Question

Three consecutive vertices of a parallelogram are A(1.2) B(2.3) and C(8,5). Find the fourth vertex.


ANSWER THE QUESTION I WILL MARK THEM AS BRAINLIST​

Answer 1

Three consecutive of a parallelogram A(1,2),B(2,3) and C(8,5)
∴ From the graph
the fourth vertex =(7,4)

Answer 2

★Given:-

• Three consecutive vertices of a parallelogram are A(1.2) B(2.3) and C(8,5).

★To find:-

• The fourth vertex.

★Solution:-

Let  A(1,2) B(2,3) and C(8,5) and D(x,y) be the given points taken in order.

✦In a parallelogram,Diagonals bisect each other.

Hence,

Coordinates of mid-point of AC = Coordinates of midpoint of BD

We know:

Mid point formula,

$\leadsto \underline{\boxed{\sf (x,y) = (\frac{x_1 +x_2}{2},\frac{y_1 + y_2 }{2} )}}$

Where,

(x,y) = coordinates of midpoint

(x1,y1)=coordinates of first point

(x2,y2)=coordinates of second point

Now,

Substituting given values in the formula,

Coordinates of mid-point of AC = Coordinates of midpoint of BD

$\mapsto \sf (\dfrac{8+1}{2} ,\dfrac{5+2}{2} )=(\dfrac{x+2}{2} ,\dfrac{3+y}{2} )\\\\\\\mapsto \sf (\dfrac{9}{2} ,\dfrac{7}{2} )=(\dfrac{x+2}{2} ,\dfrac{3+y}{2} )$

Therefore,

$\\ \mapsto \sf \dfrac{x+2}{\cancel{2}} = \dfrac{9}{\cancel{2}} \\\\\mapsto \sf x +2=9\\\\\mapsto \sf x=9-2\\\\\mapsto \boxed{\bold{x=7}}$

And,

$\\ \mapsto \sf \dfrac{3+y}{\cancel{2}} =\dfrac{7}{\cancel{2}} \\\\\\\mapsto \sf 3+y=7\\\\\mapsto \sf y = 7-3\\\\\mapsto \boxed{\bold{y=4.}}$

Hence,

The fourth vertex is (7,4)

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