Question

State whether the following statement is true or false.

Enter $1$ for true and $0$ for false

The points A (1, 2), B (3, 4), C (2, 5) and D (0, 3) in that order form a rectangle.

Answer 1

$LetfourpointsA(x_1,y_1)=(1,2),B(x_2,y_2)=(3,4),C(x_3,y_3)=(2,5)$

$\&D(x_4,y_4)=(0,3),takeninorder,formarectangle.$

$ThenAC\&BDwillbethediagonalswhichwillbisecteachother.$

$AlsoAC=BD.$

$LetusfindoutthemidpointsofAC\&BD.$

$ForAC(x_m,y_m)=(\frac{x_1+x_3}{2},\frac{y_2+y_3}{2})=(\frac{1+2}{2},\frac{2+5}{2})=(1.5,3.5).$

$ForBD(x_m,y_m)=(\frac{x_2+x_4}{2},\frac{y_2+y_4}{2})=(\frac{3+0}{2},\frac{4+3}{2})=(1.5,3.5).$

$SothemidpointsofAC\&BDaresame...........\left(i\right)$

$\therefore AC\&BDbisecteachother.$

$AgainAC=\sqrt{{(x_3-x_1)}^2+{(y_3-y_1)}^2=}\sqrt{{(2-1)}^2+{(5-2)}^2}units.=\sqrt{10}units.$

$AndBD=\sqrt{{(x_4-x_2)}^2+{(y_4-y_2)}^2}=\sqrt{{(3-4)}^2+{(0-3)}^2}units.=\sqrt{10}units.$

$\therefore AC=BD..........\left(ii\right)$

$Sofrom\left(i\right)\&\left(ii\right)weconcludethatABCDisarectangle.$

$Ans-TRUE.$