Question

Column II gives distance between pair of points given in column I match them correctly.

Answer 1

$A)Lettheco-ordinatesofA\&BareA(x_1,y_1)=(-5,7)\&B(x_2,y_2)=(-1,3)respectively.ThenAB=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}=\sqrt{{(-5+1)}^2+{(7-3)}^2}units=4\sqrt{2}units.⟹Matcheswith4.B)Lettheco-ordinatesofA\&BareA(x_1,y_1)=(5,4)\&B(x_2,y_2)=(1,3)respectively.ThenAB=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}=\sqrt{{(1-5)}^2+{(4-3)}^2}units=\sqrt{17}units.⟹Matcheswith1.C)Lettheco-ordinatesofA\&BareA(x_1,y_1)=(\sqrt{3}+1,1)\&B(x_2,y_2)=(0,\sqrt{3})respectively.ThenAB=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}=\sqrt{{(\sqrt{3}+1-0)}^2+{(1-\sqrt{3})}^2}units=\sqrt{8}units.⟹Matcheswith2.D)Lettheco-ordinatesofA\&BareA(x_1,y_1)=(0,0)\&B(x_2,y_2)=(-\sqrt{3},\sqrt{3})respectively.ThenAB=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}=\sqrt{{(\sqrt{3}-0)}^2+{(-\sqrt{3-0})}^2}units=\sqrt{6}units.⟹Matcheswith3.Ans-A⟶4B⟶1C⟶2D⟶3$