To solve this problem, we aim to determine the equations of the straight lines that pass through a given point and have intercepts whose sum equals a specified value. Specifically, the lines must pass through the point \((4, 3)\) and have intercepts on the coordinate axes whose sum is \(-1\).
We use the intercept form of a line equation:
\(\frac{x}{a} + \frac{y}{b} = 1\)
where \(a\) and \(b\) are the x-intercept and y-intercept, respectively. We know:
\(a + b = -1\)
Since the line passes through \((4, 3)\), substitute \(x = 4\) and \(y = 3\) into the intercept form:
\(\frac{4}{a} + \frac{3}{b} = 1\)
Now, we have two equations:
Let's solve these equations. From the first equation, express \(b\) in terms of \(a\):
\(b = -1 - a\)
Substitute this into the second equation:
\(\frac{4}{a} + \frac{3}{-1 - a} = 1\)
Multiply throughout by \(a(-1-a)\) to clear fractions:
\(4(-1-a) + 3a = a(-1-a)\)
Simplify this equation:
\(-4a - 4 + 3a = -a - a^2\) \(-a - 4 = -a - a^2\)
Rearrange to get:
\(a^2 - 4 = 0\)
Solve for \(a\):
\(a^2 = 4 \implies a = 2 \text{ or } a = -2\)
Substituting these values back, for \(a = 2\):
\(b = -1 - 2 = -3\)
Equation: \(\frac{x}{2} - \frac{y}{3} = 1\)
For \(a = -2\):
\(b = -1 + 2 = 1\)
Equation: \(\frac{x}{-2} + \frac{y}{1} = 1\)
Therefore, the equations of the lines are:
Thus, the correct answer is:
\(\frac{x}{2} - \frac{y}{3} = 1\) and \(\frac{x}{-2} + \frac{y}{1} = 1\)