Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are $1$ and $-6$, respectively.

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> [!TLDR]
> The key idea is to use the intercept form of a line: $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ and $b$ are the $x$- and $y$-intercepts. Given $a+b = 1$ and $ab = -6$, we solve for $a$ and $b$ to get two pairs: $(3, -2)$ and $(-2, 3)$. The required lines are $2x - 3y = 6$ and $3x - 2y = -6$.

When a line cuts intercepts on the axes, the intercept form is the most natural tool. The intercepts are the points where the line meets the $x$-axis and $y$-axis — call them $a$ and $b$. The line's equation is then $\frac{x}{a} + \frac{y}{b} = 1$, provided neither intercept is zero. The problem gives us the sum and product of these intercepts, so we can find $a$ and $b$ by solving a quadratic. Once we have the intercepts, we write the equations.

Let's work through it step by step.

1. **Set up the intercepts.**  
   Let the $x$-intercept be $a$ and the $y$-intercept be $b$. The line is  
   $$\frac{x}{a} + \frac{y}{b} = 1.$$  
   We are told:  
   $$a + b = 1 \quad 	ext{and} \quad ab = -6.$$

2. **Form a quadratic equation.**  
   If two numbers have sum $S$ and product $P$, they are the roots of $t^2 - S t + P = 0$. Here $S = 1$ and $P = -6$, so  
   $$t^2 - t - 6 = 0.$$

3. **Solve for $a$ and $b$.**  
   Factor the quadratic:  
   $$t^2 - t - 6 = (t - 3)(t + 2) = 0.$$  
   So $t = 3$ or $t = -2$.  
   This gives two possible ordered pairs for $(a, b)$:  
   - If $a = 3$, then $b = -2$ (since $a+b=1$).  
   - If $a = -2$, then $b = 3$.

> [!WARNING]
   > A common mistake is to assume $a$ and $b$ are both positive. Here, the product is negative, so one intercept is positive and the other negative. That's perfectly fine — the line will cross one axis on the positive side and the other on the negative side.

4. **Write the equations.**  
   For $(a, b) = (3, -2)$:  
   $$\frac{x}{3} + \frac{y}{-2} = 1 \quad \Rightarrow \quad \frac{x}{3} - \frac{y}{2} = 1.$$  
   Multiply through by 6:  
   $$2x - 3y = 6.$$

For $(a, b) = (-2, 3)$:  
   $$\frac{x}{-2} + \frac

Q.Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are $1$ and $-6$ , respectively.

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Q.Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 111 and −6-6−6, respectively.

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Q.Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are $1$ and $-6$ , respectively.

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