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 though the point (0, 1). If the two lines intersects 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 and then determine their point of intersection. Finally, we will calculate the value of \( p + q \).

### Step 1: Find the equation of the first line
The first line passes through the points \( (5, -5) \) and \( (1, 3) \). We can use the point-slope form of the line equation:

\[ \frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1} \]

Let \( (x_1, y_1) = (1, 3) \) and \( (x_2, y_2) = (5, -5) \).

Calculating the slope:

\[ \text{slope} = \frac{-5 - 3}{5 - 1} = \frac{-8}{4} = -2 \]

Now, we can write the equation using the point-slope form:

\[ y - 3 = -2(x - 1) \]

Expanding this:

\[ y - 3 = -2x + 2 \]
\[ y = -2x + 5 \]

### Step 2: Find the equation of the second line
The second line has a slope of 6 and passes through the point \( (0, 1) \). We can use the slope-intercept form of the line equation:

\[ y = mx + c \]

Substituting \( m = 6 \) and using the point \( (0, 1) \) to find \( c \):

\[ 1 = 6(0) + c \implies c = 1 \]

Thus, the equation of the second line is:

\[ y = 6x + 1 \]

### Step 3: Find the intersection of the two lines
To find the intersection point \( (p, q) \), we set the two equations equal to each other:

\[ -2x + 5 = 6x + 1 \]

Rearranging gives:

\[ 5 - 1 = 6x + 2x \]
\[ 4 = 8x \]
\[ x = \frac{4}{8} = \frac{1}{2} \]

Now, substituting \( x = \frac{1}{2} \) back into either equation to find \( y \). We will use the second line's equation:

\[ y = 6\left(\frac{1}{2}\right) + 1 = 3 + 1 = 4 \]

Thus, the intersection point is \( \left(\frac{1}{2}, 4\right) \).

### Step 4: Calculate \( p + q \)
Now we can find \( p + q \):

\[ p + q = \frac{1}{2} + 4 = \frac{1}{2} + \frac{8}{2} = \frac{9}{2} \]

### Final Answer
The value of \( p + q \) is:

\[ \boxed{\frac{9}{2}} \]