Question

A line passes through the points (5, 17) and (12, 38). A second line passes through the points (5, 26) and (8, 44). At what point do the two lines intersect?


(3, 11)


(4, 14)


(0, 0)


(2, 8)

Answer 1

Let's find the equations of the two lines first.

**Line 1** passes through the points (5, 17) and (12, 38).

To find its slope \( m \):
\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{38 - 17}{12 - 5} = \frac{21}{7} = 3
\]

Now, we use the point-slope form of the equation of a line, which is \( y - y_1 = m(x - x_1) \).
Using the point (5, 17):
\[
y - 17 = 3(x - 5)
\]
\[
y - 17 = 3x - 15
\]
\[
y = 3x + 2
\]

So, the equation of Line 1 is:
\[
y = 3x + 2
\]

**Line 2** passes through the points (5, 26) and (8, 44).

To find its slope \( m \):
\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{44 - 26}{8 - 5} = \frac{18}{3} = 6
\]

Using the point-slope form of the equation of a line with the point (5, 26):
\[
y - 26 = 6(x - 5)
\]
\[
y - 26 = 6x - 30
\]
\[
y = 6x - 4
\]

So, the equation of Line 2 is:
\[
y = 6x - 4
\]

To find the intersection of the two lines, we need to set their equations equal to each other:
\[
3x + 2 = 6x - 4
\]

Now, rearranging the equation:
\[
2 + 4 = 6x - 3x
\]
\[
6 = 3x
\]
\[
x = 2
\]

Now, we substitute \( x = 2 \) back into either equation to find the corresponding \( y \).

Using Line 1's equation:
\[
y = 3(2) + 2 = 6 + 2 = 8
\]

Thus, the point of intersection is \( (2, 8) \).

Therefore, the two lines intersect at:
\[
\boxed{(2, 8)}
\]