To find the intersection point of lines A and B, we first need to determine the equations for both lines.
Line A:
Line A passes through points (0, 5) and (4, 13). We can find the slope (m) of line A using the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the points (0, 5) and (4, 13):
\[ m = \frac{13 - 5}{4 - 0} = \frac{8}{4} = 2 \]
Using point-slope form, the equation of line A is:
\[ y - y_1 = m(x - x_1) \]
Using point (0, 5):
\[ y - 5 = 2(x - 0) \]
This simplifies to:
\[ y = 2x + 5 \]
Line B:
Line B passes through points (0, 2) and (5, 17). We can similarly find the slope of line B:
\[ m = \frac{17 - 2}{5 - 0} = \frac{15}{5} = 3 \]
Using point-slope form for line B:
\[ y - 2 = 3(x - 0) \]
This simplifies to:
\[ y = 3x + 2 \]
Now we have the equations of both lines:
- Line A: \(y = 2x + 5\)
- Line B: \(y = 3x + 2\)
To find the intersection, we set the equations equal to each other:
\[ 2x + 5 = 3x + 2 \]
Solving for \(x\):
\[ 5 - 2 = 3x - 2x \] \[ 3 = x \]
Now that we have \(x = 3\), we can substitute it back into either equation to find the corresponding \(y\) value. We’ll use line A:
\[ y = 2(3) + 5 = 6 + 5 = 11 \]
Therefore, the point of intersection of the two lines is:
\[ \boxed{(3, 11)} \]