Respuesta :
Answer:
Part 1) The equation of line p is [tex]y=4x+16[/tex]
Part 2) The equation of line q is [tex]y=-\frac{1}{4}x+\frac{13}{4}[/tex]
Part 3) The equation of line r is [tex]y=4[/tex]
The graph in the attached figure
Step-by-step explanation:
step 1
Find the equation of the line p
we have
[tex]m=4\\point\ (-3,4)[/tex]
Find the equation in point slope form
[tex]y-y1=m(x-x1)[/tex]
substitute
[tex]y-4=4(x+3)[/tex] ----> equation in point slope form
Convert to slope intercept form
[tex]y-4=4x+12[/tex]
[tex]y=4x+12+4[/tex]
[tex]y=4x+16[/tex]
step 2
Find the equation of line q
we know that
If two lines are perpendicular, then their slope are opposite reciprocal (the product of their slopes is equal to -1)
we have that
Line q is perpendicular to line q
so
[tex]m_q*m_p=-1[/tex]
we have
[tex]m_p=4[/tex]
therefore
[tex]m_q=-\frac{1}{4}[/tex]
we have
[tex]m=-\frac{1}{4}\\point\ (-3,4)[/tex]
Find the equation in point slope form
[tex]y-y1=m(x-x1)[/tex]
substitute
[tex]y-4=-\frac{1}{4}(x+3)[/tex] ----> equation in point slope form
Convert to slope intercept form
[tex]y-4=-\frac{1}{4}x-\frac{3}{4}[/tex]
[tex]y=-\frac{1}{4}x-\frac{3}{4}+4[/tex]
[tex]y=-\frac{1}{4}x+\frac{13}{4}[/tex]
step 3
Find the equation of line r
we know that
Line r passes through Quadrants 1 and Quadrants 2 only and passes through the point (-3,4)
therefore
The equation of Line r must be equal to the y-coordinate of the given point
[tex]y=4[/tex] ----> is a horizontal line
step 4
Graph the three lines
using a graphing tool
see the attached figure
