# How do you write the equation of the line in the form AX+BY=C if (9, -8) and (0, 3)?

Sep 20, 2015

$\frac{11}{9} x + y = 3$

#### Explanation:

Okay, the line passes through these two points, and the equation of a line is usually in the form $y = a x + b$.

Then the point (9, -8) is because:
$- 8 = a \cdot 9 + b$
and the point (0, 3) is because:
$3 = a \cdot 0 + b$

We just need to solve for a and b to find the equation of the line.
From the second equation we readily see that $b = 3$.

Now substitute this $b$ into the equation of the first point:
$- 8 = a \cdot 9 + 3$
Rearranging, we have:
$9 a = - 11$
so $a = - \frac{11}{9}$

So the equation of the line is: $y = - \frac{11}{9} x + 3$
Rearranging, we get:
$\frac{11}{9} x + y = 3$
which is in the desired form $A x + B y = C$
where $A = \frac{11}{9}$, $B = 1$, and $C = 3$