Question

If $A(6,1),B(8,2),C(9,4)$ and $D(7,3)$ are the vertices of $\square ABCD$ show that $\square ABCD$ is a parallelogram.

Answer 1

i) Given $A(6,1)B(8,2)C(9,4)D(7,3)$

$\overline{AB}=\sqrt{(8-6{)}^2+(2-1{)}^2}$

$=\sqrt{2^2+1^2}=\sqrt{5}$

$[\because$ Distance =$\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$\overline{BC}=\sqrt{(9-8{)}^2+(4-2{)}^2}$

$=\sqrt{1^2+2^2}=\sqrt{5}$

$\overline{CD}=\sqrt{(7-9{)}^2+(3-4{)}^2}$

$=\sqrt{(2{)}^2+(-1{)}^2}=\sqrt{5}$

$\overline{DA}=\sqrt{(7-6{)}^2+(3-1{)}^2}$

$=\sqrt{1^2+2^2}=\sqrt{5}$

$\therefore AB=BC=CA=DA$

ABCD is a parallelogram.