Prove that the points $(3, 0), (4, 5), (-1, 4)$ and $(-2, -1)$, taken in order, form a rhombus. Also, find its area.

Given:

Given points are $(3, 0), (4, 5), (-1, 4)$ and $(-2, -1)$.

To do:

We have to prove that the points $(3, 0), (4, 5), (-1, 4)$ and $(-2, -1)$, taken in order, form a rhombus and find its area.

Solution:

Let \( \mathrm{ABCD} \) is a quadrilateral whose vertices are \( \mathrm{A}(3,0), \mathrm{B}(4,5), \mathrm{C}(-1,4) \) and \( \mathrm{D}(-2,-1) \).

We know that,

The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).

Therefore,

 \( A B=\sqrt{(4-3)^{2}+(5-0)^{2}} \)

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

Squaring on both sides, we get,

\( AB^{2}=(1)^{2}+(5)^{2} \)

\( =1+25=26 \)
Similarly,

\( B C^{2}=(-1-4)^{2}+(4-5)^{2} \)
\( =(-5)^{2}+(-1)^{2} \)

\( =25+1 \)

\( =26 \)
\( C D^{2}=(-2+1)^{2}+(-1-4)^{2} \)
\( =(-1)^{2}+(-5)^{2} \)

\( =1+25 \)

\( =26 \)
\( D A^2=(3+2)^{2}+(0+1)^{2} \)
\( =(5)^{2}+(1)^{2} \)

\( =25+1 \)

\( =26 \)
\( AC^{2}=(-1-3)^{2}+(4-0)^{2} \)
\( =(-4)^{2}+(4)^{2} \)

\( =16+16 \)

\( =32 \)

\( \Rightarrow AC=\sqrt{32}=4\sqrt2 \)
\( B D^{2}=(-2-4)^{2}+(-1-5)^{2} \)
\( =(-6)^{2}+(-6)^{2} \)

\( =36+36 \)

\( =72 \)

\( \Rightarrow BD=\sqrt{72}=6\sqrt2 \)

Here,

$A B=B C=C D=D A=\sqrt{26}$

Sides are equal but diagonal are not equal. \( \therefore \mathrm{ABCD} \) is a rhombus.

We know that,

Area of rhombus \( =\frac{\text { Product of diagonals }}{2} \)

\( =\frac{4 \sqrt{2} \times 6 \sqrt{2}}{2} \) \( =\frac{24\times2}{2} \)

\( =24 \) sq. units

The area of the rhombus is $24$ sq. units.