The expression Start Absolute Value negative 2 minus left-parenthesis negative 3 right-parenthesis End Absolute Value can be used to find the distance between point A and which other point?
A coordinate grid is shown with five plotted points.Both the x- and y-axes are between negative 5 and 5. Each interval mark represents one unit.
· Point A is located two units to the left of the y-axis and two units above the x-axis.
· Point B is located three units to the right of the y-axis and two units above the x-axis.
· Point C is located three units to the left of the y-axis and one unit above the x-axis.
· Point D is located three units to the left of the y-axis and two units above the x-axis.
· Point E is located two units to the right of the y-axis and three units above the x-axis.
A. point B
B. point C
C. point D
D. point E
The expression Start Absolute Value negative 2 minus left-parenthesis negative 3 right-parenthesis End Absolute Value represents the distance between point A, which is located at (-2,2), and the point located at (-3,y), where y is unknown. To find this point, we can use the distance formula, which is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Plugging in the values for point A and the unknown point, we get:
$\sqrt{(-3-(-2))^2+(y-2)^2}$
Simplifying this gives:
$\sqrt{1+(y-2)^2}$
We now have the expression Start Absolute Value negative 2 minus left-parenthesis negative 3 right-parenthesis End Absolute Value, which is equivalent to $\sqrt{1+(y-2)^2}$. To find the point that is at this distance from A, we need to find the point that satisfies this equation. Taking the absolute value of the difference between the x-coordinates of A and the unknown point gives us:
$|-2-(-3)| = 1$
Therefore, the point we are looking for must be 1 unit to the left of point A. Looking at the options, we can see that the only point that meets this criteria is point C, which is located at (-3,3). Checking the distance between A and C using the distance formula confirms that the expression Start Absolute Value negative 2 minus left-parenthesis negative 3 right-parenthesis End Absolute Value is indeed equivalent to the distance between A and C:
$\sqrt{(-3-(-2))^2+(3-2)^2} = \sqrt{1+1} = \sqrt{2}$
Therefore, the answer is B. point C.
Which of the following points are more than 5 vertical units away from the point left-parenthesis 0 comma negative 2 right-parenthesis? Choose all that apply.
A. left-parenthesis 6 comma negative 2 right-parenthesis
B. left-parenthesis negative 8 comma negative 2 right-parenthesis
C. left-parenthesis 0 comma negative 8 right-parenthesis
D. left-parenthesis 0 comma 4 right-parenthesis
The point (0,-2) is located at 2 units below the origin on the y-axis. To determine which points are more than 5 vertical units away from this point, we need to calculate the vertical distance between each point and the point (0,-2).
A. The point (6,-2) is located to the right of the origin and at the same height as (0,-2). Therefore, it is not more than 5 vertical units away from (0,-2).
B. The point (-8,-2) is located to the left of the origin and at the same height as (0,-2). Therefore, it is not more than 5 vertical units away from (0,-2).
C. The point (0,-8) is located directly below (0,-2). Therefore, it is more than 5 vertical units away from (0,-2).
D. The point (0,4) is located directly above (0,-2). Therefore, it is more than 5 vertical units away from (0,-2).
Therefore, the points that are more than 5 vertical units away from the point (0,-2) are C and D. The answer is (C) and (D).
Name the shape that will result from connecting the following points: left-parenthesis negative 4 comma 1 right-parenthesis, left-parenthesis negative 4 comma negative 4 right-parenthesis, left-parenthesis 0 comma 3 right-parenthesis, and left-parenthesis 0 comma negative 6 right-parenthesis.
A. square
B. rectangle
C. trapezoid
D. quadrilateral
A quadrilateral will result from connecting the given points since a quadrilateral is any four-sided polygon. None of the given options are more specific types of quadrilaterals, so the answer is (D) quadrilateral.
A square has a perimeter of 12 units. One vertex is at the point left-parenthesis negative 1 comma 1 right-parenthesis, and another vertex is at the point left-parenthesis 2 comma 4 right-parenthesis. Which of the following points could be another vertex?
A. left-parenthesis 1 comma 2 right-parenthesis
B. left-parenthesis 2 comma 1 right-parenthesis
C. left-parenthesis 1 comma negative 2 right-parenthesis
D. left-parenthesis 2 comma negative 1 right-parenthesis
The distance between the two given vertices of the square can be found using the distance formula as follows:
$\sqrt{(2-(-1))^2+(4-1)^2} = \sqrt{3^2+3^2} = 3\sqrt{2}$
Since a square has four sides of equal length, the length of each side of the square is $\frac{1}{4}$ of the perimeter, which is $\frac{1}{4}(12)=3$ units.
We can use this information to determine whether each of the given points could be another vertex of the square. We can calculate the distance between each given point and one of the known vertices, (2,4), using the distance formula.
A. $\sqrt{(1-2)^2+(2-4)^2} = \sqrt{2^2+(-2)^2} = 2\sqrt{2} \neq 3$. Therefore, this point cannot be another vertex of the square.
B. $\sqrt{(2-2)^2+(1-4)^2} = \sqrt{(-3)^2+(-3)^2} = 3\sqrt{2}$. Therefore, this point could be another vertex of the square.
C. $\sqrt{(1-2)^2+(-2-4)^2} = \sqrt{1^2+(-6)^2} = \sqrt{37} \neq 3$. Therefore, this point cannot be another vertex of the square.
D. $\sqrt{(2-(-1))^2+(-1-4)^2} = \sqrt{3^2+(-5)^2} = \sqrt{34} \neq 3$. Therefore, this point cannot be another vertex of the square.
Therefore, the answer is (B) left-parenthesis 2 comma 1 right-parenthesis.