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

Question

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

vocaloid · Answer

use the slope formula to calculate the slope between those two points

anonymous · Answer

i still can't get it @Vocaloid

vocaloid · Answer

slope = m = (y2-y1)/(x2-x1)

vocaloid · Answer

(x1,y1) = (-3,11)
(x2,y2) = (1,-3)

vocaloid · Answer

just plug in the values for x1,x2,y1, and y2, then find the slope.

anonymous · Answer

i got c @Vocaloid

vocaloid · Answer

show me how you got that answer

anonymous · Answer

I'm on a ezam i can't show you i just need to know is that right I'm timed @Vocaloid

vocaloid · Answer

not allowed to help on exams, sorry :(

best of luck~

anonymous · Answer

here this helps me...

anonymous · Answer

http://www.freemathhelp.com/combining-like-terms.html