Question

Prove that (-5,-3), (1,-11), (7,-6), (1, 2) coordinates are the vertices of parallelograms.

Answer 1

Hint: Here in this question basic properties of parallelograms have to be used and distance formula is being used which are mentioned below: -
Property of a parallelogram is as follow:-
Opposite sides of a parallelogram are parallel and equal in length.
Distance formula: -We will use distance formula between the two points ${x_1}, {y_1}$ and ${x_2}, {y_2}$ that is mentioned below: -
$d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $ d= distance between two points.

Complete step-by-step answer:
Let the four given points be named as A (-5,-3), B (1,-11), C (7,-6), D (1, 2). So in order to prove these points as parallelograms we have to prove opposite sides equal in length.
To prove: - AB=CD and BC=AD
Proof: - We will first find length of AB
Points for AB are A (-5,-3) and B (1,-11)
$ \Rightarrow AB = \sqrt {{{[1 + 5]}^2} + {{[ - 11 + 3]}^2}} $ (Putting values in distance formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $ )
$ \Rightarrow AB = \sqrt {{{[6]}^2} + {{[ - 8]}^2}} $
$ \Rightarrow AB = \sqrt {36 + 64} $
$ \Rightarrow AB = \sqrt {100} $
$\therefore AB = 10$
 Points for BC are B (1,-11) and C (7,-6)
$ \Rightarrow BC = \sqrt {{{[7 - 1]}^2} + {{[ - 6 + 11]}^2}} $ (Putting values in distance formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $ )
$ \Rightarrow BC = \sqrt {{{[6]}^2} + {{[5]}^2}} $
$ \Rightarrow BC = \sqrt {36 + 25} $
$ \Rightarrow BC = \sqrt {61} $ (Finding Square root of 61)
$\therefore BC = 7.81$
 Points for CD are C (7,-6) and D (1, 2)
$ \Rightarrow CD = \sqrt {{{[1 - 7]}^2} + {{[2 + 6]}^2}} $ (Putting values in distance formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $ )
$ \Rightarrow CD = \sqrt {{{[ - 6]}^2} + {{[8]}^2}} $
$ \Rightarrow CD = \sqrt {36 + 64} $
$ \Rightarrow CD = \sqrt {100} $ (Finding Square root of 100)
$\therefore CD = 10$
 Points for AD are A (-5,-3) and D (1, 2)
$ \Rightarrow AD = \sqrt {{{[1 + 5]}^2} + {{[2 + 3]}^2}} $ (Putting values in distance formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $ )
$ \Rightarrow AD = \sqrt {{{[6]}^2} + {{[5]}^2}} $
$ \Rightarrow AD = \sqrt {36 + 25} $
$ \Rightarrow AD = \sqrt {61} $ (Finding Square root of 61)
$\therefore AD = 7.81$
As we can see that AB=CD=10 and BC=AD=7.81 therefore the points given in the question which are A (-5,-3), B (1,-11), C (7,-6), D (1, 2) are of parallelogram.

Note: Students must apply distance formula carefully as the common mistake which is done by most of the students is that they get confused between subtraction sign instead of addition sign in distance formula.