coordinate geometry

a)A triangle has vertices at A(1,-3),B(3,3) and C(-3,1)

i) Find coordinates of the midpoint (labeled P) of AB and the midpoint (labeled Q) of BC

\(\vec{P}=\frac{ \vec{A}+\vec{B}}{2} =\frac{ \begin{pmatrix}1\\-3\end{pmatrix} +\begin{pmatrix}3\\3\end{pmatrix} } {2} =\frac{ \begin{pmatrix}4\\0\end{pmatrix} }{2} =\begin{pmatrix}2\\0\end{pmatrix}\\\\ \vec{Q}=\frac{ \vec{B}+\vec{C}}{2} =\frac{ \begin{pmatrix}3\\3\end{pmatrix} +\begin{pmatrix}-3\\13\end{pmatrix} } {2} =\frac{ \begin{pmatrix}0\\4\end{pmatrix} }{2} =\begin{pmatrix}0\\2\end{pmatrix}\)

ii) show that PQ=½AC

\(\small{ \overline{PQ}=|\vec{P}-\vec{Q}|\\ =\sqrt{ ( \vec{P}-\vec{Q} )\cdot ( \vec{P}-\vec{Q} ) }\\ =\sqrt{ \left[ \begin{pmatrix}2\\0\end{pmatrix} -\begin{pmatrix}0\\2\end{pmatrix} \right] \cdot \left[\begin{pmatrix}2\\0\end{pmatrix} -\begin{pmatrix}0\\2\end{pmatrix} \right] }\\ =\sqrt{ \begin{pmatrix}2\\-2\end{pmatrix} \cdot \begin{pmatrix}2\\-2\end{pmatrix} }\\ =\sqrt{ 4+4 }\\ \overline{PQ}=\sqrt{ 8 }\\ \overline{AC}=|\vec{A}-\vec{C}|\\ =\sqrt{ (\vec{A}-\vec{C})\cdot (\vec{A}-\vec{C}) }\\ =\sqrt{ \left[ \begin{pmatrix}1\\-3\end{pmatrix} -\begin{pmatrix}-3\\1\end{pmatrix} \right] \cdot \left[\begin{pmatrix}1\\-3\end{pmatrix} -\begin{pmatrix}-3\\1\end{pmatrix} \right] }\\ =\sqrt{ \begin{pmatrix}4\\-4\end{pmatrix} \cdot \begin{pmatrix}1\\-4\end{pmatrix} }\\ =\sqrt{ 4\cdot8 }\\ \overline{AC}=2\cdot \sqrt{ 8 }\\ } \)

\(\dfrac{ \overline{PQ} }{\overline{AC} } = \dfrac{ \sqrt{ 8 } }{ 2\cdot \sqrt{ 8 } } = \dfrac{ 1 }{ 2 }\\ \overline{PQ} = \dfrac{ 1 }{ 2 }\cdot \overline{AC}\)

iii) show that PQ||AC

\((\vec{P}-\vec{Q} )\cdot ( \vec{A}-\vec{C} ) = \overline{PQ}\cdot \overline{AC} \cdot \cos{(\alpha)}\\ \cos{(\alpha)} = \frac{ (\vec{P}-\vec{Q} )\cdot ( \vec{A}-\vec{C} ) } {\overline{PQ}\cdot \overline{AC} }\\ \cos{(\alpha)}= \frac{ \left[ \begin{pmatrix}2\\0\end{pmatrix}-\begin{pmatrix}0\\2\end{pmatrix} \right]\cdot \left[ \begin{pmatrix}1\\-3\end{pmatrix}-\begin{pmatrix}-3\\1\end{pmatrix} \right] } { \sqrt{ 8 }\cdot 2\cdot \sqrt{ 8 } }\\ \cos{(\alpha)}= \frac{ \begin{pmatrix}2\\-2\end{pmatrix}\cdot \begin{pmatrix}4\\-4\end{pmatrix} } {16}\\ \cos{(\alpha)}= \frac{ 16 }{16}\\ \cos{(\alpha)}= 1 \qquad \Rightarrow \qquad \alpha = 0\qquad \Rightarrow \qquad PQ||AC \)

b)THEOREM:the interval joining the midpoints of two sides of a triangle is parallel to the base and half its length

Prove this theorem for any triangle by placing its vertices at A(2a,0),B(2b,2c) and C(0,0) , where a>0, and proceeding as in part (a)

\(\vec{P}=\frac{ \vec{A}+\vec{B}}{2} =\frac{ \begin{pmatrix}2a\\0\end{pmatrix} +\begin{pmatrix}2b\\2c\end{pmatrix} } {2} =\frac{ \begin{pmatrix}2a+2b\\2c\end{pmatrix} }{2} =\begin{pmatrix}a+b\\c\end{pmatrix} \\\\ \vec{Q}=\frac{ \vec{B}+\vec{C}}{2} =\frac{ \begin{pmatrix}2b\\2c\end{pmatrix} +\begin{pmatrix}0\\0\end{pmatrix} } {2} =\frac{ \begin{pmatrix}2b\\2c\end{pmatrix} }{2} =\begin{pmatrix}b\\c\end{pmatrix}\)

\(\small{ \overline{PQ}=|\vec{P}-\vec{Q}|\\ =\sqrt{ ( \vec{P}-\vec{Q} )\cdot ( \vec{P}-\vec{Q} ) }\\ =\sqrt{ \left[ \begin{pmatrix}a+b\\c\end{pmatrix} -\begin{pmatrix}b\\c\end{pmatrix} \right] \cdot \left[\begin{pmatrix}a+b\\c\end{pmatrix} -\begin{pmatrix}b\\c\end{pmatrix} \right] }\\ =\sqrt{ \begin{pmatrix}a\\0\end{pmatrix} \cdot \begin{pmatrix}a\\0\end{pmatrix} }\\ =\sqrt{ a^2 }\\ \overline{PQ}=a\\ \overline{AC}=|\vec{A}-\vec{C}|\\ = |\begin{pmatrix}2a\\0\end{pmatrix}-\begin{pmatrix}0\\0\end{pmatrix}|\\ = |\begin{pmatrix}2a\\0\end{pmatrix}|\\ \overline{AC}=2a }\)

\((\vec{P}-\vec{Q} )\cdot ( \vec{A}-\vec{C} ) = \overline{PQ}\cdot \overline{AC} \cdot \cos{(\alpha)}\\ \cos{(\alpha)} = \frac{ (\vec{P}-\vec{Q} )\cdot ( \vec{A}-\vec{C} ) } {\overline{PQ}\cdot \overline{AC} }\\ \cos{(\alpha)}= \frac{ \left[ \begin{pmatrix}a+b\\c\end{pmatrix}-\begin{pmatrix}b\\c\end{pmatrix} \right]\cdot \left[ \begin{pmatrix}2a\\0\end{pmatrix}-\begin{pmatrix}0\\0\end{pmatrix} \right] } { a\cdot 2a }\\ \cos{(\alpha)}= \frac{ \begin{pmatrix}a\\0\end{pmatrix}\cdot \begin{pmatrix}2a\\0\end{pmatrix} } {2a^2}\\ \cos{(\alpha)}= \frac{ 2a^2 }{2a^2}\\ \cos{(\alpha)}= 1 \qquad \Rightarrow \qquad \alpha = 0\qquad \Rightarrow \qquad PQ||AC\)