Question

show that (-3,-4),(12,5),(14,12),(-1,3) are the vertices of the parallelogram ​

Answer 1

Answer:-

Let , A = ( -3 , -4)

B = ( 12 , 5)

C = ( 14 , 12)

D = ( -1 , 3 )

We know that, Opposite sides of a parallelogram are equal.

Hence, AB = CD & BC = DA.

We know that, distance between two points is,

$\sqrt{ (x_{2} - x _{1})^{2} + {(y _{2} - y _{1}) }^{2} }$

AB = CD,

$\sqrt{ {(12 - ( - 3))}^{2} + (5 - ( - 4)) ^{2} } = \sqrt{ {( - 1 - 14)}^{2} + ( {3 - 12)}^{2} } \\ \\ \sqrt{ {(15)}^{2} + (9) ^{2} } = \sqrt{ {( - 15)}^{2} + ( { - 9)}^{2} } \\ \\ \sqrt{225 + 81} = \sqrt{225 + 81} \\ \\ \sqrt{306} = \sqrt{306} \\ \\ 3 \sqrt{34} = 3 \sqrt{34}$

Hence, AB = CD.

Now, BC = DA

$\sqrt{ {(14 - 12)}^{2} + {(12 - 5)}^{2} } = \sqrt{ {( - 1 + 3)}^{2} + {(3 + 4)}^{2} } \\ \\ \sqrt{ {2}^{2} + {7}^{2} } = \sqrt{ {2}^{2} + {7}^{2} } \\ \\ \sqrt{4 + 49} = \sqrt{4 + 49} \\ \\ \sqrt{53} = \sqrt{53}$

Hence, BC = DA.

As, opposite sides are equal, the given points are vertices of parallelogram.