The equation of the straight line passing through the point $$(4, 3)$$ and making intercepts on the co-ordinate axes whose sum is $$-1$$ is

Let the equation of the straight line in intercept form be:

$$ \frac{x}{a} + \frac{y}{b} = 1 $$

In this form, a represents the x-intercept and b represents the y-intercept. The problem states that the sum of these intercepts is -1:

$$ a + b = -1 $$

Isolate the variable b in terms of a:

$$ b = -1 - a $$

Substitute this expression for b back into the intercept form of the line:

$$ \frac{x}{a} + \frac{y}{-1 - a} = 1 $$

$$ \frac{x}{a} - \frac{y}{a + 1} = 1 $$

The line passes through the point (4, 3). Therefore, substituting $$ x = 4 $$ and $$ y = 3 $$ into the equation must yield a true statement:

$$ \frac{4}{a} - \frac{3}{a + 1} = 1 $$

Find a common denominator on the left side to merge the fractions:

$$ \frac{4(a + 1) - 3a}{a(a + 1)} = 1 $$

Expand the numerator and simplify the expression:

$$ \frac{4a + 4 - 3a}{a^2 + a} = 1 $$

$$ \frac{a + 4}{a^2 + a} = 1 $$

Cross-multiply to eliminate the fraction and form a quadratic equation:

$$ a + 4 = a^2 + a $$

Subtract a from both sides of the equation:

$$ 4 = a^2 $$

$$ a^2 = 4 $$

Taking the square root gives two possible values for a:

$$ a = 2 $$

$$ a = -2 $$

Now calculate the corresponding value of b for each value of a using the relationship $$ b = -1 - a $$.

Case 1: When $$ a = 2 $$, the value of b is:

$$ b = -1 - 2 $$

$$ b = -3 $$

Substitute $$ a = 2 $$ and $$ b = -3 $$ back into the intercept equation:

$$ \frac{x}{2} + \frac{y}{-3} = 1 $$

$$\implies \frac{x}{2} - \frac{y}{3} = 1 $$

Case 2: When $$ a = -2 $$, the value of b is:

$$ b = -1 - (-2) $$

$$ b = -1 + 2 $$

$$ b = 1 $$

Substitute $$ a = -2 $$ and $$ b = 1 $$ back into the intercept equation:

$$ \frac{x}{-2} + \frac{y}{1} = 1 $$

Final Answer:

The equations of the straight lines are $$\frac{x}{2} - \frac{y}{3} = 1 $$ and $$ \frac{x}{-2} + \frac{y}{1} = 1 $$.