Question

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?

Answer 1

The right answer would be B).

Answer 2

First, we find the slope of the line using the points.

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

$m = \dfrac{-3 - 11}{1 - (-3)}$

$m = \dfrac{-14}{4}$

$m = -\dfrac{7}{2}$

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

$y - y_1 = m(x - x_1)$

$y - 11 = -\dfrac{7}{2}(x - (-3))$

$y - 11 = -\dfrac{7}{2}(x + 3)$

$2y - 22 = -7x - 21$

$7x + 2y = 1$

Answer: Choice B)