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.
Solution
The given points are A(6,1), B(8,2), C(9,4) and D(7,3) .
`AB = sqrt ((6-8)^2 +(1-2)^2) = sqrt((-2)^2 +(-1)^2)`
`= sqrt(4+1) = sqrt(5) `
`BC = sqrt((8-9)^2 +(2-4)^2) = sqrt((-1)^2+(-2)^2)`
`= sqrt(1+4) = sqrt(5)`
`CD= sqrt((9-7) ^2 + (4-3)^2) = sqrt((2)^2 +(1)^2)`
`= sqrt(4+1) = sqrt(5)`
`AD = sqrt((7-6)^2 +(3-1)^2 ) = sqrt((1)^2 +(2)^2)`
`=sqrt (1+4) = sqrt(5)`
`AC = sqrt((6-9)^2 +(1-4)^2) = sqrt((-3)^2+(-3)^2)`
`= sqrt(9+9) = 3 sqrt(2)`
`=BD = sqrt(( 8-7)^2 +(2-3)^2) = sqrt((1)^2 +(-1)^2)`
`= sqrt(1+1) = sqrt(2)`
`∵ AB =BC = CD=AD = sqrt(5) and AC ≠ BD`
Therefore, the given points are the vertices of a rhombus. Now
Area` ( ΔABCD ) = 1/2 xx AC xx BD`
` = 1/2 xx 3 sqrt(2) xx sqrt(2) = 3 ` sq. units
Hence, the area of the rhombus is 3 sq. units