The graph of a line in the xy-plane passes through the points `(5, -5) and (1, 3)`. The graph of a second line has a slope of 6 and passes through the point `(-1, 15)`. If the two lines intersect at `(p, q)`, what is the value of `p+q`?

Answer

To solve the problem, we need to find the equations of the two lines based on the given points and slope, and then determine their intersection point. Finally, we will calculate \( p + q \) from the intersection point \( (p, q) \).

### Step 1: Find the equation of the first line

The first line passes through the points \( (5, -5) \) and \( (1, 3) \).

1. **Calculate the slope (m)**:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-5)}{1 - 5} = \frac{3 + 5}{1 - 5} = \frac{8}{-4} = -2 \]

2. **Use the point-slope form of the line equation**:
The point-slope form is given by:
\[ y - y_1 = m(x - x_1) \]
Using point \( (5, -5) \):
\[ y - (-5) = -2(x - 5) \]
Simplifying:
\[ y + 5 = -2x + 10 \]
\[ y = -2x + 5 \]

### Step 2: Find the equation of the second line

The second line has a slope of \( 6 \) and passes through the point \( (-1, 15) \).

1. **Use the slope-intercept form**:
The slope-intercept form is given by:
\[ y = mx + b \]
Substituting the slope and point:
\[ 15 = 6(-1) + b \]
\[ 15 = -6 + b \implies b = 15 + 6 = 21 \]
Thus, the equation of the second line is:
\[ y = 6x + 21 \]

### Step 3: Find the intersection point of the two lines

We have the equations:
1. \( y = -2x + 5 \)
2. \( y = 6x + 21 \)

Set them equal to find \( x \):
\[ -2x + 5 = 6x + 21 \]
Rearranging gives:
\[ 5 - 21 = 6x + 2x \]
\[ -16 = 8x \implies x = -2 \]

Now, substitute \( x = -2 \) back into one of the equations to find \( y \):
Using the first line's equation:
\[ y = -2(-2) + 5 = 4 + 5 = 9 \]

Thus, the intersection point is \( (-2, 9) \).

### Step 4: Calculate \( p + q \)

Here, \( p = -2 \) and \( q = 9 \):
\[ p + q = -2 + 9 = 7 \]

### Final Answer
The value of \( p + q \) is \( \boxed{7} \).