Question

Let A, B and C be three points whose coordinates are (2, 3), (3, b) and (5, 7) respectively

Answer 1

$ThepointsA,B\&CareA(x_1,y_1)=(2,3),B(x_2,y_2)=(3,b)\&C(x_3,y_3)=(5,7)respectively..\left(A\right)Nowar\Delta ABCwithverticesA(x_1,y_1),B(x_2,y_2)\&C(x_3,y_3)isar\Delta ABC=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)].\therefore ar\Delta ABC=\frac{1}{2}[2(b-7)+3(7-3)+5(3-b)]=\frac{1}{2}\left(given\right)⟹-3b+13=1⟹3b=12⟹b=4.\left(B\right)IfA,B\&Carecollinearthenar\Delta ABC=0-3b+13=0⟹3b=13\left(C\right)BdividesthelineABCintheratio\lambda :1.Usingsectionformula(3,b)=(\frac{5\times \lambda +2\times 1}{\lambda +1},\frac{7\times \lambda +3\times 1}{\lambda +1})=(\frac{5\lambda +2}{\lambda +1},\frac{7\lambda +3}{\lambda +1}).\therefore \frac{5\lambda +2}{\lambda +1}=3⟹5\lambda +2=3\lambda +3⟹2\lambda -1=0⟹4\lambda -2=0⟹4\lambda +4-4-2=0⟹4(\lambda +1)=6.\left(D\right)b=4.Sotheverticesof\Delta ABCareA(x_1,y_1)=(2,3),B(x_2,y_2)=(3,b)\&C(x_3,y_3)=(5,7).Usingdistanceformula.AB=\sqrt{{(2-3)}^2+{(3-4)}^2}units=\sqrt{2}units,BC=\sqrt{{(3-5)}^2+{(4-7)}^2}units=\sqrt{13}unitsandAC=\sqrt{{(2-5)}^2+{(4-7)}^2}units=\sqrt{25}=5units.So5units>\sqrt{13}units>\sqrt{2}units.\therefore Thegreatestsideis5units.ListIListII\therefore A⟶3\Rightarrow 4B⟶4\Rightarrow 13C⟶1\Rightarrow 6D⟶2\Rightarrow 5$