Intercepts Sum and Product: Deriving Line Equations

This problem asks us to find the equations of straight lines based on specific conditions related to the intercepts they make on the coordinate axes. The conditions given are that the sum of the intercepts is 1, and their product is -6.

Understanding the Intercept Form of a Line

Recall the intercept form of a straight line equation, which is used when the points where the line crosses the x-axis and y-axis are known. The equation is given by:

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

Here, '$a$' represents the x-intercept (the x-coordinate where the line crosses the x-axis) and '$b$' represents the y-intercept (the y-coordinate where the line crosses the y-axis).

Applying Conditions to Find Intercept Values

The problem states two conditions:

• The sum of the intercepts is 1: $a + b = 1$
• The product of the intercepts is -6: $a \times b = -6$

We can find the values of '$a$' and '$b$' by relating these conditions to the properties of a quadratic equation. If we consider a quadratic equation whose roots are '$a$' and '$b$', it can be written as:

$$ t^2 - (\text{sum of roots})t + (\text{product of roots}) = 0 $$

Substituting the given values:

$$ t^2 - (1)t + (-6) = 0 $$

This simplifies to:

$$ t^2 - t - 6 = 0 $$

Now, we need to solve this quadratic equation for '$t$'. We can factorize the equation:

$$ (t - 3)(t + 2) = 0 $$

The possible values for '$t$' are $t=3$ and $t=-2$. This means we have two possible pairs for the intercepts $(a, b)$:

• Case 1: $a = 3$ and $b = -2$
• Case 2: $a = -2$ and $b = 3$

Deriving the Equations of the Lines

Let's use the intercept form $\frac{x}{a} + \frac{y}{b} = 1$ for each case:

• For Case 1 ($a = 3, b = -2$):
  The equation is $\frac{x}{3} + \frac{y}{-2} = 1$.
  To eliminate the fractions, we multiply the entire equation by the least common multiple of 3 and -2, which is 6:
  $$ 6 \times \left( \frac{x}{3} \right) + 6 \times \left( \frac{y}{-2} \right) = 6 \times 1 $$ $$ 2x - 3y = 6 $$
• For Case 2 ($a = -2, b = 3$):
  The equation is $\frac{x}{-2} + \frac{y}{3} = 1$.
  Multiply by the least common multiple of -2 and 3, which is 6:
  $$ 6 \times \left( \frac{x}{-2} \right) + 6 \times \left( \frac{y}{3} \right) = 6 \times 1 $$ $$ -3x + 2y = 6 $$

Therefore, the two equations of the lines satisfying the given conditions are $2x - 3y = 6$ and $-3x + 2y = 6$.

Comparing Derived Equations with Options

• Option 1 ($2x+3y=6, -3x+2y=6$): The first equation $2x+3y=6$ has intercepts $a=3, b=2$. Sum = 5, Product = 6. Incorrect.
• Option 2 ($2x+3y=6, 3x+2y=6$): Both equations have intercepts with sum 5 and product 6. Incorrect.
• Option 3 ($2x-3y=6, -3x+2y=6$): This matches the equations we derived. The first line has intercepts $a=3, b=-2$ (Sum=1, Product=-6). The second line has intercepts $a=-2, b=3$ (Sum=1, Product=-6). Correct.
• Option 4 ($2x+3y=0, 3x+2y=0$): These equations represent lines passing through the origin $(0,0)$. They do not have finite intercepts in the context of the intercept form equation $x/a + y/b = 1$. Incorrect.