Question

The coordinates of A, B, and C in the diagram are A(p,4), B(6,1), and C(9,q). Which equation correctly relates p and q?

Answer 1

Answer:

$q+p=7$

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

$m_A_B*m_B_C=-1$

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

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{1-4}{6-p}$

$m_A_B=-\frac{3}{6-p}$

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

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{q-1}{9-6}$

$m_B_C=\frac{q-1}{3}$

step 3

Find the equation that relates p and q

we know that

$m_A_B*m_B_C=-1$

we have

$m_A_B=-\frac{3}{6-p}$

$m_B_C=\frac{q-1}{3}$

substitute

$(-\frac{3}{6-p})(\frac{q-1}{3})=-1$

$(\frac{q-1}{6-p})=1$

$q-1=6-p\\q+p=7$