Question

show that the points (-3,1),(-6,-7),(3,-9),(6,-1) take n in order form a parallelogram.​

Answer 1

Topic - Mid point formula

Explanation:

Here in this question, we need to prove that the given coordinates if taken in order are the coordinates of a parallelogram.

We know that diagonals of a quadrilateral bisects each other i.e. the mid point of diagonals will meet at a certain point.

This implies that the mid point of (-3, 1) and (3, -9) will be equal to the mid point of (-6, -7) and (6, -1) respectively.

Mid point formula is given by,

$\boxed{\sf (x, y) = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right)}$

Here $\sf(x, y)$ is the coordinate of mid point of $\sf(x_1, y_1)$ and $\sf(x_2, y_2)$.

Equating the mid point of $(-3, 1)$ and $(3, -9)$ with the mid point of $(-6, -7)$ and $(6, -1)$.

${ \implies \left(\dfrac{ - 3 + 3}{2}, \dfrac{1 - 9}{2} \right)= \left( \dfrac{ - 6 + 6}{2} ,\dfrac{ - 7 - 1}{2} \right)}$

${ \implies \left(\dfrac{0}{2}, \dfrac{ - 8}{2} \right)= \left( \dfrac{0}{2} ,\dfrac{ - 8}{2} \right)}$

${ \implies \left(0, - 4 \right)= \left( 0 , - 4 \right)}$

Hence the required result is proved.