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?

A) 7x + 2y = –1
B) 7x + 2y = 1
C) 14x + 4y = –1
D) 14x + 4y = 1

Answer 1

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


`m = (y_2 - y_1)/(x_2 - x_1)`


`m = (-3 - 11)/(1 - (-3))`


`m = (-14)/(4)`


`m = -(7)/(2)`

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


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


`y - 11 = -(7)/(2)(x - (-3))`


`y - 11 = -(7)/(2)(x + 3)`


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


`7x + 2y = 1`

Answer: Choice B)

Answer 2

The right answer would be B).