**Topic**

In the coordinate plane *Oxygen*for 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* = 2*MN*.

**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* = 2*MN *\( \Rightarrow \overrightarrow {PQ} = 2\overrightarrow {NM} \)

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)