Question

line A passes through (0. 5) and (4,13). Line B passes through (0,2) and (5, 17)

At what point do the two lines intersect?

Answer 1

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:

\[
\boxed{(3, 11)}
\]