Equation of the line that has $x$ and $y$ intercepts at $a$ and $b$.

Question

Please can anyone help me with proving the following problem:

Show that the line that crosses the $X$-axis at $a \neq 0$ and the $Y$-axis at $b \neq 0$ has the equation $$\dfrac{x}a + \dfrac{y}b -1=0$$

I don't have an idea of where I should start from for proving the problem.

Answer 1

The $x$ intercept is given by $(a, 0)$, and the $y$ intercept is given by $(0, b)$. And we know $a \neq 0, b\neq 0$, so the points are distinct.

All you need to do is substitute these values into your equation to see that equality holds: i.e., that they lie on the line defined by $$\frac xa + \frac yb - 1 = 0$$

$$(a, 0): x = a, y = 0 \implies \frac aa + \frac 0b - 1 = 0\implies \frac aa = 1\implies 1 = 1\quad \checkmark$$

$$(0, b): x = 0, y = b \implies \frac 0a + \frac bb - 1 = 0 \implies \frac bb = 1 \implies 1 = 1 \quad \checkmark$$

Indeed: These are unique points lying on the line given by $$\frac xa + \frac yb - 1 = 0$$

Answer 2

• Clearly, the equation $\dfrac{x}a + \dfrac{y}b - 1 =0 $ defines a straight line.
• Also, two distinct points uniquely define a straight line.
• Now check that $(a,0)$ and $(0,b)$ satisfy the equation.
• This would imply $(a,0)$ and $(0,b)$ lie on the straight line.
• Since $a \neq 0 \neq b$, these are two distinct distinct points.

Answer 3

Hint: you have two points from the problem description: $(a, 0)$, and $(0, b)$. Can you write the equation of a line connecting them?

If you are still stuck:

Using slope-intercept form, we have:

$$y = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}x + b$$ where $b$ is the $y$-intercept and $(x_{1}, y_{1}), (x_{2}, y_{2})$ are two points on the line. Evaluating, then, we see:

$$y = \frac{b - 0}{0-a}x + b = \frac{-b}{a}x + b$$ $$\implies y + \frac{b}{a}x - b = 0$$ $$\implies \frac{1}{b}y + \frac{1}{a}x - 1 = 0$$ $$\implies \frac{x}{a} + \frac{y}{b} - 1 = 0$$

Answer 4

Actually there are set of 4 lines. because intercepts are length and its coordinates have set of 4 coordinates.(a,0)(0,b),(a,0)(0,-b),(-a,0)(0,b),(-a,0)(0,-b). so lines equation will be.. $$ \frac xa + \frac yb -1 =0$$

$$ \frac xa - \frac yb -1 =0$$

$$ -\frac xa + \frac yb -1 =0$$ $$ -\frac xa - \frac yb -1 =0$$