**Question:**

Let M(2; 0), N(2; 2), P(-1; 3) be the midpoints of sides BC, CA, AB of triangle ABC. The coordinates of point B are:

**Reference explanation:**

**Correct Answer:**

Triangle ABC has M; N; P is the midpoint of BC respectively; AC ; AB so PN and MN are the medians of the triangle.

Infer: PN // BC and MN // AB.

Then, quadrilateral PNMB is a parallelogram.

Thus, \(\overrightarrow {PB} = \overrightarrow {NM} \) with \(\overrightarrow {PB} \left( {x + 1;\,\,y – 3} \right);\overrightarrow {NM } \left( {0;\,\, – 2} \right)\).

\( \Rightarrow \left\{ {\begin{array}{*{20}{c}}

{x + 1 = 0}\\

{y – 3 = – 2}

\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}

{x = – 1}\\

{y = 1}

\end{array} \Rightarrow B\left( { – 1;1} \right)} \right.\)

ANSWER C

ADSENSE

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