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Determinants

Question

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Using determinants, show that the points (11, 7), (5, 5) and (– 1, 3) are collinear.

Solution

Let ∆ be the area of the triangle whose vertices are (11, 7), (5, 5), (– 1, 3).

therefore space space increment space equals space 1 half open vertical bar table row 11 cell space space 7 end cell cell space space 1 end cell row 5 cell space space 5 end cell cell space space 1 end cell row cell negative 1 end cell cell space space 3 end cell cell space space 1 end cell end table close vertical bar space space space equals space 1 half open vertical bar table row 11 cell space space space space 7 end cell cell space space space space 1 end cell row cell negative 6 end cell cell space minus 2 end cell cell space space space space 0 end cell row cell negative 12 end cell cell space minus 4 end cell cell space space space space 0 end cell end table close vertical bar comma space space straight R subscript 2 space minus straight R subscript 1 comma space space straight R subscript 3 space minus straight R subscript 1 space space space space space equals space 1 half open vertical bar table row cell negative 6 end cell cell space space space minus 2 end cell row cell negative 12 end cell cell space space space minus 4 end cell end table close vertical bar space equals space 1 half left parenthesis 24 minus 24 right parenthesis space equals space 0. therefore space space space given space points space left parenthesis 11 comma space 7 right parenthesis comma space left parenthesis 5 comma space 5 right parenthesis space and space left parenthesis negative 1 comma space 3 right parenthesis space are space collinear. space space space space space

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