contestada

WORTH 50 POINTS Line JK passes through points J(–3, 11) and K(1, –3). What is the equation of line JK in standard form?
A) 7x + 2y = –1
B) 7x + 2y = 1
C) 14x + 4y = –1
D) 14x + 4y = 1

Respuesta :

itzvini06
The right answer would be B).
mathstudent55
First, we find the slope of the line using the points.

[tex] m = \dfrac{y_2 - y_1}{x_2 - x_1} [/tex]

[tex] m = \dfrac{-3 - 11}{1 - (-3)} [/tex]

[tex] m = \dfrac{-14}{4} [/tex]

[tex] m = -\dfrac{7}{2} [/tex]

Now we use the point-slope equation of a line.

[tex] y - y_1 = m(x - x_1) [/tex]

[tex] y - 11 = -\dfrac{7}{2}(x - (-3)) [/tex]

[tex] y - 11 = -\dfrac{7}{2}(x + 3) [/tex]

[tex] 2y - 22 = -7x - 21 [/tex]

[tex] 7x + 2y = 1 [/tex]

Answer: Choice B)