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Topic
In the coordinate plane Oxygenfor three noncollinear points USA(twelfth), WOMEN(thirty first), P(− 1 ; 2). Find the coordinates of the point Q such that quadrilateral MNPQ is a trapezoid with MN // PQ and PQ = 2MN.
Solution method – See details
From the assumption find the coordinates of the point Q satisfy \(\overrightarrow {PQ} = 2\overrightarrow {NM} \)
Detailed explanation
We have: MN // PQ so \(\overrightarrow {MN} \) and \(\overrightarrow {PQ} \) have the same direction
Other way, PQ = 2MN \( \Rightarrow \overrightarrow {PQ} = 2\overrightarrow {NM} \)
Call point coordinates Q is \(Q(a;b)\). We have: \(\overrightarrow {PQ} = (a + 1;b – 2)\) and \(\overrightarrow {NM} = ( – 2; – 3)\)
\( \Rightarrow \overrightarrow {PQ} = 2\overrightarrow {NM} \Leftrightarrow \left\{ \begin{array}{l}a + 1 = 2.( – 2)\\b – 2 = 2.( – 3)\end{array} \right \Leftrightarrow \left\{ \begin{array}{l}a + 1 = – 4\\b – 2 = – 6\end{array} \right \Leftrightarrow \left \{ \begin{array}{l}a = – 5\\b = – 4\end{array} \right.\) . So Q(-5 ; -4)
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