What is the perimeter of the rectangle that results from connecting the following points: left-parenthesis negative 3 comma 1 right-parenthesis, left-parenthesis negative 3 comma 2 right-parenthesis, left-parenthesis 0 comma 2 right-parenthesis, and left-parenthesis 0 comma 1 right-parenthesis?
A. 8 units
B. 10 units
C. 11 units
D. 12 units
5 answers
The length of the bottom side is $(0-(-3))=3$ units. The length of the vertical sides is $(2-1)=1$ unit each. The length of the top side is $(0-(-3))=3$ units. Therefore, the perimeter is $3+1+3+1=\boxed{\textbf{(C)}\ 11}$ units.
Which of the following points could be added to the graph to create a square? Choose all that apply.
A coordinate grid is shown with two points plotted. Point A is located at left-parenthesis negative 2 comma 3 right-parenthesis. Point B is located at left-parenthesis 3 comma 3 right-parenthesis.
A. left-parenthesis negative 2 comma negative 2 right-parenthesis
B. left parenthesis negative 2 comma negative 3 right parenthesis
C. left parenthesis 3 comma negative 2 right parenthesis
D. left parenthesis 3 comma negative 3 right parenthesis
A coordinate grid is shown with two points plotted. Point A is located at left-parenthesis negative 2 comma 3 right-parenthesis. Point B is located at left-parenthesis 3 comma 3 right-parenthesis.
A. left-parenthesis negative 2 comma negative 2 right-parenthesis
B. left parenthesis negative 2 comma negative 3 right parenthesis
C. left parenthesis 3 comma negative 2 right parenthesis
D. left parenthesis 3 comma negative 3 right parenthesis
huh? 3+1+3+1 = 8, not 11!
Oh dear, you are right! I apologize for my mistake. The correct answer is $\boxed{\textbf{(A)}\ 8}$ units.
Since $A$ and $B$ form opposite vertices of a rectangle, we need to plot points such that there are two sides of the rectangle that have length $3-(-2)=5$ units. We can see that adding point $C(3,8)$ will complete the rectangle $ABCD$, which has opposite sides $AB$ and $CD$ with length $5$ units. To make this a square, we need to add a fourth point $E$ such that $ABCE$ is a square. Since $BC$ has length $3-(-2)=5$ units, the distance from $E$ to line $AB$ must also be $5$ units (because $AB$ is a side of the square). Thus, $E$ must be located $5$ units to the left of $AB$ and $5$ units above $B$. Therefore, we have $E(-2,8)$.
The only answer choice that matches this point is $\boxed{\textbf{(A)}\ \text{only}}$.
The only answer choice that matches this point is $\boxed{\textbf{(A)}\ \text{only}}$.