Question

Show that $A( - 3,2)$, $B( - 5, - 5)$, $C(2, - 3)$ and $D(4,4)$ are the vertices of a rhombus.

Answer 1

Hint : We will prove that given points are vertices of a rhombus using the distance formula. We will find the distance $AB$, $BC$, $CD$ and $DA$, if all these lengths are equal then the given points will form a rhombus if all the sides of a rhombus are equal.
Formula Used :
We will use the formula to find the distance between two given points $({x_1},{y_1})$ and $({x_2},{y_2})$which is given by $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$.

Complete step-by-step answer :
We have to show that given points $A( - 3,2)$, $B( - 5, - 5)$, $C(2, - 3)$ and $D(4,4)$ are the vertices of a rhombus.
Let the point $A( - 3,2)$, $B( - 5, - 5)$, $C(2, - 3)$ and $D(4,4)$ be the vertices of a quadrilateral $ABCD$.
Let’s draw the diagram showing all the given points.

As we know that the length of all the sides of a rhombus is equal and the length of its diagonal are not equal.
We will find the length of each side to prove that points $A( - 3,2)$, $B( - 5, - 5)$, $C(2, - 3)$ and $D(4,4)$ are the vertices of a rhombus.
Now, we will find the distance $AB$ using the distance formula.
We have $A( - 3,2)$ and $B( - 5, - 5)$.Therefore, using the distance formula we get
$\Rightarrow AB = \sqrt {{{\left( {\left( { - 5} \right) - ( - 3)} \right)}^2} + {{\left( {\left( { - 5} \right) - \left( 2 \right)} \right)}^2}}$
Subtracting the terms in the bracket, we get
$\Rightarrow AB = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 7} \right)}^2}}$
On solving,
$\Rightarrow AB = \sqrt {4 + 49}$
On further simplification we get
$\Rightarrow AB = \sqrt {53}$
Now, we will find the distance $BC$ using the distance formula.
We have $B( - 5, - 5)$ and $C(2, - 3)$.Therefore, using the distance formula we get
$\Rightarrow BC = \sqrt {{{\left( {\left( 2 \right) - ( - 5)} \right)}^2} + {{\left( {\left( { - 3} \right) - \left( { - 5} \right)} \right)}^2}}$
Subtracting the terms in the bracket, we get
$\Rightarrow BC = \sqrt {{{\left( 7 \right)}^2} + {{\left( 2 \right)}^2}}$
On solving,
$\Rightarrow BC = \sqrt {49 + 4}$
On further simplification we get
$\Rightarrow BC = \sqrt {53}$
Now, we will find the distance $CD$ using the distance formula.
We have $C(2, - 3)$ and $D(4,4)$.Therefore, using the distance formula we get
$\Rightarrow CD = \sqrt {{{\left( {\left( 4 \right) - (2)} \right)}^2} + {{\left( {\left( 4 \right) - \left( { - 3} \right)} \right)}^2}}$
Subtracting the terms in the bracket, we get
$\Rightarrow CD = \sqrt {{{\left( 2 \right)}^2} + {{\left( 7 \right)}^2}}$
On solving,
$\Rightarrow CD = \sqrt {4 + 49}$
On further simplification we get
$\Rightarrow CD = \sqrt {53}$
Now, we will find the distance $DA$ using the distance formula.
We have $D(4,4)$ and $A( - 3,2)$.Therefore, using the distance formula we get
$\Rightarrow DA = \sqrt {{{\left( {\left( { - 3} \right) - (4)} \right)}^2} + {{\left( {\left( 2 \right) - \left( 4 \right)} \right)}^2}}$
Subtracting the terms in the bracket, we get
$\Rightarrow DA = \sqrt {{{\left( { - 7} \right)}^2} + {{\left( { - 2} \right)}^2}}$
On solving,
$\Rightarrow DA = \sqrt {49 + 4}$
On further simplification we get
$\Rightarrow DA = \sqrt {53}$
Also, we have length of diagonal $AC$ as
$\Rightarrow AC = \sqrt {{{\left( {\left( 2 \right) - ( - 3)} \right)}^2} + {{\left( {\left( { - 3} \right) - \left( 2 \right)} \right)}^2}}$
Subtracting the terms in the bracket, we get
$\Rightarrow AC = \sqrt {{{\left( 5 \right)}^2} + {{\left( { - 5} \right)}^2}}$
On solving,
$\Rightarrow AC = \sqrt {25 + 25}$
On further simplification we get
$\Rightarrow AC = \sqrt {50}$
Now, we have length of diagonal $BD$ as
$\Rightarrow BD = \sqrt {{{\left( {\left( 4 \right) - ( - 5)} \right)}^2} + {{\left( {\left( 4 \right) - \left( { - 5} \right)} \right)}^2}}$
Subtracting the terms in the bracket, we get
$\Rightarrow BD = \sqrt {{{\left( 9 \right)}^2} + {{\left( 9 \right)}^2}}$
On solving,
$\Rightarrow BD = \sqrt {81 + 81}$
On further simplification we get
$\Rightarrow BD = \sqrt {162}$
Therefore, we can see that $AB = BC = CD = DA$ and $AC \ne BD$ i.e., length of all the sides is equal and diagonals are not equal.
Hence, points $A( - 3,2)$, $B( - 5, - 5)$, $C(2, - 3)$ and $D(4,4)$ are the vertices of a rhombus.

Note : We can also prove that the given points are vertices of a rhombus by first finding the midpoint of both the diagonals $AC$ and $BD$ using the midpoint formula. If the midpoint of both the diagonals lies on the same point then it forms a parallelogram. Then we will find the length of opposite sides, if the length of opposite sides is equal then it is a rhombus.