Question

The line l1 passes through the points P(-1,2) and Q(11,8).
Find the equation for l1 in the form y=mx+c where m and c are constants

Answer 1

To find the equation of the line \( l_1 \) that passes through the points \( P(-1, 2) \) and \( Q(11, 8) \), we first need to calculate the slope \( m \) of the line.

The formula for the slope \( m \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is:

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

Here, we can assign:

- \( (x_1, y_1) = (-1, 2) \)
- \( (x_2, y_2) = (11, 8) \)

Now, substituting the coordinates into the slope formula:

\[
m = \frac{8 - 2}{11 - (-1)} = \frac{6}{11 + 1} = \frac{6}{12} = \frac{1}{2}
\]

Now that we have the slope \( m = \frac{1}{2} \), we can use the point-slope form of the equation of a line to find the equation in the form \( y = mx + c \). The point-slope form is given by:

\[
y - y_1 = m(x - x_1)
\]

Using point \( P(-1, 2) \):

\[
y - 2 = \frac{1}{2}(x - (-1))
\]

Simplifying this:

\[
y - 2 = \frac{1}{2}(x + 1)
\]

Now distribute on the right side:

\[
y - 2 = \frac{1}{2} x + \frac{1}{2}
\]

Next, we add \( 2 \) to both sides to solve for \( y \):

\[
y = \frac{1}{2} x + \frac{1}{2} + 2
\]

\[
y = \frac{1}{2} x + \frac{5}{2}
\]

Thus, the equation of the line \( l_1 \) in the form \( y = mx + c \) is:

\[
y = \frac{1}{2}x + \frac{5}{2}
\]

This is the final equation for the line passing through the points \( P(-1,2) \) and \( Q(11,8) \).