Question

the three vertices of a parallelogram taken in order (-1,0) (3,1) and (2,2) respectively find the coordinates of the fourth vertex​

Answer 1

Given:-

• Three vertices of a parallelogram taken in order (-1,0) (3,1) and (2,2) respectively.

To find:-

• The coordinates of the fourth vertex.

Solution:-

$\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(6,1)\qbezier(1,1)(1,1)(1.6,4)\qbezier(1.6,4)(1.6,4)(6.6,4)\qbezier(6,1)(6,1)(6.6,4)\qbezier(6.6,4)(6.6,4)(1,1)\qbezier(1.6,4)(1.6,4)(6,1)\put(0.7,0.5){\sf A(-1,0)}\put(6,0.5){\sf B(3,1)}\put(1.4,4.3){\sf D(x,y)}\put(6.6,4.3){\sf C(2,2)}\end{picture}$

❥ The diagonals of a parallelogram bisect each other.

Hence,

Coordinate of midpoint of AC = Coordinate of midpoint of BD.

We know :-

Mid point formula ,

$\mapsto \orange{\boxed {\sf{(\frac{x_1 + x_2}{2})},(\frac{y_1 + y_2}{2})}}$

Now ,

Coordinate of midpoint of AC = Coordinate of midpoint of BD.

=>( -1+2/2 , 0+2/2 )= (3+x/2 ,1+y/2)

=> (-1/2 , 1) = (3+x/2 , 1+y/2 )

=> 3+x/2 = 1/2   & y+1/4=1

=> x = -2  & y = 1.

Hence, coordinates of fourth vertex are -2,1 .

___________________

Answer 2

Answer:

Solution:-

$\rm \frac{x + 3}{2} = \frac{1}{x} \\ \rm \: x = - 2 \\ \rm \frac{y + 1}{2} \\ \rm \: y = 1 \\ \\ \rm \: The \: fourth \: vertex \: is \: (-2,1)$