Solving for (a) and (b) in the Line Equation (x/a y/b 1)
Introduction: This article provides a detailed solution to finding the product ab for a line equation given two specific points. The line equation x/a y/b 1 is examined through algebraic solving techniques that involve point substitution and equation manipulation. The process is fully explained here for clarity and to ensure each step is understandable for educational purposes.
Given Points and the Line Equation
Given the line equation x/a y/b 1 passing through two points: (2, -3) and (4, -5), we need to find the product ab.
Step 1: Setting up Equations for the Given Points
We establish two equations based on the given points (2, -3) and (4, -5).
For the point (2, -3): [frac{2}{a} frac{-3}{b} 1] For the point (4, -5): [frac{4}{a} frac{-5}{b} 1]Step 2: Solving for b
First, let's find the value of b by manipulating the equations.
[2frac{1}{a} - 3frac{1}{b} 1] [4frac{1}{a} - 5frac{1}{b} 1]Multiplying the first equation by 2 and subtracting the second equation from it:
[2cdot(2frac{1}{a} - 3frac{1}{b}) - (4frac{1}{a} - 5frac{1}{b}) 2 - 1] [4frac{1}{a} - 6frac{1}{b} - 4frac{1}{a} 5frac{1}{b} 1] [-frac{1}{b} 1] [frac{1}{b} -1] [b -1]Step 3: Solving for a
Now that we have the value of b, we substitute b -1 into one of the original equations. We use the first equation to find a.
[2frac{1}{a} - 3frac{1}{-1} 1] [2frac{1}{a} 3 1] [2frac{1}{a} 1 - 3] [2frac{1}{a} -2] [frac{1}{a} -1] [a -1]Step 4: Calculating the Product ab
Finally, we calculate the product ab:
[ab (-1) cdot (-1) 1]Therefore, the product ab is 1.
Alternative Path to Solving the Equation
Now, let's look at the alternative path to simplifying the given equation through the least common denominator (LCD).
[frac{x}{a} frac{y}{b} 1] The LCD ab helps combine the fractions into a single denominator.We can set up the equation using the points (2, -3) and (4, -5) and solve for a and b in a systematic way.
Step A: Using Rational Form to Simplify
Point (2, -3): [frac{2}{a} frac{-3}{b} 1] Converting to common denominator: [frac{2b - 3a}{ab} 1] Point (4, -5): [frac{4}{a} frac{-5}{b} 1] Converting to common denominator: [frac{4b - 5a}{ab} 1]Step B: Solving the Simplified Equations
By setting the numerators equal and solving the system of equations:
[2b - 3a 4b - 5a] [2b - 3a - 4b 5a 0] [-2b 2a 0] [a b]Substitute a b back into one of the original simplified equations:
[2b - 3b 0] [-b 0] [b -1] Thus, [a -1] And the product [ab (-1) cdot (-1) 1]To verify, we can substitute a -1 and b -1 back into the original simplified equations and check for consistency:
[frac{2}{-1} frac{-3}{-1} -2 3 1] [frac{4}{-1} frac{-5}{-1} -4 5 1]The results are consistent, confirming our solution.
Conclusion
In conclusion, through both algebraic manipulation and using the LCD method, we find that the product ab is 1 for the line equation given the points (2, -3) and (4, -5). This process demonstrates the importance of verifying solutions by substitution.