Question

Write an equation of the line, in standard form, containing the points (−6, 19) and (−15, 28).

Answer 1

The equation of the line that contains (−6, 19) and (−15, 28), in standard form, is: x + y = 13

Recall:

• Equation of a line can be written in standard from as Ax + By = C, where Ax and By are all terms of variable x and y, and C is a constant.

• The equation of a line in point-slope, $y - y_1 = m(x - x_1)$, can be rewritten in the standard form.

• Slope (m) = $\frac{y_2 - y_1}{x_2 - x_1}$

Given: (−6, 19) and (−15, 28)

Find the slope (m):

$m = \frac{28 - 19}{-15 -(-6)} = \frac{9}{-9} = -1$

Write the equation in point-slope form by substituting m = -1 and $(x_1, y_1) = (-6, 19)$ into $y - y_1 = m(x - x_1)$.

$y - 19 = -1(x - (-6))\\\\y - 19 = -1(x + 6)$

• Rewrite in standard form

$y - 19 = -1(x + 6)\\\\y - 19 = -x - 6\\\\y = -x - 6 + 19\\\\y = -x + 13\\\\\mathbf{x + y = 13}$

Therefore, the equation of the line that contains (−6, 19) and (−15, 28), in standard form, is: x + y = 13

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