anairaanidem98
anairaanidem98 anairaanidem98
  • 15-06-2018
  • Mathematics
contestada

Line JK passes through points J(–3, 11) and K(1, –3). What is the equation of line JK in standard form?

Respuesta :

gmany
gmany gmany
  • 13-08-2018

The standard form of a linear equation: Ax + By = C.

The slope-point formula:

[tex]y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

We have:

[tex]J(-3,\ 11)\to x_1=-3,\ y_1=11\\K(1,\ -3)\to x_2=1,\ y_2=-3[/tex]

Substitute:

[tex]m=\dfrac{-3-11}{1-(-3)}=\dfrac{-14}{4}=-\dfrac{7}{2}\\\\y-11=-\dfrac{7}{2}(x-(-3))\\\\y-11=-\dfrac{7}{2}x-\dfrac{21}{2}\ \ \ \ |\cdot2\\\\2y-22=-7x-21\ \ \ |+22\\\\2y=-7x+1\ \ \ \ |+7x\\\\7x+2y=1[/tex]

Answer: JK: 7x + 2y = 1

Answer Link
msblackberrybaby8
msblackberrybaby8 msblackberrybaby8
  • 14-05-2021

Answer:

7x + 2y= -1 this is the answer

Answer Link

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