Question

$\pink{ TOO \: HARD}$

$\red{CHALLENGE}$


Show that the points (-2, -1) (1, 0) (4, 3) and (1, 2) taken in order and vertices of parallelogram?

$‎\sf Distance \ Formula = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}‎‎$
‎

Solve it with the help of formula​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that, the given vertices as

Coordinates of A (- 2, - 1)

Coordinates of B (1, 0)

Coordinates of C (4, 3)

Coordinates of D (1, 2).

In order, to show that, ABCD are the vertices of parallelogram using distance formula, we have to show that, pair of opposite sides are equal. i.e. AB = CD and BC = AD.

We know,

Distance between two Coordinates A and B is given as

$\green{ \boxed{ \sf{ \:\rm \: AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}}}$

So, Consider,

$\rm :\longmapsto\:AB = \sqrt{ {(1 + 2)}^{2} + {(0 + 1)}^{2}} = \sqrt{9 + 1} = \sqrt{10}$

$\rm :\longmapsto\:BC = \sqrt{ {(4 - 1)}^{2} + {(3 - 0)}^{2} } = \sqrt{9 + 9} = \sqrt{18}$

$\rm :\longmapsto\:CD = \sqrt{ {(1 - 4)}^{2} + {(1 - 2)}^{2}} = \sqrt{9 + 1} = \sqrt{10}$

$\rm :\longmapsto\:AD = \sqrt{ {(2 + 1)}^{2} + {(1 + 2)}^{2} } = \sqrt{9 + 9} = \sqrt{18}$

$\rm :\implies\:AB = CD = \sqrt{10} \: \: and \: \: BC = AD = \sqrt{18}$

So, In a quadrilateral, if opposite pair of sides are equal, then quadrilateral is a parallelogram.

Hence,

$\bf\implies \:A,B,C,D \: are \: the \: vertices \: of \: parallelogram$

Additional Information :-

1. If we have to show that 4 points A, B, C, D are the vertices of rectangle, we have to show that

$\red{ \boxed{ \sf{ \:AB = CD, \: BC = AD, \: AC = BD}}}$

2. If we have to show that 4 points A, B, C, D are the vertices of square, we have to show that

$\red{ \boxed{ \sf{ \:AB = CD = BC = AD \: and \: AC = BD}}}$

3. If we have to show that 4 points A, B, C, D are the vertices of rhombus, we have to show that

$\red{ \boxed{ \sf{ \:AB = CD = BC = AD \: and \: AC \ne BD}}}$