To find the intersection of lines P and Q, we need to start by determining the equations of both lines.
Step 1: Find the equation of Line P
Line P passes through points (2, 9) and (4, 13).
First, we calculate the slope \( m \) of line P:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{13 - 9}{4 - 2} = \frac{4}{2} = 2 \]
Using the point-slope form of the line equation \( y - y_1 = m(x - x_1) \), we can use point (2, 9):
\[ y - 9 = 2(x - 2) \]
Expanding this:
\[ y - 9 = 2x - 4 \ y = 2x + 5 \]
Step 2: Find the equation of Line Q
Line Q passes through points (1, -13) and (3, -19).
Calculating the slope \( m \) of line Q:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-19 - (-13)}{3 - 1} = \frac{-19 + 13}{2} = \frac{-6}{2} = -3 \]
Using point-slope form with point (1, -13):
\[ y - (-13) = -3(x - 1) \]
Expanding this:
\[ y + 13 = -3x + 3 \ y = -3x - 10 \]
Step 3: Finding the intersection of Line P and Line Q
Now we have the equations of the lines:
- Line P: \( y = 2x + 5 \)
- Line Q: \( y = -3x - 10 \)
To find the point of intersection, set the equations equal to each other:
\[ 2x + 5 = -3x - 10 \]
Solving for \( x \):
\[ 2x + 3x = -10 - 5 \ 5x = -15 \ x = -3 \]
Now, substitute \( x = -3 \) back into one of the line equations to find \( y \). We can use Line P:
\[ y = 2(-3) + 5 \ y = -6 + 5 \ y = -1 \]
Conclusion
The lines intersect at the point \((-3, -1)\).
Thus, we have:
- \( x = -3 \)
- \( y = -1 \)
So, the complete sentence is:
Line P and Line Q intersect at the point (x,y), where \( x = -3 \) and \( y = -1 \).
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