Math, asked by 1098439, 1 month ago

if A (5,0) B (-5,5) C (5,5) FIND D if abcd is a rectangle i will mark u steps​

Answers

Answered by tennetiraj86
2

Step-by-step explanation:

Given :-

ABCD is a rectangle whose vertices are A (5,0), B (-5,5) , C (5,5)

To find :-

Find the fourth vertex D ?

Solution :-

Given that

ABCD is a rectangle whose vertices are A (5,0), B (-5,5) , C (5,5)

Let the coordinates of D be (x,y)

We know that

Opposite sides are equal in a rectangle.

So, AB = CD and BC = DA

Length of AB :-

Let (x1, y1) = (5,0) => x1 = 5 and y1 = 0

Let (x2, y2) = (-5,5) => x2 = -5 and y2 = 5

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)²+(y2-y1)²] units

The distance between A and B

=> AB =√[(-5-5)²+(5-0)²]

=> AB = √[(-10)²+5²]

=> AB = √(100+25)

=> AB = √125

=> AB = √(5×5×5)

=> AB = 5√5 units

Length of BC :-

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

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

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)²+(y2-y1)²] units

The distance between A and B

=> BC =√[(5+5)²+(5-5)²]

=> BC = √[(10)²+0²]

=> BC = √(100+0)

=> BC = √100

=> BC = 10 units

Length of CD :-

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

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

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)²+(y2-y1)²] units

The distance between A and B

=> CD =√[(x-5)²+(y-5)²]

=> CD = √[x²-10x+25+y²-10y+25]

=> CD = √(x²-10x-10y+y²+50) units

Length of DA :-

Let (x1, y1) = (x,y) => x1 = x and y1 = y

Let (x2, y2) = (5,0) => x2 = 5 and y2 = 0

We know that

The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)²+(y2-y1)²] units

The distance between A and B

=> DA =√[(5-x)²+(0-y)²]

=> DA =√[(5-x)²+(-y)²]

=> DA = √[25-10x+x²+y²] units

Now,

AB = CD

=> 5√5 = √(x²-10x-10y+y²+50)

On squaring both sides then

=> (5√5)² = (√(x²-10x-10y+y²+50))²

=> 125 = x²-10x-10y+y²+50

=> x²-10x-10y+y² = 125-50

=> x²-10x-10y+y² = 75 -----------(1)

and BC = DA

=> 10= √[25-10x+x²+y²]

On squaring both sides then

=> 10² = (√[25-10x+x²+y²])²

=> 100 = 25-10x+x²+y²

=> x²+y²-10x = 100-25

=> x²+y²-10x = 75 -----------------(2)

From (1)&(2)

x²-10x-10y+y² = x²+y²-10x

=> -10y = 0

=> y = 0/-10

=> y = 0

On substituting the value of y in (2)

=> x²+0²-10x = 75

=> x²-10x = 75

=> x²-10x-75 = 0

=> x²+5x-15x-75 = 0

=> x(x+5)-15(x+5) = 0

=> (x+5)(x-15) = 0

=> x+5 = 0 or x-15 = 0

=> x = -5 or x = 15

The point = (-5,0) or (15,0)

Answer :-

The fourth vertex is D(-5,0) or D(15,0)

Used formulae:-

→ The distance between two points (x1, y1) and

(x2, y2) is √[(x2-x1)²+(y2-y1)²] units

→ Opposite sides are equal in a rectangle.

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