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Question Number 54025 by pieroo last updated on 28/Jan/19
The points A, B, C, D have coordinates  (−7,9), (3,4), (1,12), and (−2,−9).  find the length of the linePQ where P  devides AB in the ratio 2:3 and devides  CD in the ratio 1:−4.
$$\mathrm{The}\:\mathrm{points}\:\mathrm{A},\:\mathrm{B},\:\mathrm{C},\:\mathrm{D}\:\mathrm{have}\:\mathrm{coordinates} \\ $$$$\left(−\mathrm{7},\mathrm{9}\right),\:\left(\mathrm{3},\mathrm{4}\right),\:\left(\mathrm{1},\mathrm{12}\right),\:\mathrm{and}\:\left(−\mathrm{2},−\mathrm{9}\right). \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{linePQ}\:\mathrm{where}\:\mathrm{P} \\ $$$$\mathrm{devides}\:\mathrm{AB}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{2}:\mathrm{3}\:\mathrm{and}\:\mathrm{devides} \\ $$$$\mathrm{CD}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{1}:−\mathrm{4}. \\ $$
Commented by pieroo last updated on 28/Jan/19
I need some help, especially with the  diagram.
$$\mathrm{I}\:\mathrm{need}\:\mathrm{some}\:\mathrm{help},\:\mathrm{especially}\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{diagram}. \\ $$
Answered by ajfour last updated on 28/Jan/19
Commented by ajfour last updated on 28/Jan/19
x_P = ((mx_B +nx_A )/(m+n)) = ((2(3)+3(−7))/5)=−3  y_P =((my_B +ny_A )/(m+n)) =((2(4)+3(9))/5)= 7  P (−3,7)  x_Q =((1(−2)+(−4)(1))/(−3)) = 2  y_Q = ((1(−9)+(−4)(12))/(−3)) = 19  Q(2,19)  PQ = (√((−3−2)^2 +(7−19)^2 ))           = (√(25+144)) = 13 units.
$${x}_{{P}} =\:\frac{{mx}_{{B}} +{nx}_{{A}} }{{m}+{n}}\:=\:\frac{\mathrm{2}\left(\mathrm{3}\right)+\mathrm{3}\left(−\mathrm{7}\right)}{\mathrm{5}}=−\mathrm{3} \\ $$$${y}_{{P}} =\frac{{my}_{{B}} +{ny}_{{A}} }{{m}+{n}}\:=\frac{\mathrm{2}\left(\mathrm{4}\right)+\mathrm{3}\left(\mathrm{9}\right)}{\mathrm{5}}=\:\mathrm{7} \\ $$$${P}\:\left(−\mathrm{3},\mathrm{7}\right) \\ $$$${x}_{{Q}} =\frac{\mathrm{1}\left(−\mathrm{2}\right)+\left(−\mathrm{4}\right)\left(\mathrm{1}\right)}{−\mathrm{3}}\:=\:\mathrm{2} \\ $$$${y}_{{Q}} =\:\frac{\mathrm{1}\left(−\mathrm{9}\right)+\left(−\mathrm{4}\right)\left(\mathrm{12}\right)}{−\mathrm{3}}\:=\:\mathrm{19} \\ $$$${Q}\left(\mathrm{2},\mathrm{19}\right) \\ $$$${PQ}\:=\:\sqrt{\left(−\mathrm{3}−\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{7}−\mathrm{19}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\:\sqrt{\mathrm{25}+\mathrm{144}}\:=\:\mathrm{13}\:{units}. \\ $$
Commented by pieroo last updated on 28/Jan/19
thanks senior
$$\mathrm{thanks}\:\mathrm{senior} \\ $$

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