Question

If the points $( - 2, - 1),(1,0),(x,3){\text{ and (1,y) }}$ form a parallelogram, find the values of $x$ and $y.$

Answer 1

Hint: In this question first we have to let the coordinates name and then using distance formula and properties of parallelogram we have to find the relationship between the coordinates and then solve them to get the values of $x$ and $y.$

Complete step-by-step solution -
Let the points of the given parallelogram be $A( - 2, - 1),B(1,0),C(x,3){\text{ and D(1,y)}}$.

We know, distance formula=$\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
Consider AB side of parallelogram ABCD
$
   \Rightarrow AB = \sqrt {{{(1 + 2)}^2} + {{(0 + 1)}^2}} \\
   \Rightarrow AB = \sqrt {10} $ ………………………….eq.(1)
 Consider CD side of parallelogram ABCD
$ \Rightarrow CD = \sqrt {{{(1 - x)}^2} + {{(y - 3)}^2}} $......................... eq.2
We know, opposite sides of a parallelogram are equal.
Then CD=AB
Now on squaring on both side, we get
$
   \Rightarrow C{D^2} = A{B^2} \\
   \Rightarrow 1 + {x^2} - 2x + {y^2} + 9 - 6y = 10{\text{ \{ from eq}}{\text{.1, eq}}{\text{.2\} }} \\
   \Rightarrow {x^2} + {y^2} - 2x - 6y = 0 ………………………\text{eq.(3)}
$

Now consider BC side of parallelogram ABCD
$ \Rightarrow BC = \sqrt {{{(x - 1)}^2} + {3^2}} $ ………………..eq.(4)
Now, consider AD side of parallelogram ABCD
$ \Rightarrow AD = \sqrt {{{(1 + 2)}^2} + {{(y + 1)}^2}} $ ……………….eq.5
We know, opposite sides of a parallelogram are equal.
Then BC=AD
Now on squaring on both side , we get
$
   \Rightarrow B{C^2} = A{D^2} \\
   \Rightarrow {x^2} + 2x + 10 = 9 + {y^2} + 1 + 2y \\
   \Rightarrow {x^2} - 2x = {y^2} + 2y{\text{ }} \\
   \Rightarrow {x^2} - 2x - {y^2} - 2y = 0 $ …………………..eq.(6)
On subtracting eq. 6 from eq.3 we get

  $ \Rightarrow 2{y^2} - 4y = 0 \\$
   $\Rightarrow y(y - 2) = 0 \\$
   $\Rightarrow y = 0{\text{ or } } 2 $ …………………….. eq.(7)

Since , y-coordinate of B is already zero, we reject this value
 So $y = 2$
On adding eq.3 and eq.6 we get
$ \Rightarrow 2{x^2} - 4x - 8y = 0$
Now putting $y = 2$ in above equation we get
$
   \Rightarrow 2{x^2} - 4x - 16 = 0 \\
   \Rightarrow {x^2} - 2x - 8 = 0 \\
   \Rightarrow (x - 4)(x + 2) = 0 \\
   \Rightarrow x = 4{\text{ or }} - 2{\text{ …………………….. eq}}{\text{.8}} \\
$
Since, x-coordinate of A is already -2, we reject this value.
$\therefore x = 4$
Hence, $x = 4{\text{ and y = 2}}$

Note: Whenever you get this type of question the key concept to solve this is to learn about different shapes and their properties like in this question we require the property of parallelogram in which length of opposite sides are equal and about how to apply distance formula($\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $) on given coordinates.