Question

Show that the points (-3,5),(3,1),(1,-3)and (-5,1) form a quadrilateral or not?

Answer 1

Let A(- 3, 5), B(3, 1), C (1, - 3) and D(- 5, 1)  be the four points of quadrilateral.

We have to show that the given points are quadrilateral or not.

Solution:

We know that:

Area of triangle = $\dfrac{1}{2} [{x_{1} (y_{2}-y_{3})+x_{2} (y_{3}-y_{1})+x_{3} (y_{1}-y_{2})}]$

∴ Area of triangle ABC = $\dfrac{1}{2} [(-3)(1-(-3))+3 (-3-5)+1 (5-1)}]$

                                       = $\dfrac{1}{2} [(-3)(1+3)+3 (-8)+1 (4)}]$

                                      = $\dfrac{1}{2} [-12-24+4]$

                                     = $\dfrac{1}{2} [-36+4]$

                                    = $\dfrac{1}{2} [-32]$

                                   = 16

Area of triangle BCD = $\dfrac{1}{2} [3(-3-1)+1 (1-1)-5(1+3)}]$

= $\dfrac{1}{2} [3(-4)+1 (0)-5(4)}]$

= $\dfrac{1}{2} [-12+0-20]$

= $\dfrac{1}{2} [-32]$

= -16

Area of triangle ACD = $\dfrac{1}{2} [-3(-3-1)+1 (1-5)-5(5+3)}]$

= $\dfrac{1}{2} [-3(-4)+1 (-4)-5(8)}]$

= $\dfrac{1}{2} [12-4-40}]$

= $\dfrac{1}{2} [-32}]$

= 16

Area of triangle ABD = $\dfrac{1}{2} [-3(1-1)+3 (1-5)-5(5-1)}]$

= $\dfrac{1}{2} [-3(0)+3 (-4)-5(4)}]$

= $\dfrac{1}{2} [0-12-20]$

= 16

∵ The area of triangle are not collinear, then the given points are quadrilateral.

Thus, the points (- 3, 5), (3, 1),(1, - 3) and (- 5, 1) form a quadrilateral.

Answer 2

$\textbf{Given:}$

$\textsf{Points (-3,5),(3,1),(1,-3) and (-5,1)}$

$\textbf{To check:}$

$\textsf{The given points form a quadrilateral or not}$

$\textbf{Solution:}$

$\textsf{Let the given points be A(-3,5), B(3,1),C(1,-3) and D(-5,1)}$

$\textsf{First we fiind length of the sides}$

$\mathsf{AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}$

$\mathsf{AB=\sqrt{(-3-3)^2+(5-1)^2}}$

$\mathsf{AB=\sqrt{36+16}}$

$\mathsf{AB=\sqrt{52}}$

$\mathsf{AB=\sqrt{4{\times}13}}$

$\mathsf{AB=2\sqrt{13}}$

$\mathsf{BC=\sqrt{(3-1)^2+(1+3)^2}}$

$\mathsf{BC=\sqrt{4+16}}$

$\mathsf{BC=\sqrt{20}}$

$\mathsf{BC=\sqrt{4{\times}5}}$

$\mathsf{BC=2\sqrt{5}}$

$\mathsf{CD=\sqrt{(1+5)^2+(-3-1)^2}}$

$\mathsf{CD=\sqrt{36+16}}$

$\mathsf{CD=\sqrt{52}}$

$\mathsf{CD=\sqrt{4{\times}13}}$

$\mathsf{CD=2\sqrt{13}}$

$\mathsf{AD=\sqrt{(-3+5)^2+(5-1)^2}}$

$\mathsf{AD=\sqrt{4+16}}$

$\mathsf{AD=\sqrt{20}}$

$\mathsf{AD=\sqrt{4{\times}5}}$

$\mathsf{AD=2\sqrt{5}}$

$\mathsf{AB=CD\;and\;BC=AD}$

$\implies\textsf{opposite sides are equal}$

$\mathsf{AC=\sqrt{(-3-1)^2+(5+3)^2}}$

$\mathsf{AC=\sqrt{16+64}}$

$\mathsf{AC=\sqrt{80}}$

$\mathsf{Now,}$

$\mathsf{AB^2+BC^2=52+20=72{\neq}AC^2}$

$\mathsf{\angle{B}{\neq}90^\circ}$

$\implies\textsf{ABCD is not  a rectangle}$

$\textsf{Hence ABCD is  a parallelogram}$