Question

Show that the points A(- 4, -1), B(-2, - 4), C(4, 0) and D(2, 3) are the vertices points of a rectangle.

Answer 1

Step-by-step explanation:

• Co-ordinate of vertices is given as

        $\mathbf{A(x_{1},y_{1})=(-4,-1)}$

        $\mathbf{B(x_{2},y_{2})=(-2,-4)}$

        $\mathbf{C(x_{3},y_{3})=(4,0)}$

        $\mathbf{D(x_{4},y_{4})=(2,3)}$

• Magnitude of side AB by distance formula

        $\mathbf{AB=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}}$

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

• Magnitude of side BC by distance formula

        $\mathbf{BC=\sqrt{(x_{2}-x_{3})^{2}+(y_{2}-y_{3})^{2}}}$

        $\mathbf{BC=\sqrt{(-2-4)^{2}+(-4-0)^{2}}=\sqrt{36+16}=\sqrt{52}}$

• Magnitude of side CD by distance formula

        $\mathbf{CD=\sqrt{(x_{3}-x_{4})^{2}+(y_{3}-y_{4})^{2}}}$

        $\mathbf{CD=\sqrt{(4-2)^{2}+(0-3)^{2}}=\sqrt{4+9}=\sqrt{13}}$

• Magnitude of side DA by distance formula

        $\mathbf{DA=\sqrt{(x_{4}-x_{1})^{2}+(y_{4}-y_{1})^{2}}}$

        $\mathbf{DA=\sqrt{(2+4)^{2}+(3+1)^{2}}=\sqrt{36+16}=\sqrt{52}}$

• Magnitude of diagonal AC by distance formula

        $\mathbf{AC=\sqrt{(x_{1}-x_{3})^{2}+(y_{1}-y_{3})^{2}}}$

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

• Magnitude of diagonal BD by distance formula

        $\mathbf{BD=\sqrt{(x_{2}-x_{4})^{2}+(y_{2}-y_{4})^{2}}}$

        $\mathbf{BD=\sqrt{(-2-2)^{2}+(-4-3)^{2}}=\sqrt{16+49}=\sqrt{65}}$

• AB = CD

         BC =DA

         AC = BD

• Here we see that opposite side are equal and both diagonal are equal, hence given vertices are vertices of rectangle.