Practice 2 Solutions

Find the distance between $(3.6, - 9.1), (8.2, - 7.4)$ . Round to the nearest hundredth.

$d = \sqrt{{(3.6 - 8.2)}^{2} + {(- 9.1 - (- 7.4))}^{2}}$

$d = 4.90$

Find the midpoint between $(3.6, - 9.1), (8.2, - 7.4)$ .

$(\frac{3.6 + 8.2}{2}, \frac{- 9.1 + - 7.4}{2})$

$(5.9, - 8.25)$

Find the midpoint between $(7, 14)$ and $(19, 3)$ .

$(\frac{7 + 19}{2}, \frac{14 + 3}{2})$

$(13, 8.5)$

For problems 4–6, use the given graph of a miniature golf course hole.

What is the distance between the hole in quadrant 2 and the hole on the axis?

$(- 3, 8) to (3, 0) d = \sqrt{{(- 3 - 3)}^{2} + {(8 - 0)}^{2}}$

10 units

What is the midpoint between the holes in quadrants one and two?

$(1, 5), (- 3, 8) (\frac{1 + - 3}{2}, \frac{5 + 8}{2})$

$(- 1, 6.5)$

How much further would a player using tee #3 have to putt to the hole in quadrant 2 than a player using tee #1?

Tee #1

$1 (- 1, - 6) to (- 3, 8) d_{1} = \sqrt{{(- 1 - (- 3))}^{2} + {(- 6 - 8)}^{2}} d_{1} = 14.14$

Tee #3

$(1, - 6) to (- 3, 8) d_{3} = \sqrt{{(1 - (- 3))}^{2} + {(- 6 - 8)}^{2}} d_{3} = 14.56$

$d_{3} - d_{1} = 14.56 - 14.14$

Sample: The player using tee #3 would have to putt 0.42 units further than a player using tee #1.

The definition of a rhombus is a parallelogram with four equal sides. Susan drew a figure with the points $A (- 3, - 2), B (0, 2), C (3, - 2), D (0, - 5)$ and determined it was a rhombus. Show and explain your work to prove or disprove this claim.

$A B = \sqrt{{(- 3 - 0)}^{2} + {(- 2 - 2)}^{2}} = 5 C D = \sqrt{{(3 - 0)}^{2} + {(- 2 - (- 5))}^{2}} = 4.24$

Sample: Susan is incorrect. A rhombus has 4 equal sides and this figure does not have 4 equal sides.

Marie drew the figure and created a quadrilateral with the points $A (- 3, - 2), B (0, 2), C (3, - 2), D (0, - 6)$ then found the midpoints of diagonals $A C and B D$ . Describe the relationship between the midpoints of $A C and B D$ .

$midpoint of A C (\frac{- 3 + 3}{2}, \frac{- 2 + - 2}{2}) = (0, - 2) midpoint of B D (\frac{0 + 0}{2}, \frac{2 + - 6}{2}) = (0, - 2)$

Sample: The midpoints of the diagonals of the figure are both at $(0, - 2)$ .

Note

Marie drew a rhombus. The diagonals of a rhombus are perpendicular bisectors and can be proven by calculating the slope of each diagonal.

For problems 9–11, use the table.

Diameter of a Circle	Radius of a Circle
Has two endpoints on the circle Goes through the center of the circle	Is the distance from the center to an endpoint on the circle

The endpoints of the diameter of a circle are at $(- 4, - 5)$ and $(- 7, 1)$ . Find the center.

$The center is the midpoint (\frac{- 4 + - 7}{2}, \frac{- 5 + 1}{2}) (- 5.5, - 2)$

The center is at the point $(- 5.5, - 2)$ .

Note

The center is the midpoint of the two endpoints of the diameter

Find the length of the radius. Round to the nearest hundredth.

$radius = distance from (- 5.5, - 2) to (- 4, - 5) r = \sqrt{{(- 5.5 - (- 4))}^{2} + {(- 2 - (- 5))}^{2}}$

The radius is 3.35 units long.

Find the area of the circle using 3.14 for π. Round your answer to the nearest hundredth.

$r = 3.35 A = π r^{2} A = π {(3.35)}^{2} A = 3.14 {(3.35)}^{2}$

The area is $35.24 {units}^{2}$ .

For problems 12–13, use the given graph.

Find the area of the rectangle WXYZ. Round the final solution to a whole number.

$A = l w l = Y Z, w = X Y Y (6, - 1), Z (- 1, - 8) Y Z = \sqrt{{(6 - (- 1))}^{2} + {(- 1 - (- 8))}^{2}} = 9.90 X (1, 4), Y (6, - 1) X Y = \sqrt{{(1 - 6)}^{2} + {(4 - (- 1))}^{2}} = 7.07 A = (9.90) (7.07) = 69.99$

$A = 70 {units}^{2}$

Note

The box in the corner of XY and YZ indicates that the lines form a right angle.

Find the perimeter of the rectangle WXYZ. Round to the nearest hundredth.

$P = 2 l + 2 w P = 2 (9.90) + 2 (7.07)$

The perimeter is 33.94 units.

Given the equation $f (x) = {(x + 4)}^{2} - 3$ , determine the distance between the vertex and the y-intercept. Round to the nearest hundredth.

$f (x) = x^{2} + 8 x + 16 - 3 f (x) = x^{2} + 8 x + 13 c = 13 y ‐ int : (0, 13) vertex : (- 4, - 3) d = \sqrt{{(- 4 - 0)}^{2} + {(- 3 - 13)}^{2}}$

$d = 16.49$

Given the equation $g (x) = {(x - 2)}^{2} - 9$ , find the distance between the vertex and one of the roots. Round to the nearest hundredth.

$vertex : (2, - 9) g (x) = x^{2} - 4 x + 4 - 9 g (x) = x^{2} - 4 x - 5 0 = (x + 1) (x - 5) x = - 1, 5 (2, - 9) to (0, - 1) d = \sqrt{{(2 - 0)}^{2} + {(- 9 - (- 1))}^{2}}$

$d = 8.25$

Note

Because the vertex is on the axis of symmetry, it is equidistant from either root.