Equation of the line that has $x$ and $y$ intercepts at $a$ and $b$. 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.
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 $$\frac xa + \frac yb - 1 = 0$$
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..
$$ \frac xa + \frac yb -1 =0$$

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

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