Question
Line JK passes through points J(–3, 11) and K(1, –3). What is the equation of line JK in standard form?
Asked by: USER6148
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103 Answers
Answer (103)
(-3,11) & (1,-3)
x1 y1 x2 y2
-3-11 -14 -7
1+3 4 2
y=-7/2x+b
11=-7/2(-3) +b
11=21/2+b
22/2=21/2+b (subtract 21/2 from each side)
1/2=b
so it would be y=-7/2x + 1/2
x1 y1 x2 y2
-3-11 -14 -7
1+3 4 2
y=-7/2x+b
11=-7/2(-3) +b
11=21/2+b
22/2=21/2+b (subtract 21/2 from each side)
1/2=b
so it would be y=-7/2x + 1/2