Question

Nikki drew a rectangle with a perimeter of 18 units on a coordinate plane. Two of the vertices were (4,-3) and (-1,-3). Which could be coordinates of the other two vertices of the rectangle?

Answer 1

(4 , - 7)  ( -1 , - 7)   or   (4 , 1)  ( -1 , 1)  could be coordinates of the other two vertices of the rectangle

Step-by-step explanation:

options should be given to find the answer

as multiple co-ordinate possible

Assuming Two of the vertices given (4,-3) and (-1,-3) are one side of a rectangle

then length of side = 5

Perimeter = 18

So length of other side = 4     as Perimeter = 2 * ( 5 + 4) = 18

as Length of other side = 4

so Possible co-ordinates are

(4 , - 7)  ( -1 , - 7)   or   (4 , 1)  ( -1 , 1)

but if we consider  (4,-3) and (-1,-3) as one of diagonal then so many possibilities

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Answer 2

The coordinates of other vertices of rectangle (4,1) and (-1,1) or (4,-7)and (-1,-7)

Step-by-step explanation:

Perimeter of rectangle=18 units

Length of side which passing through the points (4,-3) and (-1,-3)

Length of side=$\sqrt{(4+1)^2+(-3+3)^2}=5 units$

By using the distance formula:$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Perimeter of rectangle=$2(l+b)$

Where l=Length of rectangle

b=Breadth of rectangle

Substitute the values then we get

$18=2(5+b)$

$\frac{18}{2}=5+b$

$5+b=9$

$b=9-5=4$

Breadth of rectangle=4 units

Let  coordinates  of other two  vertices of rectangle are $(4,y_1)\;and\;(-1,y_1}$

Using again distance formula

$4=\sqrt{(-1+1)^2+(-3-y_1)^2$

Squaring on both sides then we get

$16=(y_1+3^2$

$(y_1+3)^2=(4)^2$

$y_1+3\pm 4$

$y_1+3=4$

$y_1=4-3=1$

$y_1+3=-4$

$y_1=-4-3=-7$

Therefore, the  coordinates of other two  vertices of rectangle (4,1) and (-1,1) or (4,-7)and (-1,-7)

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