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

A7x + 2y = –1
B7x + 2y = 1
C14x + 4y = –1
D14x + 4y = 1

Verified Answer :

Nirina7

if the Line JK passes through points J(–3, 11) and K(1, –3), so the the equation of line JK in standard form is b. 7x + 2y = 1

proof:
for J(–3, 11),  7(-3) +2(11)=-21+22=1
for K(1, –3),  7(1) +2(-3)= 7-6=1

Answer:

b

Step-by-step explanation:


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