Answer:
[tex]q+p=7[/tex]
Step-by-step explanation:
we know that
If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)
In this problem line AB and line BC are perpendicular
so
[tex]m_A_B*m_B_C=-1[/tex]
step 1
Find the slope AB
we have
A(p,4), B(6,1)
The formula to calculate the slope between two points is equal to
[tex]m=\frac{y2-y1}{x2-x1}[/tex]
substitute the values
[tex]m=\frac{1-4}{6-p}[/tex]
[tex]m_A_B=-\frac{3}{6-p}[/tex]
step 2
Find the slope BC
we have
B(6,1), and C(9,q)
The formula to calculate the slope between two points is equal to
[tex]m=\frac{y2-y1}{x2-x1}[/tex]
substitute the values
[tex]m=\frac{q-1}{9-6}[/tex]
[tex]m_B_C=\frac{q-1}{3}[/tex]
step 3
Find the equation that relates p and q
we know that
[tex]m_A_B*m_B_C=-1[/tex]
we have
[tex]m_A_B=-\frac{3}{6-p}[/tex]
[tex]m_B_C=\frac{q-1}{3}[/tex]
substitute
[tex](-\frac{3}{6-p})(\frac{q-1}{3})=-1[/tex]
[tex](\frac{q-1}{6-p})=1[/tex]
[tex]q-1=6-p\\q+p=7[/tex]