# Can we simplify it further using determinant properties?

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?

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.
• 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? – Mary Star Jul 13 '17 at 6:46
• 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 $$\begin{vmatrix}a & b & c\\ d & e & f\\ h & i &j\end{vmatrix}= c\begin{vmatrix} d & e\\ h & i\end{vmatrix}+f\begin{vmatrix} h & i\\a & b\end{vmatrix}+j\begin{vmatrix} a & b\\ d & e\end{vmatrix}$$ – Hamed Jul 13 '17 at 9:12
• 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. – Hamed Jul 13 '17 at 9:16
• Oh I'm sorry in derivation of the first equation I'm also using $\det M=\det M^T$ for a matrix $M$. – Hamed Jul 13 '17 at 9:17
• 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? – Mary Star Jul 13 '17 at 16:37