Answer :
Solution:
Given:
[tex](3,4)\text{ and }(-6,5)[/tex]Using the formula;
[tex]\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}[/tex]where;
[tex]\begin{gathered} x_1=3,y_1=4 \\ x_2=-6,y_2=5 \end{gathered}[/tex]Hence,
[tex]\begin{gathered} \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1} \\ \frac{y-4}{x-3}=\frac{5-4}{-6-3} \\ \frac{y-4}{x-3}=\frac{1}{-9} \\ Cross\text{ multiplying;} \\ y-4=-\frac{1}{9}(x-3) \\ y-4=-\frac{1}{9}x+\frac{1}{3} \\ y=-\frac{1}{9}x+\frac{1}{3}+4 \\ y=-\frac{1}{9}x+4\frac{1}{3} \\ y=-\frac{1}{9}x+\frac{13}{3} \end{gathered}[/tex]Therefore, the equation of the line in slope-intercept form is;
[tex]y=-\frac{1}{9}x+\frac{13}{3}[/tex]