Answer :
Certainly! Let's solve these exercises step-by-step.
### Exercise 14: Endpoints [tex]\( A(2,5) \)[/tex] and [tex]\( B(4,3) \)[/tex]
1. Length of the Segment:
To find the length of segment [tex]\( AB \)[/tex], use the distance formula:
[tex]\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
where [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] are the coordinates of points [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
[tex]\[ AB = \sqrt{(4 - 2)^2 + (3 - 5)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} \approx 2.8 \][/tex]
2. Midpoint of the Segment:
To find the midpoint [tex]\( M \)[/tex] of segment [tex]\( AB \)[/tex], use the midpoint formula:
[tex]\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
[tex]\[ M = \left( \frac{2 + 4}{2}, \frac{5 + 3}{2} \right) = \left( \frac{6}{2}, \frac{8}{2} \right) = (3.0, 4.0) \][/tex]
### Exercise 15: Endpoints [tex]\( F(1,7) \)[/tex] and [tex]\( G(6,0) \)[/tex]
1. Length of the Segment:
[tex]\[ FG = \sqrt{(6 - 1)^2 + (0 - 7)^2} = \sqrt{5^2 + (-7)^2} = \sqrt{25 + 49} = \sqrt{74} \approx 8.6 \][/tex]
2. Midpoint of the Segment:
[tex]\[ M = \left( \frac{1 + 6}{2}, \frac{7 + 0}{2} \right) = \left( \frac{7}{2}, \frac{7}{2} \right) = (3.5, 3.5) \][/tex]
### Exercise 16: Endpoints [tex]\( H(-3,9) \)[/tex] and [tex]\( I(5,4) \)[/tex]
1. Length of the Segment:
[tex]\[ HI = \sqrt{(5 + 3)^2 + (4 - 9)^2} = \sqrt{8^2 + (-5)^2} = \sqrt{64 + 25} = \sqrt{89} \approx 9.4 \][/tex]
2. Midpoint of the Segment:
[tex]\[ M = \left( \frac{-3 + 5}{2}, \frac{9 + 4}{2} \right) = \left( \frac{2}{2}, \frac{13}{2} \right) = (1.0, 6.5) \][/tex]
### Exercise 17: Endpoints [tex]\( K(10,6) \)[/tex] and [tex]\( L(0,-7) \)[/tex]
1. Length of the Segment:
[tex]\[ KL = \sqrt{(0 - 10)^2 + (-7 - 6)^2} = \sqrt{(-10)^2 + (-13)^2} = \sqrt{100 + 169} = \sqrt{269} \approx 16.4 \][/tex]
2. Midpoint of the Segment:
[tex]\[ M = \left( \frac{10 + 0}{2}, \frac{6 + (-7)}{2} \right) = \left( \frac{10}{2}, \frac{-1}{2} \right) = (5.0, -0.5) \][/tex]
So, the results are:
1. For [tex]\( A(2,5) \)[/tex] and [tex]\( B(4,3) \)[/tex]:
- Length: [tex]\( 2.8 \)[/tex]
- Midpoint: [tex]\( (3.0, 4.0) \)[/tex]
2. For [tex]\( F(1,7) \)[/tex] and [tex]\( G(6,0) \)[/tex]:
- Length: [tex]\( 8.6 \)[/tex]
- Midpoint: [tex]\( (3.5, 3.5) \)[/tex]
3. For [tex]\( H(-3,9) \)[/tex] and [tex]\( I(5,4) \)[/tex]:
- Length: [tex]\( 9.4 \)[/tex]
- Midpoint: [tex]\( (1.0, 6.5) \)[/tex]
4. For [tex]\( K(10,6) \)[/tex] and [tex]\( L(0,-7) \)[/tex]:
- Length: [tex]\( 16.4 \)[/tex]
- Midpoint: [tex]\( (5.0, -0.5) \)[/tex]
### Exercise 14: Endpoints [tex]\( A(2,5) \)[/tex] and [tex]\( B(4,3) \)[/tex]
1. Length of the Segment:
To find the length of segment [tex]\( AB \)[/tex], use the distance formula:
[tex]\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
where [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] are the coordinates of points [tex]\( A \)[/tex] and [tex]\( B \)[/tex].
[tex]\[ AB = \sqrt{(4 - 2)^2 + (3 - 5)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} \approx 2.8 \][/tex]
2. Midpoint of the Segment:
To find the midpoint [tex]\( M \)[/tex] of segment [tex]\( AB \)[/tex], use the midpoint formula:
[tex]\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \][/tex]
[tex]\[ M = \left( \frac{2 + 4}{2}, \frac{5 + 3}{2} \right) = \left( \frac{6}{2}, \frac{8}{2} \right) = (3.0, 4.0) \][/tex]
### Exercise 15: Endpoints [tex]\( F(1,7) \)[/tex] and [tex]\( G(6,0) \)[/tex]
1. Length of the Segment:
[tex]\[ FG = \sqrt{(6 - 1)^2 + (0 - 7)^2} = \sqrt{5^2 + (-7)^2} = \sqrt{25 + 49} = \sqrt{74} \approx 8.6 \][/tex]
2. Midpoint of the Segment:
[tex]\[ M = \left( \frac{1 + 6}{2}, \frac{7 + 0}{2} \right) = \left( \frac{7}{2}, \frac{7}{2} \right) = (3.5, 3.5) \][/tex]
### Exercise 16: Endpoints [tex]\( H(-3,9) \)[/tex] and [tex]\( I(5,4) \)[/tex]
1. Length of the Segment:
[tex]\[ HI = \sqrt{(5 + 3)^2 + (4 - 9)^2} = \sqrt{8^2 + (-5)^2} = \sqrt{64 + 25} = \sqrt{89} \approx 9.4 \][/tex]
2. Midpoint of the Segment:
[tex]\[ M = \left( \frac{-3 + 5}{2}, \frac{9 + 4}{2} \right) = \left( \frac{2}{2}, \frac{13}{2} \right) = (1.0, 6.5) \][/tex]
### Exercise 17: Endpoints [tex]\( K(10,6) \)[/tex] and [tex]\( L(0,-7) \)[/tex]
1. Length of the Segment:
[tex]\[ KL = \sqrt{(0 - 10)^2 + (-7 - 6)^2} = \sqrt{(-10)^2 + (-13)^2} = \sqrt{100 + 169} = \sqrt{269} \approx 16.4 \][/tex]
2. Midpoint of the Segment:
[tex]\[ M = \left( \frac{10 + 0}{2}, \frac{6 + (-7)}{2} \right) = \left( \frac{10}{2}, \frac{-1}{2} \right) = (5.0, -0.5) \][/tex]
So, the results are:
1. For [tex]\( A(2,5) \)[/tex] and [tex]\( B(4,3) \)[/tex]:
- Length: [tex]\( 2.8 \)[/tex]
- Midpoint: [tex]\( (3.0, 4.0) \)[/tex]
2. For [tex]\( F(1,7) \)[/tex] and [tex]\( G(6,0) \)[/tex]:
- Length: [tex]\( 8.6 \)[/tex]
- Midpoint: [tex]\( (3.5, 3.5) \)[/tex]
3. For [tex]\( H(-3,9) \)[/tex] and [tex]\( I(5,4) \)[/tex]:
- Length: [tex]\( 9.4 \)[/tex]
- Midpoint: [tex]\( (1.0, 6.5) \)[/tex]
4. For [tex]\( K(10,6) \)[/tex] and [tex]\( L(0,-7) \)[/tex]:
- Length: [tex]\( 16.4 \)[/tex]
- Midpoint: [tex]\( (5.0, -0.5) \)[/tex]