10.

A line passes through (2, –1) and (8, 4).
a. Write an equation for the line in point-slope form.
b. Rewrite the equation in standard form using integers.

1 answer

To find the equation of the line that passes through the points (2, -1) and (8, 4), we will follow the steps below.

a. Write an equation for the line in point-slope form.

  1. Calculate the slope (m) of the line: \[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-1)}{8 - 2} = \frac{4 + 1}{8 - 2} = \frac{5}{6} \]

  2. Use the point-slope form of the line equation: The point-slope form is given by: \[ y - y_1 = m(x - x_1) \] Using point (2, -1): \[ y - (-1) = \frac{5}{6}(x - 2) \] Simplifying this gives: \[ y + 1 = \frac{5}{6}(x - 2) \] Thus, the equation in point-slope form is: \[ y + 1 = \frac{5}{6}(x - 2) \]

b. Rewrite the equation in standard form using integers.

  1. Distributing and rearranging: Starting from the point-slope form: \[ y + 1 = \frac{5}{6}(x - 2) \]

    Distribute: \[ y + 1 = \frac{5}{6}x - \frac{5}{3} \]

    Subtract 1 from both sides: \[ y = \frac{5}{6}x - \frac{5}{3} - 1 \]

    Convert -1 to a fraction with a common denominator (-3): \[ y = \frac{5}{6}x - \frac{5}{3} - \frac{3}{3} \] This simplifies to: \[ y = \frac{5}{6}x - \frac{8}{3} \]

  2. Clear the fractions by multiplying the entire equation by 6 (the denominator): \[ 6y = 5x - 16 \]

  3. Rearranging to standard form (Ax + By = C): To get it into standard form, bring \(5x\) to the left side: \[ -5x + 6y = -16 \]

    To convert to the traditional standard form (with A as a positive integer), multiply through by -1: \[ 5x - 6y = 16 \]

Thus, the final equations are:

  • Point-slope form: \( y + 1 = \frac{5}{6}(x - 2) \)
  • Standard form: \( 5x - 6y = 16 \)