Question

2 The three vertices of a parallelogram are (1,1), (5,5) and (5,8). Find the fourth vertex
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Answer 1

Step-by-step explanation:

Given :-

The three vertices of a parallelogram are (1,1), (5,5) and (5,8).

To find :-

Find the fourth vertex?

Solution :-

Given that :

The three vertices of a parallelogram are (1,1), (5,5) and (5,8).

Let A = (1,1)

Let B = (5,5)

Let C = (5,8)

Let the fourth vertex be D(x,y)

We know that

In a Parallelogram ABCD , the diagonals AC and BD are bisect to each other

=> Midpointbof AC = Mid point of BD

Finding Mid Point of AC:-

Let (x1, y1)=A(1,1)=>x1=1 and y1 = 1

Let (x2, y2)=C(5,8)=>x2=5 and y2 = 8

We know that

The mid point of the linesegment joining the points (x1, y1) and (x2,y2) is ({x1+x2}/2 ,{y1+y2}/2)

=>Mid point of AC=({1+5}/2 ,{1+8}/2)

=> Mid point of AC =( 6/2 , 9/2)

Mid Point of AC = (3,9/2)-------(1)

Finding Mid Point of BD:-

Let (x1, y1)=B(5,5)=>x1=5 and y1 = 5

Let (x2, y2)=D(x,y)=>x2=x and y2 = y

We know that

The mid point of the linesegment joining the points (x1, y1) and (x2,y2) is ({x1+x2}/2 ,{y1+y2}/2)

=>Mid point of BD=({5+x}/2 ,{5+y}/2)----(2)

We have

Midpointbof AC = Mid point of BD

=> (1) = (2)

=> (3,9/2) = ({5+x}/2 ,{5+y}/2)

On Comparing both sides then

=> (5+x)/2 = 3 and (5+y)/2 = 9/2

=> 5+x= 3×2

=> 5+x = 6

=> x = 6-5

=> x = 1

and

(5+y)/2 = 9/2

=>5+y = 9

=>y= 9-5

=> y = 4

Therefore, x = 1 and y = 4

(x,y) = (1,4)

Answer :-

The fourth vertex for the given problem is (1,4)

Used Concept :-

In a Parallelogram ABCD , the diagonals AC and BD are bisect to each other

=> Midpointbof AC = Mid point of BD

Used formulae:-

The mid point of the linesegment joining the points (x1, y1) and (x2,y2) is ({x1+x2}/2 ,{y1+y2}/2)

Points to know :-

In a Parallelogram ,

• The opposite sides are parallel and equal.

• The adjacent angles are supplementary.

• The opposite angles are equal.

• The diagonals bisect each other.