Glencoe Algebra 2, Chapter 2, 2.3, Section 2.3, Problem 49

Given a line 3x - 2y = 24
we need to find a line which is perpendicular to the line 3x - 2y = 24 and passes through the x- intercept of the line 3x - 2y = 24
let us first find the slope of the line 3x - 2y = 24
so, getting the line in the standard form as follows
3x - 2y = 24
=> -2y = 24 -3x
=> 2y = 3x -24
=> y = (3/2)x - 24/2
=> y=(3/2)x - 12
so the slope of the line 3x - 2y = 24 is m_1 = 3/2
let the slope of the line which we need to find be m_2
as the product of the slopes of perpendicular lines is -1
so the equation is given as
(m_1)(m_2) = -1
=> (3/2)(m_2) = -1
=> m_2 = -2/3
now let us find the x- intercept of the line3x - 2y = 24
x- intercept(y=0)
3x- 2(0) = 24
=> 3x = 24
=> x = 8
so the desired line which we wanted passes through the point (8,0) and the slope of the line is -2/3
so the equation of the line is
y = (-2/3)x +c
=> as it passes through (8,0) so,
0 =(-2/3)(8) +c
=> c= 16/3
so the equation of the line is y = (-2/3)x + 16/3
the graph is plotted as below