Question

The points (0,0) and (3,2) are two vertices of a polygon with integer coordinates What are the answers 8-10?

Answer 1

For a square, you can use the points, (1, 5) and (–2, 3). To find these, think about how you move from point (0, 0) to (3, 2). Rise = 2 and run = 3 for a slope of 2/3. The perpendicular line segments will have a slope of –3/2 or rise = 3 and run = –2. Starting at (0, 0) move up 3 and left 2 to get the point (–2, 3). Do the same movements from (3, 2) to get the point (1, 5).For a rectangle you could use the points (3, 0) and (0, 2). To make it a NON-rectangular parallelogram, just scoot each of those points over a bit in opposite directions ... from (3, 0) to (4, 0) and from (0, 2) to (–1, 2).For a right triangle, just use either (3, 0) or (0, 2). Combined with the other 2 given points, this will form a right triangle

Answer 2

Answer: The required answers are

(8) C(5, -1) and D(2, -3),

(9) C(0, -2) and D(3, 0)

(10) C(3, 0).

Step-by-step explanation:  Given that the points A(0, 0) and B(3, 2) are the vertices of a polygon with integers co-ordinates.

(8) We are to find the other two vertices if the polygon is a square.

We know that

a square has the length of all the four sides equal and the adjacent sides are perpendicular to each other.

From the figure (1) attached below, we can see that if the co-ordinates of the other two vertices are C(5, -1) and D(2, -3), then

the length of all the four sides AB, BC, CD and DA can be calculate using distance formula as follows :

$AB=\sqrt{(0-3)^2+(0-2)^2}=\sqrt{13},\\\\BC=\sqrt{(3-5)^2+(2+1)^2}=\sqrt{13},\\\\CD=\sqrt{(2-5)^2+(-3+1)^2}=\sqrt{13},\\\\DA=\sqrt{(2-0)^2+(-3-0)^2}=\sqrt{13}.$

Also, the slopes of AB, BC, CD and DA are

$\textup{slope of AB}=\dfrac{2-0}{3-0}=\dfrac{2}{3},\\\\\\\textup{slope of BC}=\dfrac{-1-2}{5-3}=-\dfrac{3}{2},\\\\\\\textup{slope of CD}=\dfrac{-1+3}{5-2}=\dfrac{2}{3},\\\\\\\textup{slope of DA}=\dfrac{-3-0}{2-0}=-\dfrac{3}{2}.$

So, the product of the slopes of any two adjacent sides is -1. That is, all the sides are perpendicular to the sides adjacent to them and the lengths of all sides are equal.

Hence, ABCD is a square.

(9) We are to find the other two vertices if the polygon is a non rectangular parallelogram.

We know that

the length of the opposite sides of a non rectangular parallelogram are equal and the adjacent sides are not perpendicular to each other.

We see from figure 2 attached below that if the other two vertices are C(0, -2) and D(3, 0), then

$AB=\sqrt{(0-3)^2+(0-2)^2}=\sqrt{13},\\\\BC=\sqrt{(3-3)^2+(2-0)^2}=\sqrt{4}=2,\\\\CD=\sqrt{(3-0)^2+(0+2)^2}=\sqrt{13},\\\\DA=\sqrt{(0-0)^2+(-2-0)^2}=\sqrt{4}=2.$

Also, the slopes of AB, BC, CD and DA are

$\textup{slope of AB}=\dfrac{2-0}{3-0}=\dfrac{2}{3},\\\\\\\textup{slope of BC}=\dfrac{0-2}{3-3}=\infty,\\\\\\\textup{slope of CD}=\dfrac{-2-0}{0-3}=\dfrac{2}{3},\\\\\\\textup{slope of DA}=\dfrac{-2-0}{0-0}=\infty.$

So, the slopes of opposite sides are equal and the product of the slopes of adjacent sides are not -1.

That is, the polygon ABCD is a non rectangular parallelogram.

(10) We are to find the other vertex if the polygon is a right triangle.

We know that

in a right-angled triangle, the sum of the squares of lengths of smaller sides is equal to the square of the length of the largest side (hypotenuse).

We see in figure 3 attached below, if the third vertex is C(3, 0), then

$AB^2=(3-0)^2+(2-0)^2=13,\\\\BC^2=(3-3)^2+(2-0)^2=4,\\\\CA=(3-0)^2+(0-0)^2=9\\\\\Rightarrow AB^2=BC^2+CA^2.$

So, ABC is a right-angled triangle.

Thus, the required answers are

(8) C(5, -1) and D(2, -3),

(9) C(0, -2) and D(3, 0)

(10) C(3, 0).