Glencoe Algebra 2, Chapter 2, 2.3, Section 2.3, Problem 44

The line which we needed is perpendicular to a line whose slope is (-3/2) .
the product of the slopes of two lines which are perpendicular is equal to -1
let the slope of the line which we need to find be m_1 and the slope of the other line be m_2 = (-3/2).
so ,
m_1 * m_2 = -1
=> m_1 = -1/(m_2) = -1 /(-3/2) = 1/(3/2) = 2/3
so , m_1 = 2/3
and the line of slope m_1 passes through the point (-4,1)
then the line is
y = mx+c
=> y = (2/3)x +c
as it passes through (x,y)=(-4,1) so
=> 1= (2/3)(-4) +c
=> 1= -8/3 +c
=> 1+8/3 = c
=> c = 11/3
so the equation of the line is y= (2/3)x+ 11/3 and the graph plotted is as follows in the attachments. the point (-4,1) is spotted with a green dot.