Answer to Question #157060 in Geometry for Kelvin Sowah

Question #157060

ABCD is a square and P, Q are the mid points of BC, CD respectively. If AP = a and AQ = b, find in terms of a and b, the directed line segment AB, AD, BD, and AC.


1
Expert's answer
2021-02-24T07:34:01-0500

"ABCD is a square: \\overrightarrow{AB}=\\overrightarrow{DC}, \\overrightarrow{BC}=\\overrightarrow{AD}"

"P, Qare the midpoints of BC, CD respectively:"

"\\overrightarrow{BP} =\\frac{1} {2} \\overrightarrow{BC}, \\overrightarrow{DQ} =\\frac{1} {2} \\overrightarrow{DC}"


"\\overrightarrow{AP} =\\overrightarrow{AB} +\\overrightarrow{BP} =\\overrightarrow{AB} +\\frac{1} {2} \\overrightarrow{BC}"


"\\overrightarrow{AQ} =\\overrightarrow{AD} +\\overrightarrow{DQ} =\\overrightarrow{AD} +\\frac{1} {2} \\overrightarrow{DC}"



If "\\overrightarrow{AP} =\\overrightarrow{a}" and "\\overrightarrow{AQ} =\\overrightarrow{b}"



"\\overrightarrow{AB} +\\frac{1} {2} \\overrightarrow{AD} =\\overrightarrow{a}"

"\\overrightarrow{AD} +\\frac{1} {2} \\overrightarrow{AB} =\\overrightarrow{b}"


i) "\\overrightarrow{AD} =\\overrightarrow{b} - \\frac {1} {2} \\overrightarrow{AB}"

"\\overrightarrow{AB} +\\frac{1} {2} \\overrightarrow{b} - \\frac {1} {2} \\overrightarrow{AB}=\\overrightarrow{a}"


"=\\overrightarrow{AB} =\\frac{4}{3} \\overrightarrow{a} - \\frac {2}{3} \\overrightarrow{b}"




ii) "\\overrightarrow{AD} =\\overrightarrow{b} - \\frac {1} {2} (\\frac{4}{3} \\overrightarrow{a} - \\frac {2}{3} \\overrightarrow{b})"


"=\\overrightarrow{AD} =\\frac {4} {3} \\overrightarrow{b} - \\frac{2}{3} \\overrightarrow{a}"




iii) "\\overrightarrow{AD}-\\overrightarrow{AB }=\\overrightarrow{BD}"


"\\overrightarrow{BD} =" "\\frac {4} {3} \\overrightarrow{b} - \\frac{2}{3} \\overrightarrow{a}" "-" "(\\frac{4}{3} \\overrightarrow{a} - \\frac {2}{3} \\overrightarrow{b})"


"\\overrightarrow{BD} =2\\overrightarrow{b} -2\\overrightarrow{a}"



iv) "\\overrightarrow{AD}+\\overrightarrow{AB }=\\overrightarrow{AC}"


"\\overrightarrow{AC} =" "\\frac {4} {3} \\overrightarrow{b} - \\frac{2}{3} \\overrightarrow{a}" "+(\\frac{4}{3} \\overrightarrow{a} - \\frac {2}{3} \\overrightarrow{b})"


"\\overrightarrow{AC} =\\frac {2}{3} \\overrightarrow{a} + \\frac {2}{3} \\overrightarrow{b}"


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