Prove that the point A(1, 3, 0), B(–5, 5, 2), C(–9, –1, 2) and D(–3, –3, 0) taken in order are the vertices of a parallelogram. Also, show that ABCD is not a rectangle.

Open in App

Solution

Let A(1,3,0),B( $-$ 5,5,2), C( $-$ 9, $-$ 1,2) and D( $-$ 3, $-$ 3,0) be the coordinates of quadrilateral $□ A B C D$

$A B = \sqrt{{(- 5 - 1)}^{2} + {(5 - 3)}^{2} + {(2 - 0)}^{2}} = \sqrt{36 + 4 + 4} = \sqrt{44} B C = \sqrt{{(- 9 + 5)}^{2} + {(- 1 - 5)}^{2} + {(2 - 2)}^{2}} = \sqrt{16 + 36 + 0} = \sqrt{52} C D = \sqrt{{(- 3 + 9)}^{2} + {(- 3 + 1)}^{2} + {(0 - 2)}^{2}} = \sqrt{36 + 4 + 4} = \sqrt{44} D A = \sqrt{{(1 + 3)}^{2} + {(3 + 3)}^{2} + {(0 - 0)}^{2}} = \sqrt{16 + 36 + 0} = \sqrt{52} Here, we see that A B = C D & B C = D A$
Since, opposite pair of sides are equal .

Therefore, $□ A B C D$ is a parallelogram .

$A C = \sqrt{{(- 9 - 1)}^{2} + {(- 1 - 3)}^{2} + {(2 - 0)}^{2}} = \sqrt{{(- 10)}^{2} + {(- 4)}^{2} + {(2)}^{2}} = \sqrt{100 + 16 + 4} = \sqrt{120}$

$B D = \sqrt{{(- 3 + 5)}^{2} + {(- 3 - 5)}^{2} + {(0 - 2)}^{2}} = \sqrt{{(2)}^{2} + {(- 8)}^{2} + {(- 2)}^{2}} = \sqrt{4 + 64 + 4} = \sqrt{72}$

$∴ A C \neq B D$
∴ ABCD is not a rectangle.

Suggest Corrections

Similar questions

Prove that the point A(1, 3, 0), B(-5, 5, 2), C(-9, -1, 2) and D(-3, -3, 0) taken in order are the vertices of a parallelogram. Also, show that ABCD is not a rectangle.

Show that A(1, 2), B(4, 3), C(6, 6) and D(3, 5) are the vertices parallelogram. Show that ABCD is not a rectangle.

Q. Show that A(1, 2), B(4, 3), C(6, 6) and D(3, 5) are the vertices of a parallelogram. Show that ABCD is not a rectangle.

Show that the points A(1, 2, 3), B(-1, -2, -1), C(2, 3, 2) and D (4, 7, 6) are the vertices of a parallelogram ABCD, but not a rectangle.

Q. Show that the points A(3, 0), B(6, 4) and C(−1, 3) are the vertices of a right-angled triangle. Also, prove that these are the vertices of an isosceles triangle.

Join BYJU'S Learning Program