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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? a. 7x + 2y = –1 b. 7x + 2y = 1 c. 14x + 4y = –1 d. 14x + 4y = 1
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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?

a. 7x + 2y = –1
b. 7x + 2y = 1
c. 14x + 4y = –1
d. 14x + 4y = 1

User Dexture
by
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2 Answers

3 votes

Answer:

b

Explanation:


User Liondancer
by
8.0k points
6 votes
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
User Chankruze
by
8.0k points
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