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We have the points $A'(a_1'\mid a_2')$, $B'(b_1'\mid b_2')$ and $C'(c_1'\mid c_2')$. These points are collinear iff \begin{equation*}\begin{vmatrix} a_1' & a_2' & 1 \\ b_1' & b_2' & 1 \\ c_1' & c_2' & 1 \end{vmatrix}=0\end{equation*}

It is given that $\vec{C'}=\vec{A}+\lambda(\vec{B}-\vec{A})=(1-\lambda )\vec{A}+\lambda \vec{B}$, $\vec{A'}=\vec{B}+\mu(\vec{C}-\vec{B})=(1-\mu )\vec{B}+\mu \vec{C}$ and $\vec{B'}=\vec{C}+\nu(\vec{A}-\vec{C})=(1-\nu )\vec{C}+\nu \vec{A}$.

Let $A(a_1 \mid a_2)$, $B(b_1 \mid b_2)$ uand $C(c_1 \mid c_2)$.

We get the following \begin{align*}&\begin{vmatrix} a_1' & a_2' & 1 \\ b_1' & b_2' & 1 \\ c_1' & c_2' & 1 \end{vmatrix}=0 \iff \begin{vmatrix} (1-\mu )b_1+\mu c_1 & (1-\mu )b_2+\mu c_2 & 1 \\ (1-\nu )c_1+\nu a_1 & (1-\nu )c_2+\nu a_2 & 1 \\ (1-\lambda )a_1+\lambda b_1 & (1-\lambda )a_2+\lambda b_2 & 1 \end{vmatrix}=0 \\ & \iff \begin{vmatrix} (1-\mu )b_1 & (1-\mu )b_2+\mu c_2 & 1 \\ (1-\nu )c_1 & (1-\nu )c_2+\nu a_2 & 1 \\ (1-\lambda )a_1 & (1-\lambda )a_2+\lambda b_2 & 1 \end{vmatrix}+\begin{vmatrix} \mu c_1 & (1-\mu )b_2+\mu c_2 & 1 \\ \nu a_1 & (1-\nu )c_2+\nu a_2 & 1 \\ \lambda b_1 & (1-\lambda )a_2+\lambda b_2 & 1 \end{vmatrix}=0 \\ & \iff \begin{vmatrix} (1-\mu )b_1 & (1-\mu )b_2 & 1 \\ (1-\nu )c_1 & (1-\nu )c_2 & 1 \\ (1-\lambda )a_1 & (1-\lambda )a_2 & 1 \end{vmatrix}+\begin{vmatrix} (1-\mu )b_1 & \mu c_2 & 1 \\ (1-\nu )c_1 & \nu a_2 & 1 \\ (1-\lambda )a_1 & \lambda b_2 & 1 \end{vmatrix}+\begin{vmatrix} \mu c_1 & (1-\mu )b_2 & 1 \\ \nu a_1 & (1-\nu )c_2 & 1 \\ \lambda b_1 & (1-\lambda )a_2 & 1 \end{vmatrix}+\begin{vmatrix} \mu c_1 & \mu c_2 & 1 \\ \nu a_1 & \nu a_2 & 1 \\ \lambda b_1 & \lambda b_2 & 1 \end{vmatrix}=0 \end{align*}

Is it correct so far? Can we simplify it further using properties of the determinant or do we have to calculate the determinants?

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You can simplify this problem from the beginning. Note that $$ \begin{vmatrix} a_1 & a_2 & 1\\ b_1 & b_2 & 1\\ c_1 & c_2 & 1 \end{vmatrix}= \det (B,C)+\det(C,A)+\det(A,B) $$ where $A=(a_1, a_2)^T$ and $B,C$ are defined similarly. Now given you definitions of $A', B', C'$ $$ \begin{aligned} \det (A',B') &=\det [(1-\mu) B+\mu C,(1-\nu)C+\nu A]\\ &=(1-\mu)(1-\nu)\det(B,C)-(1-\mu)\nu\det (A,B)+\mu\nu\det (C,A)\\\\ \det (B',C') &=\det [(1-\nu) C+\nu A,(1-\lambda)A+\lambda B]\\ &=(1-\lambda)(1-\nu)\det(C,A)-(1-\nu)\lambda\det (B,C)+\lambda\nu\det (A,B) \\\\ \det (C',A') &=\det [(1-\lambda) A+\lambda B,(1-\mu)B+\mu C]\\ &=(1-\lambda)(1-\mu)\det(A,B)-(1-\lambda)\mu\det (C,A)+\lambda\mu\det (B,C) \end{aligned} $$ Therefore, after a bit of calculation $$ \begin{aligned} &\det(A',B')+\det(B',C')+\det(C',A')=\\ &\big[ \lambda \mu \nu +(1-\lambda)(1-\mu)(1-\nu)\big]\big[\det(A,B)+\det(B,C)+\det (C,A)\big]\\ \end{aligned} $$ In your original form of the question $$ \begin{vmatrix} a'_1 & a'_2 & 1\\ b'_1 & b'_2 & 1\\ c'_1 & c'_2 & 1 \end{vmatrix}= \big[ \lambda \mu \nu +(1-\lambda)(1-\mu)(1-\nu)\big] \begin{vmatrix} a_1 & a_2 & 1\\ b_1 & b_2 & 1\\ c_1 & c_2 & 1 \end{vmatrix} $$ So $A,B,C$ are colinear iff $A',B', C'$ are colinear (I'm assuming $\lambda, \mu, \nu\in (0,1)$). However, using determinants to solve this geometric problem, in my humble opinion, is a bit odd and unneccasrily complicated.

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  • $\begingroup$ Why does the first relation hold? How did you calculated the the right part from the left one? $$$$ how else could we solve the problem besides using determinants? $\endgroup$ – Mary Star Jul 13 '17 at 6:46
  • $\begingroup$ Regarding your first question I'm just using $$\begin{vmatrix}a & b & c\\ d & e & f\\ h & i &j\end{vmatrix}= c\begin{vmatrix} d & e\\ h & i\end{vmatrix}- f\begin{vmatrix} a & b\\ h & i\end{vmatrix}+j\begin{vmatrix} a & b\\ d & e\end{vmatrix}$$ then absorbing the minus sign (behind f) to swicth the rows. Turning it into $$\begin{vmatrix}a & b & c\\ d & e & f\\ h & i &j\end{vmatrix}= c\begin{vmatrix} d & e\\ h & i\end{vmatrix}+f\begin{vmatrix} h & i\\a & b\end{vmatrix}+j\begin{vmatrix} a & b\\ d & e\end{vmatrix}$$ $\endgroup$ – Hamed Jul 13 '17 at 9:12
  • $\begingroup$ For you second question, if $A,B,C$ are collinear, the line being $L$, then a straightforward check (even a drawing) shows that $A',B', C'$ (for any choice of $\lambda, \mu, \nu$ lie on $L$ too. The converse is also similar. $\endgroup$ – Hamed Jul 13 '17 at 9:16
  • $\begingroup$ Oh I'm sorry in derivation of the first equation I'm also using $\det M=\det M^T$ for a matrix $M$. $\endgroup$ – Hamed Jul 13 '17 at 9:17
  • $\begingroup$ I understand!! Thank you so much!! There is also a second question: If $A',B',C'\neq A,B,C$ and either zero or two of the points $A',B',C'$ are at one side of triangle, then $A',B',C'$ are collinear iff $\frac{|AC'|}{|C'B|}=\frac{|BA'|}{|A'C|}=\frac{|CB'|}{|B'A|}=1$. Could you give me a hint how we could show that? $\endgroup$ – Mary Star Jul 13 '17 at 16:37

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