1)\}$

Equating the real & the imaginary parts, $$x_3-x_2=c(x_2-x_1),y_3-y_2=c(y_2-y_1)$$

So, $$\frac{y_3-y_2}{x_3-x_2}=\frac{y_2-y_1}{x_2-x_1}\implies x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)=0$$

Hence $z_1,z_2$ and $z_3$ are collinear.


the area of the triangle with vertices $z_k=x_k+iy_k$ for $k=1,2,3$

is $$\frac12\det\begin{pmatrix}x_1&y_1 &1\\x_2 & y_2 & 1 \\ x_3 & y_3&1\end{pmatrix}$$

$$=\frac12\det\begin{pmatrix}x_1-x_2&y_1-y_2 &0\\x_2 & y_2 & 1 \\ x_3-x_2 & y_3-y_2&0\end{pmatrix}$$ applying $R_1'=R_1-R_2$ and $R_3'=R_3-R_2$

$$=\frac12\det\begin{pmatrix}x_1-x_2&y_1-y_2 &0\\x_2 & y_2 & 1 \\ -c(x_1-x_2) & -c(y_1-y_2)&0\end{pmatrix}$$

$$=-\frac c2\det\begin{pmatrix}x_1-x_2&y_1-y_2 &0\\x_2 & y_2 & 1 \\ x_1-x_2&y_1-y_2&0\end{pmatrix}=0$$ as the 1st & the 3rd rows are identical.

Hence $z_1,z_2$ and $z_3$ are collinear.

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Thanks a lot for this, Lab! This is very useful! – Q.matin Jan 19 '13 at 23:57

WLOG let $z_1$ be at the origin, and $z_2 -z_1 = r_1e^{i\theta_1},$ an arbitrary vector from the origin. $z_2$ is the endpoint of this vector, and we have a third point $z_3$ and want to know if the angle of $z_3-z_2 = r_2e^{i\theta_2}$ is that of $z_2-z_1.$

But $$\frac{z_3 -z_2}{z_2-z_1}=\frac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}}= \frac{r_1}{r_2}e^{i(\theta_1-\theta_2)} = c. $$ Since c is real, $\theta_1 -\theta_2$ must equal zero, so they are equal.

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Thanks, Daniel. This is very useful ! – Q.matin Jan 19 '13 at 4:58

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