Question

write an equation of a line parallel to AB through point (1,5) if A (6,4) and B (9,6)

Answer 1

To find the equation of a line parallel to the line segment AB that passes through the point (1, 5), we first need to determine the slope of the line AB.

The slope \( m \) of a line through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula:

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

For points A(6, 4) and B(9, 6):

- \( x_1 = 6, y_1 = 4 \)
- \( x_2 = 9, y_2 = 6 \)

Now, substituting these values into the slope formula:

\[
m = \frac{6 - 4}{9 - 6} = \frac{2}{3}
\]

The slope of the line AB is \( \frac{2}{3} \).

Since we want the equation of a line parallel to AB, it will have the same slope \( \frac{2}{3} \).

Next, we can use the point-slope form of a line equation, which is given by:

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

where \((x_1, y_1)\) is the point through which the line passes. Here, we will use point (1,5):

- \( m = \frac{2}{3} \)
- \( x_1 = 1, y_1 = 5 \)

Substituting these values into the point-slope form equation:

\[
y - 5 = \frac{2}{3}(x - 1)
\]

Now let’s simplify this equation to get it into slope-intercept form \( y = mx + b \):

\[
y - 5 = \frac{2}{3}x - \frac{2}{3}
\]
\[
y = \frac{2}{3}x - \frac{2}{3} + 5
\]
\[
y = \frac{2}{3}x - \frac{2}{3} + \frac{15}{3}
\]
\[
y = \frac{2}{3}x + \frac{13}{3}
\]

Thus, the equation of the line parallel to line AB that passes through the point (1, 5) is:

\[
y = \frac{2}{3}x + \frac{13}{3}
\]