Glencoe Algebra 2, Chapter 2, 2.4, Section 2.4, Problem 9

Given the points are
(x_1,y_1) =(-3,5)
and
(x_2,y_2) =(2,2)
the slope of the line passing through the points is given as
m = (y_2 - y_1)/(x_2 - x_1)
= (2-5)/(2-(-3))
= -3/5
so the slope is -3/5
as the
slope m= -3/5
and the line passes through the point (x,y)= (2, 2)
the slope-intercept form of a line is
y= mx+b
from the above we know m = -3/5 , so the line equation is
y= (-3/5)x+b --------------(1)
we need to find the value of b, as the line passes through the point
(x,y)= (2, 2 ) , then on substituting we get
2 =(-3/5)*(2)+b
=> b = 2+ (6/5) = 16/5
so the equation of the line is
y= (-3/5)x+(16/5)