The points are given as,
[tex]\begin{gathered} (x_1,y_1)=(-5,4) \\ (x_2,y_2)=(9,-2) \end{gathered}[/tex]
The equation of a line in double point form is given as,
[tex]\begin{gathered} \frac{y-y_1}{y_2-y_1}\text{ = }\frac{x-x_1}{x_2-x_1} \\ \frac{y-4}{-2-4}\text{ = }\frac{x+5}{9+5} \\ \frac{y-4}{-6}\text{ = }\frac{x+5}{14} \end{gathered}[/tex]
Simplifying further,
[tex]\begin{gathered} 14(y-4)\text{ = -6\lparen x+5\rparen} \\ 14y\text{ - 56 = -6x -30} \\ 6x+14y\text{ = 56-30} \\ 6x\text{ + 14y = 26} \end{gathered}[/tex]
Converting the equation to slope-intercept form,slope-intercept
[tex]\begin{gathered} 6x\text{ + 14y = 26} \\ 3x\text{ + 7y = 13} \\ 7y\text{ = -3x + 13} \\ y\text{ = }\frac{-3x}{7}+\frac{13}{7} \\ \end{gathered}[/tex]
Thus the required equation of line is,
[tex]y=\frac{-3x}{7}\text{+}\frac{13}{7}[/tex]
