Answer :
Let's solve each of the problems step by step:
### 2. Equation of the Perpendicular Line
1. Find the slope of the line through points A(-1, 4) and B(5, 2):
- The formula for the slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is [tex]\( m = \frac{y_2 - y_1}{x_2 - x_1} \)[/tex].
- Here, [tex]\( m = \frac{2 - 4}{5 + 1} = \frac{-2}{6} = -\frac{1}{3} \)[/tex].
2. Find the slope of the line perpendicular to it:
- The slope of a line perpendicular to a line with slope [tex]\( m \)[/tex] is the negative reciprocal of [tex]\( m \)[/tex].
- Thus, the perpendicular slope is [tex]\( m_{\perp} = -\frac{1}{-\frac{1}{3}} = 3 \)[/tex].
3. Use the point-slope form for the perpendicular line:
- The point-slope form is [tex]\( y - y_1 = m(x - x_1) \)[/tex].
- Here, using point [tex]\( A(-1, 4) \)[/tex] and the perpendicular slope (3), the equation is [tex]\( y - 4 = 3(x + 1) \)[/tex].
- Expanding this: [tex]\( y - 4 = 3x + 3 \)[/tex] which simplifies to [tex]\( y = 3x + 7 \)[/tex].
Final Equation: [tex]\( y = 3x + 7 \)[/tex]
### 3. Distance from Point P(3,4) to Line y = -2x + 3
1. Use the distance formula from a point to a line [tex]\(Ax + By + C = 0\)[/tex]:
- For the line [tex]\( y = -2x + 3 \)[/tex], rewrite it as [tex]\( 2x + y - 3 = 0 \)[/tex].
- [tex]\( A = 2 \)[/tex], [tex]\( B = 1 \)[/tex], [tex]\( C = -3 \)[/tex].
2. Plug the values into the distance formula:
- Distance [tex]\( D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \)[/tex].
- For point [tex]\( P(3,4) \)[/tex], distance [tex]\( D = \frac{|2(3) + 1(4) - 3|}{\sqrt{2^2 + 1^2}} = \frac{|6 + 4 - 3|}{\sqrt{5}} = \frac{7}{\sqrt{5}} = 2.23606797749979 \)[/tex].
Final Distance: [tex]\( 2.23606797749979 \)[/tex]
### 4. Closest Distance and Time Between Two Ships
1. Express the positions of Ship 1 and Ship 2 as functions of time [tex]\( t \)[/tex]:
- Ship 1: [tex]\((3t - 10, t + 4)\)[/tex]
- Ship 2: [tex]\((2t - 3, -t + 13)\)[/tex]
2. Define the distance function [tex]\( D(t) \)[/tex]:
- [tex]\( D(t) = \sqrt{((2t - 3) - (3t - 10))^2 + ((-t + 13) - (t + 4))^2} \)[/tex].
- Simplified: [tex]\( D(t) = \sqrt{(-t + 7)^2 + (-2t + 9)^2} \)[/tex].
3. Minimize [tex]\( D(t) \)[/tex]:
- By using minimization techniques, the closest distance is obtained.
Closest Distance: [tex]\( 2.236067977500005 \)[/tex], at time [tex]\( t \approx 4.99999956092568 \)[/tex] hours
### 5. Intersection Points of Two Circles
1. Equations of the circles:
- Circle 1: [tex]\( (x - 3)^2 + (y + 4)^2 = 53 \)[/tex]
- Circle 2: [tex]\( (x + 2)^2 + (y - 1)^2 = 13 \)[/tex]
2. Solve the system of equations:
- Solving these simultaneously, we find the intersection points.
Intersection Points: [tex]\( (-4, -2) \)[/tex] and [tex]\( (1, 3) \)[/tex]
3. Equation of the line passing through these points:
- Slope: [tex]\( m = \frac{3 - (-2)}{1 - (-4)} = \frac{5}{5} = 1 \)[/tex]
- Using point-slope form with point [tex]\( (-4, -2) \)[/tex]:
[tex]\( y - (-2) = 1(x + 4) \)[/tex]
[tex]\( y + 2 = x + 4 \)[/tex]
[tex]\( y = x + 2 \)[/tex]
Final Line Equation: [tex]\( y = x + 2 \)[/tex]
### 2. Equation of the Perpendicular Line
1. Find the slope of the line through points A(-1, 4) and B(5, 2):
- The formula for the slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is [tex]\( m = \frac{y_2 - y_1}{x_2 - x_1} \)[/tex].
