Question

Show that the points A(6, 1), B(8, 2), C(9, 4) and D(7, 3) are the vertices of a rhombus. Find its area.

Answer 1

The given points are A (6,1), B(8,2) and C(9,4), D(7,3). Then

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

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

$=\surd 4+1$

$=\surd 5units$

$BC=\surd (9-7{)}^2+(4-3{)}^2$

$=\surd (2{)}^2+(1{)}^2$

$=\surd 4+1$

$=\surd 5units$

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

$=\surd (2{)}^2+(1{)}^2$

$=\surd 4+1$

$=\surd 5units$

$AD=\surd (7-6{)}^2+(3-1{)}^2$

$=\surd (1{)}^2+(2{)}^2$

$=\surd 1+4$

$=\surd 5units$

$AC=\surd (6-9{)}^2+(1-4{)}^2$

$=\surd (-3{)}^2+(-3{)}^2$

$=\surd 9+9$

$=\surd 18$

$=3\surd 2units$

$BD=\surd (8-7)2+(2-3)2$

$=\surd \left(1\right)2+(-1)2$

$=\surd 1+1$

$=\surd 2$

$=\surd 2units$

Therefore AB = BC = CD = DA = √5 units and diagonal AC is not equal to diagonal BD

Hence, ABCD is a rhombus

Area of a rhombus = 1/2∗(Product of its diagonals)

$=1/2\times 3\surd 2\times \surd 2$

$=6/2$

= 3 square units