Question Number 148550 by mathdanisur last updated on 29/Jul/21
$$\boldsymbol{{A}}\left(\mathrm{0};\mathrm{0}\right)\:\:;\:\:\boldsymbol{{B}}\left(−\mathrm{2};\mathrm{4}\right)\:\:{and}\:\:\boldsymbol{{C}}\left(−\mathrm{6};\mathrm{14}\right) \\ $$$${if}\:{the}\:{triangle}\:{has}\:{vertices},\:{find}\:{the} \\ $$$${lenght}\:{of}\:{the}\:{median}\:{drawn}\:{from} \\ $$$${the}\:{vertx}\:\boldsymbol{{C}} \\ $$
Commented by MJS_new last updated on 29/Jul/21
$$\mathrm{first}\:\mathrm{show}\:\mathrm{us}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{which}\:\mathrm{has}\:\boldsymbol{{no}}\:\mathrm{vertices} \\ $$
Commented by Ar Brandon last updated on 29/Jul/21
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Answered by Rasheed.Sindhi last updated on 29/Jul/21
$${Midpoint}\:{of}\:{AB}\:={M}=\left(\frac{−\mathrm{2}+\mathrm{0}}{\mathrm{2}},\frac{\mathrm{4}+\mathrm{0}}{\mathrm{2}}\right)=\left(−\mathrm{1},\mathrm{2}\right) \\ $$$${The}\:{median}\:{through}\:{C} \\ $$$${CM}=\sqrt{\left(−\mathrm{6}−\left(−\mathrm{1}\right)\right)^{\mathrm{2}} +\left(\mathrm{14}−\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:=\sqrt{\mathrm{25}+\mathrm{144}}=\sqrt{\mathrm{169}}=\mathrm{13} \\ $$
Commented by mathdanisur last updated on 29/Jul/21
$${Thank}\:{you}\:{Ser},\:{cool} \\ $$