- Here, [tex]\( m = \frac{2 - 4}{5 + 1} = \frac{-2}{6} = -\frac{1}{3} \)[/tex].
2. Find the slope of the line perpendicular to it:
- The slope of a line perpendicular to a line with slope [tex]\( m \)[/tex] is the negative reciprocal of [tex]\( m \)[/tex].
- Thus, the perpendicular slope is [tex]\( m_{\perp} = -\frac{1}{-\frac{1}{3}} = 3 \)[/tex].
3. Use the point-slope form for the perpendicular line:
- The point-slope form is [tex]\( y - y_1 = m(x - x_1) \)[/tex].
- Here, using point [tex]\( A(-1, 4) \)[/tex] and the perpendicular slope (3), the equation is [tex]\( y - 4 = 3(x + 1) \)[/tex].
- Expanding this: [tex]\( y - 4 = 3x + 3 \)[/tex] which simplifies to [tex]\( y = 3x + 7 \)[/tex].
Final Equation: [tex]\( y = 3x + 7 \)[/tex]
### 3. Distance from Point P(3,4) to Line y = -2x + 3
1. Use the distance formula from a point to a line [tex]\(Ax + By + C = 0\)[/tex]:
- For the line [tex]\( y = -2x + 3 \)[/tex], rewrite it as [tex]\( 2x + y - 3 = 0 \)[/tex].
- [tex]\( A = 2 \)[/tex], [tex]\( B = 1 \)[/tex], [tex]\( C = -3 \)[/tex].
2. Plug the values into the distance formula:
- Distance [tex]\( D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \)[/tex].
- For point [tex]\( P(3,4) \)[/tex], distance [tex]\( D = \frac{|2(3) + 1(4) - 3|}{\sqrt{2^2 + 1^2}} = \frac{|6 + 4 - 3|}{\sqrt{5}} = \frac{7}{\sqrt{5}} = 2.23606797749979 \)[/tex].
Final Distance: [tex]\( 2.23606797749979 \)[/tex]
### 4. Closest Distance and Time Between Two Ships
1. Express the positions of Ship 1 and Ship 2 as functions of time [tex]\( t \)[/tex]:
- Ship 1: [tex]\((3t - 10, t + 4)\)[/tex]
- Ship 2: [tex]\((2t - 3, -t + 13)\)[/tex]
2. Define the distance function [tex]\( D(t) \)[/tex]:
- [tex]\( D(t) = \sqrt{((2t - 3) - (3t - 10))^2 + ((-t + 13) - (t + 4))^2} \)[/tex].
- Simplified: [tex]\( D(t) = \sqrt{(-t + 7)^2 + (-2t + 9)^2} \)[/tex].
3. Minimize [tex]\( D(t) \)[/tex]:
- By using minimization techniques, the closest distance is obtained.
Closest Distance: [tex]\( 2.236067977500005 \)[/tex], at time [tex]\( t \approx 4.99999956092568 \)[/tex] hours
### 5. Intersection Points of Two Circles
1. Equations of the circles:
- Circle 1: [tex]\( (x - 3)^2 + (y + 4)^2 = 53 \)[/tex]
- Circle 2: [tex]\( (x + 2)^2 + (y - 1)^2 = 13 \)[/tex]
2. Solve the system of equations:
- Solving these simultaneously, we find the intersection points.
Intersection Points: [tex]\( (-4, -2) \)[/tex] and [tex]\( (1, 3) \)[/tex]
3. Equation of the line passing through these points:
- Slope: [tex]\( m = \frac{3 - (-2)}{1 - (-4)} = \frac{5}{5} = 1 \)[/tex]
- Using point-slope form with point [tex]\( (-4, -2) \)[/tex]:
[tex]\( y - (-2) = 1(x + 4) \)[/tex]
[tex]\( y + 2 = x + 4 \)[/tex]
[tex]\( y = x + 2 \)[/tex]
Final Line Equation: [tex]\( y = x + 2 \)[/tex]
