WAEC 2017 · Paper 2 · Q7VectorsCoordinate geometry & circles(a)A parallelogram MNQRMNQRMNQR has vertices M(4,−6)M(4, -6)M(4,−6), N(10,2)N(10, 2)N(10,2), Q(8,16)Q(8, 16)Q(8,16) and R(x,y)R(x, y)R(x,y). Find the coordinates of RRR.CheckSeparate values with commas, e.g. 3, −2Worked solution (try it first)In MNQRMNQRMNQR, MN→=RQ→\overrightarrow{MN} = \overrightarrow{RQ}MN=RQ.MN→=(10−42+6)\overrightarrow{MN} = \begin{pmatrix} 10 - 4 \\ 2 + 6 \end{pmatrix}MN=(10−42+6)=(68)= \begin{pmatrix} 6 \\ 8 \end{pmatrix}=(68) and RQ→=(8−x16−y)\overrightarrow{RQ} = \begin{pmatrix} 8 - x \\ 16 - y \end{pmatrix}RQ=(8−x16−y).So 8−x=68 - x = 68−x=6 and 16−y=816 - y = 816−y=8: R(2,8)R(2, 8)R(2,8).Report a problem with this question