Practice 1 Solutions

Calculate the distance between the points. Round answers to the nearest hundredth.

$(- 3, 5), (9, - 1)$

$d = \sqrt{{(9 - (- 3))}^{2} + {(- 1 - 5)}^{2}}$

$d = 13.42$

$(4.25, 5.6), (1.7, - 0.3)$

$d = \sqrt{{(4.25 - 1.7)}^{2} + {(5.6 - (- 0.3))}^{2}}$

$d = 6.43$

$(14, - 10), (20, - 5)$

$d = \sqrt{{(20 - 14)}^{2} + {(- 5 - (- 10))}^{2}}$

$d = 7.81$

$(6.75, 2), (0.75, - 6)$

$d = \sqrt{{(6.75 - 0.75)}^{2} + {(2 - (- 6))}^{2}}$

$d = 10.00$

For problems 5–6, use the given graph (top right).

Calculate the midpoint between points AB. Mark the midpoint, M, on the graph.

$A (- 5, 3), B (- 1, - 4) (\frac{- 5 + - 1}{2}, \frac{3 + - 4}{2})$

$M (- 3, - 0.5)$

Find the endpoint, C, with point M and point B as the midpoint.

$M (- 3, - 0.5), B (- 1, - 4) (\frac{x + - 3}{2}, \frac{y + - 0.5}{2}) = (- 1, - 4) \begin{matrix} \frac{x - 3}{2} = - 1 & \frac{y - 0.5}{2} = - 4 \\ x - 3 = - 2 & y - 0.5 = - 8 \\ x = 1 & y = - 7.5 \end{matrix}$

$C (1, - 7.5)$

For problems 7–8, use the graph of triangle ABC.

Determine the midpoint of each side of the triangle ABC.

$A (2, 6), B (8, 2) (\frac{2 + 8}{2}, \frac{6 + 2}{2}) (5, 4) A (2, 6), C (- 4, - 2) (\frac{2 + - 4}{2}, \frac{6 + - 2}{2}) (- 1, 2) C (- 4, - 2), B (8, 2) (\frac{- 4 + 8}{2}, \frac{- 2 + 2}{2}) (2, 0)$

$\bar{A B} (5, 4), \bar{A C} (- 1, 2), \bar{B C} (2, 0)$

Connect all of the midpoints to form a new triangle. Find the perimeter of the new triangle.

$(- 1, 2) to (5, 4) d = \sqrt{{(- 1 - 5)}^{2} + {(2 - 4)}^{2}} d = 6.32 (5, 4) to (2, 0) d = \sqrt{{(5 - 2)}^{2} + {(4 - 0)}^{2}} d = 5.00 (2, 0) to (- 1, 2) d = \sqrt{{(2 - (- 1))}^{2} + {(0 - 2)}^{2}} d = 3.61 Perimeter = 6.32 + 5.00 + 3.61$

$P = 14.93$

The perimeter of the new triangle is 14.93 units.

For problems 9–11, use the given graph of a miniature golf course hole.

On this mini golf course hole, Collin decided to take a straight shot from tee #3 to the hole. What is the distance to the hole?

$(- 1, - 6) to (1, 5) d = \sqrt{{(- 1 - 1)}^{2} + {(- 6 - 5)}^{2}}$

$d = 11.18$ feet

Nelly also used tee #3 and decided to bank (or deflect) her shot from the left boundary wall at the midpoint. Find the midpoint of the left boundary wall.

$(- 5, 10), (- 5, - 5) (\frac{- 5 + - 5}{2}, \frac{10 + - 5}{2})$

$(- 5, 2.5)$

Lauren took the same shot as Nelly and made a hole in one, what was the total distance for the shot?

$(1, - 6) to (- 5, 2.5) d_{1} = \sqrt{{(1 - (- 5))}^{2} + {(- 6 - 2.5)}^{2}} d_{1} = 10.40 (- 5, 2.5) to (1, 5) d_{2} = \sqrt{{(- 5 - 1)}^{2} + {(2.5 - 5)}^{2}} d_{2} = 6.5 d_{1} + d_{2} = 16.9$

Lauren shot a distance of 16.9 feet.

Each square represents 1 square foot.

For problems 12–14, use the equation $y = x^{2} - 4 x - 3$ .

A line intersects the parabola $y = x^{2} - 4 x - 3$ , at the points $(- 1, 2)$ and $(5.5, 5.25)$ . Determine the point, M, equidistant from the given points on the line.

$(- 1, 2), (5.5, 5.25) (\frac{- 1 + 5.5}{2}, \frac{2 + 5.25}{2})$

$M (2.25, 3.625)$

Note

If you need to see this problem as a graph, use technology to graph the parabola and check your work.

Calculate the distance between the y-intercept and point M found in the previous problem.

$y ‐ intercept : (0, - 3) M (2.25, 3.625) d = \sqrt{{(0 - 2.25)}^{2} + {(- 3 - 3.625)}^{2}} d = 6.996$

$d = 7.00$ units

Determine the value of the discriminant. Explain if it is possible to determine a real number distance between point M and either of the x-intercepts of the parabola.

$a = 1, b = - 4, c = - 3 b^{2} - 4 a c {(- 4)}^{2} - 4 (1) (- 3) 16 + 12 = 28$

Yes, the distance can be calculated because the value of the discriminant is greater than zero. This means that there are two real roots.