https://file.aiquickdraw.com/mathbot/file/17290987906730iv99xpe.jpg

Let's solve the problem step by step.

We have two straight paths represented by the equations:

1. \( x - y - 4 = 0 \)
2. \( 2x + y - 5 = 0 \)

A man is standing at the junction (crossing) of these two paths. The problem involves determining:
1. The coordinates where the man is standing.
2. The distance he must cover to reach a different path: \( 12x + 5y - 18 = 0 \).
3. The slope of the path he should follow.
4. The equation of the path he should follow.

### Part (i) — Finding the coordinates at the junction

The junction is the intersection point of the two lines:
1. \( x - y = 4 \) 
2. \( 2x + y = 5 \)

Solve this system of equations:

1. From \( x - y = 4 \), we have \( y = x - 4 \).
2. Substituting \( y = x - 4 \) into \( 2x + y = 5 \):
   \[
   2x + (x - 4) = 5
   \]
   \[
   3x - 4 = 5 \quad \Rightarrow \quad 3x = 9 \quad \Rightarrow \quad x = 3
   \]
3. Now substitute \( x = 3 \) into \( y = x - 4 \):
   \[
   y = 3 - 4 = -1
   \]

So, the man is standing at the point \( (3, -1) \). Thus, the answer to part (i) is:
**(a) \( (3, -1) \)**.

---

### Part (ii) — Finding the distance to the path \( 12x + 5y - 18 = 0 \)

The formula to calculate the distance from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is:
\[
d = \frac{|A x_1 + B y_1 + C|}{\sqrt{A^2 + B^2}}
\]
For the line \( 12x + 5y - 18 = 0 \), we have \( A = 12 \), \( B = 5 \), and \( C = -18 \).

Substitute the point \( (3, -1) \) into the distance formula:
\[
d = \frac{|12(3) + 5(-1) - 18|}{\sqrt{12^2 + 5^2}} = \frac{|36 - 5 - 18|}{\sqrt{144 + 25}} = \frac{|13|}{\sqrt{169}} = \frac{13}{13} = 1
\]

So, the distance is **1 unit**, making the answer to part (ii):
**(d) 1 unit**.

---

### Part (iii) — Finding the slope of the path

The slope of the line perpendicular to the given path \( 12x + 5y - 18 = 0 \) can be found by determining the slope of the line.

1. First, rewrite the equation \( 12x + 5y = 18 \) in slope-intercept form \( y = mx + c \):
   \[
   5y = -12x + 18 \quad \Rightarrow \quad y = -\frac{12}{5}x + \frac{18}{5}
   \]
   So, the slope of the line is \( -\frac{12}{5} \).

2. The slope of the line perpendicular to this would be the negative reciprocal, i.e. \( \frac{5}{12} \).

Thus, the slope is:
**(a) \( \frac{5}{12} \)**.

---

### Part (iv) — Equation of the path the man should follow

The slope of the path is \( \frac{5}{12} \) and it passes through the point \( (3, -1) \). The equation of the line can be written in point-slope form:
\[
y - y_1 = m(x - x_1)
\]
Substitute \( (x_1, y_1) = (3, -1) \) and \( m = \frac{5}{12} \):
\[
y + 1 = \frac{5}{12}(x - 3)
\]
Multiply both sides by 12 to clear the fraction:
\[
12(y + 1) = 5(x - 3)
\]
\[
12y + 12 = 5x - 15
\]
Simplify:
\[
5x - 12y = 27
\]

Thus, the equation of the path is:
**(c) \( 5x - 12y = 27 \)**.

---

Would you like a more detailed explanation for any part of this solution?

### Five Related Questions:
1. How do we derive the distance formula from a point to a line?
2. What is the point-slope form of a linear equation?
3. How do we find the slope of a line given in standard form?
4. What are the properties of perpendicular slopes in geometry?
5. How can we verify the solution for the intersection of two lines?

### Tip:
When dealing with perpendicular lines, always remember that the product of their slopes is \( -1 \).

A man is standing at the junction of two straight paths represented by the equations x - y - 4 = 0 and 2x + y - 5 = 0. He wants to reach another path represented by the equation 12x + 5y - 18 = 0 in the least time.

This problem explores finding the shortest distance from a point where two lines intersect to a third line, covering key concepts in analytical geometry including linear equations, distance calculation, and perpendicular slopes. Ideal for high school students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation