State whether the following statements are true or false. Justify your answer.
Points $A(4,3), B(6,4), C(5,-6)$ and $D(-3,5)$ are the vertices of a parallelogram.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(B) False.
To determine if the points form a parallelogram,we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$AB = \sqrt{(6-4)^2 + (4-3)^2} = \sqrt{2^2 + 1^2} = \sqrt{5}$
$BC = \sqrt{(5-6)^2 + (-6-4)^2} = \sqrt{(-1)^2 + (-10)^2} = \sqrt{1 + 100} = \sqrt{101}$
$CD = \sqrt{(-3-5)^2 + (5 - (-6))^2} = \sqrt{(-8)^2 + 11^2} = \sqrt{64 + 121} = \sqrt{185}$
$DA = \sqrt{(4 - (-3))^2 + (3-5)^2} = \sqrt{7^2 + (-2)^2} = \sqrt{49 + 4} = \sqrt{53}$
In a parallelogram,opposite sides must be equal ($AB = CD$ and $BC = DA$). Since all sides are unequal,the given points do not form a parallelogram.

Explore More

Similar Questions

Find the centre of a circle passing through the points $(6, -6), (3, -7)$ and $(3, 3)$.

Difficult
View Solution

Find $k$ if $(1, 7), (2, 4)$ and $(k, 5)$ form a right triangle.

Difficult
View Solution

If $Q(0, 1)$ is equidistant from $P(5, -3)$ and $R(x, 6)$,find the values of $x$. Also,find the distances $QR$ and $PR$.

Name the type of triangle $PQR$ formed by the points $P(\sqrt{2}, \sqrt{2})$,$Q(-\sqrt{2}, -\sqrt{2})$,and $R(-\sqrt{6}, \sqrt{6})$.

If the vertices of $\square ABCD$ are $A(1, 3), B(4, 3), C(4, 5)$ and $D(1, 5)$,then $\square ABCD$ is a $\ldots$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo