Discover a world of knowledge at Westonci.ca, where experts and enthusiasts come together to answer your questions. Explore in-depth answers to your questions from a knowledgeable community of experts across different fields. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Let's break down the problem into two main parts as requested:
### Part (i): Show that [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] bisect each other at right angles.
1. Midpoints of [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex]:
- Midpoint of [tex]\(AC\)[/tex]:
To find the midpoint of a line segment between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex], we use:
[tex]\[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
For [tex]\(A(4, 0)\)[/tex] and [tex]\(C(-10, 6)\)[/tex]:
[tex]\[ \text{Midpoint of } AC = \left( \frac{4 + (-10)}{2}, \frac{0 + 6}{2} \right) = \left( \frac{-6}{2}, \frac{6}{2} \right) = (-3.0, 3.0) \][/tex]
- Midpoint of [tex]\(BD\)[/tex]:
For [tex]\(B(14, 11)\)[/tex] and [tex]\(D(-10, -5)\)[/tex]:
[tex]\[ \text{Midpoint of } BD = \left( \frac{14 + (-10)}{2}, \frac{11 + (-5)}{2} \right) = \left( \frac{4}{2}, \frac{6}{2} \right) = (2.0, 3.0) \][/tex]
The midpoints are [tex]\((-3.0, 3.0)\)[/tex] and [tex]\((2.0, 3.0)\)[/tex], indicating they do not coincide.
Since the midpoints [tex]\(({-3.0, 3.0})\)[/tex] and [tex]\((2.0, 3.0)\)[/tex] are not the same, [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] do not bisect each other.
2. Slopes of [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex]:
- Slope of [tex]\(AC\)[/tex]:
The slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ \text{Slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
For [tex]\(A(4, 0)\)[/tex] and [tex]\(C(-10, 6)\)[/tex]:
[tex]\[ \text{Slope of } AC = \frac{6 - 0}{-10 - 4} = \frac{6}{-14} = -\frac{3}{7} = -0.42857142857142855 \][/tex]
- Slope of [tex]\(BD\)[/tex]:
For [tex]\(B(14, 11)\)[/tex] and [tex]\(D(-10, -5)\)[/tex]:
[tex]\[ \text{Slope of } BD = \frac{-5 - 11}{-10 - 14} = \frac{-16}{-24} = \frac{2}{3} = 0.6666666666666666 \][/tex]
3. Perpendicularity:
The product of the slopes of two perpendicular lines is [tex]\(-1\)[/tex]. Calculate the product of the slopes:
[tex]\[ \text{Product of slopes} = (-0.42857142857142855) \times 0.6666666666666666 \approx -0.2857142857142857 \][/tex]
Since the product is not [tex]\(-1\)[/tex], [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] are not perpendicular.
### Part (ii): Calculate the ratio [tex]\(|BD| : |AC|\)[/tex].
1. Length of [tex]\(BD\)[/tex]:
The length of a line segment between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ \text{Length} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
For [tex]\(B(14, 11)\)[/tex] and [tex]\(D(-10, -5)\)[/tex]:
[tex]\[ \text{Length of } BD = \sqrt{(14 - (-10))^2 + (11 - (-5))^2} = \sqrt{(24)^2 + (16)^2} = \sqrt{576 + 256} = \sqrt{832} \approx 28.844410203711913 \][/tex]
2. Length of [tex]\(AC\)[/tex]:
For [tex]\(A(4, 0)\)[/tex] and [tex]\(C(-10, 6)\)[/tex]:
[tex]\[ \text{Length of } AC = \sqrt{(4 - (-10))^2 + (0 - 6)^2} = \sqrt{(14)^2 + (-6)^2} = \sqrt{196 + 36} = \sqrt{232} \approx 15.231546211727817 \][/tex]
3. Ratio [tex]\(|BD| : |AC|\)[/tex]:
[tex]\[ \text{Ratio} = \frac{|BD|}{|AC|} = \frac{28.844410203711913}{15.231546211727817} \approx 1.8937283058959973 \][/tex]
### Summary:
- [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] do not bisect each other.
- [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] are not perpendicular.
- The ratio [tex]\(|BD| : |AC|\)[/tex] is approximately [tex]\(1.8937\)[/tex].
### Part (i): Show that [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] bisect each other at right angles.
1. Midpoints of [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex]:
- Midpoint of [tex]\(AC\)[/tex]:
To find the midpoint of a line segment between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex], we use:
[tex]\[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
For [tex]\(A(4, 0)\)[/tex] and [tex]\(C(-10, 6)\)[/tex]:
[tex]\[ \text{Midpoint of } AC = \left( \frac{4 + (-10)}{2}, \frac{0 + 6}{2} \right) = \left( \frac{-6}{2}, \frac{6}{2} \right) = (-3.0, 3.0) \][/tex]
- Midpoint of [tex]\(BD\)[/tex]:
For [tex]\(B(14, 11)\)[/tex] and [tex]\(D(-10, -5)\)[/tex]:
[tex]\[ \text{Midpoint of } BD = \left( \frac{14 + (-10)}{2}, \frac{11 + (-5)}{2} \right) = \left( \frac{4}{2}, \frac{6}{2} \right) = (2.0, 3.0) \][/tex]
The midpoints are [tex]\((-3.0, 3.0)\)[/tex] and [tex]\((2.0, 3.0)\)[/tex], indicating they do not coincide.
Since the midpoints [tex]\(({-3.0, 3.0})\)[/tex] and [tex]\((2.0, 3.0)\)[/tex] are not the same, [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] do not bisect each other.
2. Slopes of [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex]:
- Slope of [tex]\(AC\)[/tex]:
The slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ \text{Slope} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
For [tex]\(A(4, 0)\)[/tex] and [tex]\(C(-10, 6)\)[/tex]:
[tex]\[ \text{Slope of } AC = \frac{6 - 0}{-10 - 4} = \frac{6}{-14} = -\frac{3}{7} = -0.42857142857142855 \][/tex]
- Slope of [tex]\(BD\)[/tex]:
For [tex]\(B(14, 11)\)[/tex] and [tex]\(D(-10, -5)\)[/tex]:
[tex]\[ \text{Slope of } BD = \frac{-5 - 11}{-10 - 14} = \frac{-16}{-24} = \frac{2}{3} = 0.6666666666666666 \][/tex]
3. Perpendicularity:
The product of the slopes of two perpendicular lines is [tex]\(-1\)[/tex]. Calculate the product of the slopes:
[tex]\[ \text{Product of slopes} = (-0.42857142857142855) \times 0.6666666666666666 \approx -0.2857142857142857 \][/tex]
Since the product is not [tex]\(-1\)[/tex], [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] are not perpendicular.
### Part (ii): Calculate the ratio [tex]\(|BD| : |AC|\)[/tex].
1. Length of [tex]\(BD\)[/tex]:
The length of a line segment between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ \text{Length} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
For [tex]\(B(14, 11)\)[/tex] and [tex]\(D(-10, -5)\)[/tex]:
[tex]\[ \text{Length of } BD = \sqrt{(14 - (-10))^2 + (11 - (-5))^2} = \sqrt{(24)^2 + (16)^2} = \sqrt{576 + 256} = \sqrt{832} \approx 28.844410203711913 \][/tex]
2. Length of [tex]\(AC\)[/tex]:
For [tex]\(A(4, 0)\)[/tex] and [tex]\(C(-10, 6)\)[/tex]:
[tex]\[ \text{Length of } AC = \sqrt{(4 - (-10))^2 + (0 - 6)^2} = \sqrt{(14)^2 + (-6)^2} = \sqrt{196 + 36} = \sqrt{232} \approx 15.231546211727817 \][/tex]
3. Ratio [tex]\(|BD| : |AC|\)[/tex]:
[tex]\[ \text{Ratio} = \frac{|BD|}{|AC|} = \frac{28.844410203711913}{15.231546211727817} \approx 1.8937283058959973 \][/tex]
### Summary:
- [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] do not bisect each other.
- [tex]\(AC\)[/tex] and [tex]\(BD\)[/tex] are not perpendicular.
- The ratio [tex]\(|BD| : |AC|\)[/tex] is approximately [tex]\(1.8937\)[/tex].
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.