Question

The points $A\equiv (-4,-1)$, $B\equiv (-2,-4)$, $C\equiv (4,0)$ and $D\equiv (2,3)$ are the vertices of a

• A. parallelogram
• B. rectangle
• C. rhombus
• D. square

Answer 1

Answer: (A), (B)

The correct options are
A parallelogram
B rectangle

$A(-4,-1)\equiv (x_1,y_1)$, $C\equiv (4,0)\equiv (x_2,y_2)$

Mid point of $AC=(\frac{-4+4}{2},\frac{-1+0}{2})=(0,\frac{-1}{2})$........Using formula $X=\left(\frac{x_1+x_2}{2}\right)andY=\left(\frac{y_1+y_2}{2}\right)$

$B(-2,-4)\equiv (x_1,y_1)$, $D\equiv (2,3)\equiv (x_2,y_2)$

Mid point of $BD=(\frac{-2+2}{2},\frac{-4+3}{2})=(0,\frac{-1}{2})$........Using formula $X=\left(\frac{x_1+x_2}{2}\right)andY=\left(\frac{y_1+y_2}{2}\right)$

$\Rightarrow$ the diagonals $AC$ and $BD$ bisect each other
$\Rightarrow$ $ABCD$ is a parallelogram.

Next, $AC=\sqrt{(-4-4{)}^2+(-1+0{)}^2}=\sqrt{64+1}=\sqrt{65}$

$BD=\sqrt{(-2-2{)}^2+(-4-3{)}^2}=\sqrt{16+49}=\sqrt{65}$
Since the diagonals are also equal, it is a rectangle.

As $AB=\sqrt{(-4+2{)}^2+(-1+4{)}^2}=\sqrt{4+9}=\sqrt{13}$

and $BC=\sqrt{(-2-4{)}^2+(-4{)}^2}=\sqrt{36+16}=\sqrt{52}$

The adjacent sides are not equal.

Hence, it is not a square or rhombus.