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

Sagot :

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, [tex]y - y_1 = m(x - x_1)[/tex], can be rewritten in the standard form.
  • Slope (m) = [tex]\frac{y_2 - y_1}{x_2 - x_1}[/tex]

Given: (−6, 19) and (−15, 28)

Find the slope (m):

[tex]m = \frac{28 - 19}{-15 -(-6)} = \frac{9}{-9} = -1[/tex]

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

[tex]y - 19 = -1(x - (-6))\\\\y - 19 = -1(x + 6)[/tex]

  • Rewrite in standard form

[tex]y - 19 = -1(x + 6)\\\\y - 19 = -x - 6\\\\y = -x - 6 + 19\\\\y = -x + 13\\\\\mathbf{x + y = 13}[/tex]

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

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