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Topic
In the coordinate plane Oxygenfor three points A(1; 5), REMOVE(-1; -1), OLD(2; – 5).
a) Prove three points A, REMOVE, OLD not in line
b) Find the coordinates of the center of gravity WOOD of the triangle ABC
c) Find the coordinates of the point EASY such that quadrilateral ABCD is a trapezoid with AB // CD and CD = \(\frac{3}{2}\)AB
Solution method – See details
Step 1: Prove that 2 vectors \(\overrightarrow {AB} ,\overrightarrow {AC} \) are not in the same direction to prove A, REMOVE, OLD not in line
Step 2: Apply the result WOOD(a; b) is the centroid ofABC with \(A({x_A};{y_A}),B({x_B};{y_B}),C({x_C};{y_C})\) then \(\left\{ \begin{array}{ l}a = \frac{{{x_A} + {x_B} + {x_C}}}{3}\\b = \frac{{{y_A} + {y_B} + {y_C}}}{3}\end {array} \right.\) to calculate the coordinates of the centroid of the triangle ABC
Step 3: Find points EASY satisfy \(\overrightarrow {CD} = \frac{3}{2}\overrightarrow {BA} \)
Detailed explanation
a) We have: \(\overrightarrow {AB} = ( – 2; – 6)\); \(\overrightarrow {AC} = (1; – 10)\). Since \(\frac{{ – 2}}{1} \ne \frac{{ – 6}}{{ – 10}}\) then \(\overrightarrow {AB} \) and \(\overrightarrow {AC} \) are not in the same direction.
So three points A, REMOVE, OLD not in line
b) WOOD(a; b) is the centroid ofABC \( \Rightarrow G\left( {\frac{2}{3}; – \frac{1}{3}} \right)\)
c) Call \(D(a;b)\)
According to the assumption, ABCD is a trapezoid with AB // CD and CD = \(\frac{3}{2}\)AB \( \Rightarrow \overrightarrow {CD} = \frac{3}{2}\overrightarrow {BA} \)
We have: \(\overrightarrow {CD} = (a – 2;b + 5),\overrightarrow {AB} = ( – 2; – 6)\)
\( \Rightarrow \overrightarrow {CD} = \frac{3}{2}\overrightarrow {BA} \)\( \Leftrightarrow \left\{ \begin{array}{l}a – 2 = \frac{3} {2}.2\\b + 5 = \frac{3}{2}.6\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}a = 5\\b = 4\end{array} \right.\). So EASY(5; 4)
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