Question

The equations of the straight lines passing through the point (4, 3) and making intercepts on the coordinate axes whose sum is -1, is

• A. \(\frac{x}{2} - \frac{y}{3} = 1\) and \(\frac{x}{-2} + \frac{y}{1} = 1\)
• B. \(\frac{x}{2} - \frac{y}{3} = -1\) and \(\frac{x}{-2} + \frac{y}{1} = -1\)
• C. \(\frac{x}{2} + \frac{y}{3} = 1\) and \(\frac{x}{2} + \frac{y}{1} = 1\)
• D. None of the above

Hint:

Intercept form: \(\frac{x}{a} + \frac{y}{b} = 1\).

Answer 1

The Correct Option is A

The problem requires us to find the equations of straight lines that pass through the given point \((4, 3)\) and have the sum of their intercepts on the coordinate axes equal to -1. 

A line in the intercept form is given by the equation: \(\frac{x}{a} + \frac{y}{b} = 1\), where \(a\) and \(b\) are the x-intercept and y-intercept, respectively.

According to the problem, these intercepts have a sum equal to -1:

\(a + b = -1\)\).

Since the line passes through the point (4, 3), substituting these coordinates into the line's equation, we have:

\(\frac{4}{a} + \frac{3}{b} = 1\).

Now, we can solve these equations:

1. Equation 1: \(a + b = -1\)
2. Equation 2: \(\frac{4}{a} + \frac{3}{b} = 1\)

We assume possible values for \(a\) and \(b\) that satisfy both equations. Solving these simultaneously, we try different scenarios:

Case 1:

1. If \(a = 2\) and \(b = -3\):
   • Substituting in Equation 1: \(2 + (-3) = -1\), so satisfies Equation 1.
   • Substituting in Equation 2: \(\frac{4}{2} + \frac{3}{-3} = 2 - 1 = 1\), satisfies Equation 2.
2. Thus, the line equation is: \(\frac{x}{2} - \frac{y}{3} = 1\).

Case 2:

1. If \(a = -2\) and \(b = 1\):
   • Substituting in Equation 1: \((-2) + 1 = -1\), satisfies Equation 1.
   • Substituting in Equation 2: \(\frac{4}{-2} + \frac{3}{1} = -2 + 3 = 1\), satisfies Equation 2.
2. Thus, the line equation is: \(\frac{x}{-2} + \frac{y}{1} = 1\).

Therefore, the correct answer is Option 1: \(\frac{x}{2} - \frac{y}{3} = 1\) and \(\frac{x}{-2} + \frac{y}{1} = 1\).