Answer :
To determine the relationship between [tex]\( p \)[/tex] and [tex]\( q \)[/tex] given that line segment [tex]\( \overleftrightarrow{AB} \)[/tex] is perpendicular to line segment [tex]\( \overleftrightarrow{BC} \)[/tex], we need to calculate the slopes of these line segments and use the condition that the product of their slopes equals [tex]\(-1\)[/tex].
Here are the steps:
### Step 1: Find the slope of [tex]\( \overleftrightarrow{AB} \)[/tex]
The slope of a line passing through two points [tex]\( \left(x_1,y_1\right) \)[/tex] and [tex]\( \left(x_2,y_2\right) \)[/tex] is given by:
[tex]\[ \text{slope of } \overleftrightarrow{AB} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
For points [tex]\( A(p, 4) \)[/tex] and [tex]\( B(6, 1) \)[/tex]:
[tex]\[ \text{slope of } \overleftrightarrow{AB} = \frac{1 - 4}{6 - p} = \frac{-3}{6 - p} \][/tex]
### Step 2: Find the slope of [tex]\( \overleftrightarrow{BC} \)[/tex]
Similarly, using points [tex]\( B(6, 1) \)[/tex] and [tex]\( C(9, q) \)[/tex]:
[tex]\[ \text{slope of } \overleftrightarrow{BC} = \frac{q - 1}{9 - 6} = \frac{q - 1}{3} \][/tex]
### Step 3: Use the perpendicular slopes condition
Since [tex]\( \overleftrightarrow{AB} \)[/tex] is perpendicular to [tex]\( \overleftrightarrow{BC} \)[/tex], their slopes multiply to give [tex]\(-1\)[/tex]:
[tex]\[ \left( \frac{-3}{6 - p} \right) \times \left( \frac{q - 1}{3} \right) = -1 \][/tex]
### Step 4: Simplify the equation
Simplify the product of the slopes and solve for [tex]\( q \)[/tex]:
[tex]\[ \frac{-3(q - 1)}{3(6 - p)} = -1 \][/tex]
[tex]\[ \frac{-3(q - 1)}{3(6 - p)} = \frac{-3(q - 1)}{18 - 3p} = -1 \][/tex]
Now multiply both sides by [tex]\( 18 - 3p \)[/tex] to clear the fraction:
[tex]\[ -3(q - 1) = -1 \times (18 - 3p) \][/tex]
[tex]\[ -3(q - 1) = -(18 - 3p) \][/tex]
### Step 5: Expand and simplify
Distribute the -1 on the right side:
[tex]\[ -3q + 3 = -18 + 3p \][/tex]
Combine like terms:
[tex]\[ 3p - 3q = 18 + 3 \][/tex]
[tex]\[ 3p - 3q = 21 \][/tex]
Divide both sides by 3:
[tex]\[ p - q = 7 \][/tex]
### Conclusion
The correct equation that relates [tex]\( p \)[/tex] and [tex]\( q \)[/tex] is:
[tex]\[ p - q = 7 \][/tex]
Therefore, the correct option is:
C. [tex]\( p - q = 7 \)[/tex]
Here are the steps:
### Step 1: Find the slope of [tex]\( \overleftrightarrow{AB} \)[/tex]
The slope of a line passing through two points [tex]\( \left(x_1,y_1\right) \)[/tex] and [tex]\( \left(x_2,y_2\right) \)[/tex] is given by:
[tex]\[ \text{slope of } \overleftrightarrow{AB} = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
For points [tex]\( A(p, 4) \)[/tex] and [tex]\( B(6, 1) \)[/tex]:
[tex]\[ \text{slope of } \overleftrightarrow{AB} = \frac{1 - 4}{6 - p} = \frac{-3}{6 - p} \][/tex]
### Step 2: Find the slope of [tex]\( \overleftrightarrow{BC} \)[/tex]
Similarly, using points [tex]\( B(6, 1) \)[/tex] and [tex]\( C(9, q) \)[/tex]:
[tex]\[ \text{slope of } \overleftrightarrow{BC} = \frac{q - 1}{9 - 6} = \frac{q - 1}{3} \][/tex]
### Step 3: Use the perpendicular slopes condition
Since [tex]\( \overleftrightarrow{AB} \)[/tex] is perpendicular to [tex]\( \overleftrightarrow{BC} \)[/tex], their slopes multiply to give [tex]\(-1\)[/tex]:
[tex]\[ \left( \frac{-3}{6 - p} \right) \times \left( \frac{q - 1}{3} \right) = -1 \][/tex]
### Step 4: Simplify the equation
Simplify the product of the slopes and solve for [tex]\( q \)[/tex]:
[tex]\[ \frac{-3(q - 1)}{3(6 - p)} = -1 \][/tex]
[tex]\[ \frac{-3(q - 1)}{3(6 - p)} = \frac{-3(q - 1)}{18 - 3p} = -1 \][/tex]
Now multiply both sides by [tex]\( 18 - 3p \)[/tex] to clear the fraction:
[tex]\[ -3(q - 1) = -1 \times (18 - 3p) \][/tex]
[tex]\[ -3(q - 1) = -(18 - 3p) \][/tex]
### Step 5: Expand and simplify
Distribute the -1 on the right side:
[tex]\[ -3q + 3 = -18 + 3p \][/tex]
Combine like terms:
[tex]\[ 3p - 3q = 18 + 3 \][/tex]
[tex]\[ 3p - 3q = 21 \][/tex]
Divide both sides by 3:
[tex]\[ p - q = 7 \][/tex]
### Conclusion
The correct equation that relates [tex]\( p \)[/tex] and [tex]\( q \)[/tex] is:
[tex]\[ p - q = 7 \][/tex]
Therefore, the correct option is:
C. [tex]\( p - q = 7 \)[/tex]
