Using the "gradient, two-point form", we have:
[tex] \frac{y - y1}{x - x1} = \frac{y2 - y1}{x2 - x1} [/tex]
From the given data, we have that
[tex]y2 = - 3 \\ y1 = 3 \\ x2 = 5 \\ x1 = 0[/tex]
Therefore, substitute in the given values:
We have that:
[tex] \frac{y - (3)}{x - (0)} = \frac{( - 3) - (3)}{(5) - (0)} [/tex]
Simplify further
[tex] \frac{y - 3}{x} = \frac{ - 6}{5} [/tex]
Multiply both sides by the LCM, 5x:
[tex]5x( \frac{y - 3}{ x} ) = 5x( \frac{ - 6}{5} )[/tex]
Simplify:
[tex]5( y - 3) = x( - 6)[/tex]
[tex]5y - 15 = - 6x[/tex]
Note that the equation of a straight line is given by Ax + By = C where A and B are coefficients of x and y respectively. C is the constant. Hence:
6x+5y=15