A person standing at the junction (crossing) of two straight paths…

Question 70 of 70advanced🔧 ApplyShort Answer4 marks

A person standing at the junction (crossing) of two straight paths represented by the equations $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$ wants to reach the path whose equation is $6x - 7y + 8 = 0$ in the least time. Find equation of the path that he should follow.

Correct Answer

119x + 102y - 125 = 0

Exercise: Miscellaneous Exercise on Chapter 9 | Q: 23 | (Page 22)

For More Understanding

Explanation

This problem requires finding the shortest path from a junction point to a given line. The shortest distance from a point to a line is along the perpendicular. First, find the junction point by solving the two given equations simultaneously. Then, find the equation of a line passing through this point and perpendicular to the target path.

Solution Steps

Step 1: Find the junction point by solving $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$ using cross-multiplication method.
Step 2: Apply cross-multiplication: $\frac{x}{(-3)(-5) - (4)(4)} = \frac{y}{(4)(3) - (2)(-5)} = \frac{1}{(2)(4) - (-3)(3)}$
Step 3: Simplify: $\frac{x}{15-16} = \frac{y}{12+10} = \frac{1}{8+9}$ , giving $\frac{x}{-1} = \frac{y}{22} = \frac{1}{17}$
Step 4: Solve to get junction point: $x = -\frac{1}{17}$ and $y = \frac{22}{17}$ . So the junction point is $(-\frac{1}{17}, \frac{22}{17})$ .
Step 5: Find slope of target path $6x - 7y + 8 = 0$ . Rewriting: $y = \frac{6}{7}x + \frac{8}{7}$ , so slope = $\frac{6}{7}$ .
Step 6: For shortest distance, the required path must be perpendicular to the target path. Slope of perpendicular line = $-\frac{7}{6}$ (negative reciprocal).
Step 7: Use point-slope form for line through $(-\frac{1}{17}, \frac{22}{17})$ with slope $-\frac{7}{6}$ : $y - \frac{22}{17} = -\frac{7}{6}(x + \frac{1}{17})$
Step 8: Multiply by 102 and simplify: $102y - 132 = -119x - 7$ , giving $119x + 102y - 125 = 0$ .