Question

The points A (-4, -1), B (-2, -4), C (4, 0) and D (2, 3) are the vertices of
$\left( a \right)$ Parallelogram
$\left( b \right)$ Rectangle
$\left( c \right)$ Rhombus
$\left( d \right)$ None of these.

Answer 1

Hint: In this question use the properties of distance between two points slope of two points and if the multiplication of two slopes is (-1) then these two lines are perpendicular to each other so use these concepts to reach the solution of the question.

Complete step-by-step answer:

Let
A = ($x_1$, $y_1$) = (-4, -1)
B = ($x_2$, $y_2$) = (-2, -4)
C = ($x_3$, $y_3$) = (4, 0)
D = ($x_4$, $y_4$) = (2, 3)
As we know that the distance between two points ($x_1$, $y_1$) and ($x_2$, $y_2$) is calculated as
$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
So the distance,
$AB = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} = \sqrt {{{\left( { - 2 - \left( { - 4} \right)} \right)}^2} + {{\left( { - 4 - \left( { - 1} \right)} \right)}^2}} = \sqrt {4 + 9} = \sqrt {13} $
Similarly,
$BC = \sqrt {{{\left( {{x_3} - {x_2}} \right)}^2} + {{\left( {{y_3} - {y_2}} \right)}^2}} = \sqrt {{{\left( {4 + 2} \right)}^2} + {{\left( {0 + 4} \right)}^2}} = \sqrt {36 + 16} = \sqrt {52} $
$CD = \sqrt {{{\left( {{x_4} - {x_3}} \right)}^2} + {{\left( {{y_4} - {y_3}} \right)}^2}} = \sqrt {{{\left( {2 - \left( 4 \right)} \right)}^2} + {{\left( {3 - 0} \right)}^2}} = \sqrt {4 + 9} = \sqrt {13} $
$DA = \sqrt {{{\left( {{x_1} - {x_4}} \right)}^2} + {{\left( {{y_1} - {y_4}} \right)}^2}} = \sqrt {{{\left( { - 4 - 2} \right)}^2} + {{\left( { - 1 - 3} \right)}^2}} = \sqrt {36 + 16} = \sqrt {52} $
And the distance of diagonals are
$AC = \sqrt {{{\left( {{x_3} - {x_1}} \right)}^2} + {{\left( {{y_3} - {y_1}} \right)}^2}} = \sqrt {{{\left( {4 - \left( { - 4} \right)} \right)}^2} + {{\left( {0 - \left( { - 1} \right)} \right)}^2}} = \sqrt {64 + 1} = \sqrt {65} $
$BD = \sqrt {{{\left( {{x_4} - {x_2}} \right)}^2} + {{\left( {{y_4} - {y_2}} \right)}^2}} = \sqrt {{{\left( {2 - \left( { - 2} \right)} \right)}^2} + {{\left( {3 - \left( { - 4} \right)} \right)}^2}} = \sqrt {16 + 49} = \sqrt {65} $
As we know that the slope of two points ($x_1$, $y_1$) and ($x_2$, $y_2$) is calculated as
$m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
So the slope of AB is
$ \Rightarrow {m_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \dfrac{{ - 4 + 1}}{{ - 2 + 4}} = \dfrac{{ - 3}}{2}$
The slope of BC is
$ \Rightarrow {m_2} = \dfrac{{{y_3} - {y_2}}}{{{x_3} - {x_2}}} = \dfrac{{0 + 4}}{{4 + 2}} = \dfrac{2}{3}$
The slope of CD is
$ \Rightarrow {m_3} = \dfrac{{{y_4} - {y_3}}}{{{x_4} - {x_3}}} = \dfrac{{3 - 0}}{{2 - 4}} = \dfrac{{ - 3}}{2}$
And the slope of DA is
$ \Rightarrow {m_4} = \dfrac{{{y_1} - {y_4}}}{{{x_1} - {x_4}}} = \dfrac{{ - 1 - 3}}{{ - 4 - 2}} = \dfrac{{ - 4}}{{ - 6}} = \dfrac{2}{3}$
So the multiplication of slopes AB and BC is
$ \Rightarrow {m_1}.{m_2} = \dfrac{{ - 3}}{2} \times \dfrac{2}{3} = - 1$
Therefore AB and BC are perpendicular to each other.
So the multiplication of slopes BC and CD is
$ \Rightarrow {m_2}.{m_3} = \dfrac{2}{3} \times \dfrac{{ - 3}}{2} = - 1$
Therefore BC and CD are perpendicular to each other.
And the multiplication of slopes CD and DA is
$ \Rightarrow {m_3}.{m_4} = \dfrac{{ - 3}}{2} \times \dfrac{2}{3} = - 1$
Therefore CD and DA are perpendicular to each other.
So as we see that the length of opposite sides are equal and the length of diagonals are also equal and all the lines are perpendicular to each other which is the condition of a rectangle.
AB = CD = $\sqrt {13} $
BC = DA = $\sqrt {52} $
And AC = BD = $\sqrt {65} $
So the required answer is a rectangle. Hence option (B) is correct.

Note: Whenever we face such types of questions the key concept is the formula of distance between two points, slope of line passing through two points which is all stated above and the multiplication of slope of perpendicular lines is always (-1) so first calculate the distances as above then calculate the slope and then calculate the multiplication of two slopes which is adjacent to each other if we got -1 then the adjacent lines are perpendicular to each other.