Question

Show that the points A(-3, 2), B(-5, -5), C(-1, 4) and D(-2, -1) are the vertices of a rhombus. Find hte are of this rhombus.

Answer 1

The given points are A (-3,2), B(-5,-5) and C(-1, 4), D(-2,-1).
These given points are not the vertices of a rhombus.
if the point C(-1,4)and D(-2,-1) replace with point(2,-3) and (4,4) then these points are the vertices of an rhombus. N solution given below

$AB=\surd (-5+3{)}^2+(-5-2{)}^2$

= $\surd (-2{)}^2+(-7{)}^2$

= √4+49

=√53 units

BC =$\surd (2+5{)}^2+(-3+5{)}^2$

= $\surd (7{)}^2+(2{)}^2$

= √49+4

=√53 units

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

= $\surd (2{)}^2+(7{)}^2$

= √4+49

= √53 units

DA =$\surd (4+3{)}^2+(4-2{)}^2$

= $\surd (7{)}^2+(2{)}^2$

= √49+4

= √53 units

Therefore AB = BC = CD = DA = √53 units

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

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

= √25+25

= √50

= $\surd 25\times 2$

= 5√2 units

BD =$\surd (4+5{)}^2+(4+5{)}^2$

= $\surd (9{)}^2+(9{)}^2$

= √81+81

= √162

= $\surd 81\times 2$

= 9√2 units

Thus, diagonal AC is not equal to diagonal BD

Therefore, ABCD is a quadrilateral with equal sides and unequal diagonals

Hence, ABCD is a rhombus

Area of a rhombus = $1/2\times$(Product of its diagonals)

=$1/2\times 5\surd 2\times 9\surd 2$

= 45

= 45 square units