We check the collinearity of points. If the points A(-15, 5), B(-9, 10) and C(-3, 15) are collinear, then the lines AB and BC are the same slope.
The formula of a slope:
[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
Substitute:
[tex]AB:\ m=\dfrac{10-5}{-9-(-15)}=\dfrac{5}{-9+15}=\dfrac{5}{6}\\\\BC:\ m=\dfrac{15-10}{-3-(-9)}=\dfrac{5}{-3+9}=\dfrac{5}{6}[/tex]
A, B and C are collinear.
The slope-point form:
[tex]y-y_1=m(x-x_1)[/tex]
Substitute:
[tex]y-5=\dfrac{5}{6}(x-(-15))\\\\y-5=\dfrac{5}{6}(x+15)\qquad|\text{use distributive property}\\\\y-5=\dfrac{5}{6}x+\dfrac{75}{6}\qquad|+5=\dfrac{10}{2}\\\\y=\dfrac{5}{6}x+\dfrac{25}{2}+\dfrac{10}{2}\\\\y=\dfrac{5}{6}x+\dfrac{35}{2}[/tex]
x-intercept: put y = 0 to the equation
[tex]0=\dfrac{5}{6}x+\dfrac{35}{2}\qquad|-\dfrac{35}{2}\\\\\dfrac{5}{6}x=-\dfrac{35}{2}\qquad|\cdot6\\\\5x=-3\cdot35\qquad|:5\\\\x=-3\cdot7\\\\x=-21[/tex]
y-intercept: put x = 0 to the equation
[tex]y=\dfrac{5}{6}(0)+\dfrac{35}{2}\\\\y=\dfrac{35}{2}\to y=17.5[/tex]
