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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?
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Aug 17, 2019
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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?
Mathematics
middle-school
JoeTomks
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J(–3, 11) and K(1, –3)
Slope = (11 + 3)/(-3 - 1) = 14/-4 = -7/2
Equation
y + 3 = -7/2(x - 1)
2y + 6 = -7x + 7
2y = -7x + 1
7x + 2y = 1 ....This is standard form (Ax + By = C)
Hope it helps.
Praneeth Peiris
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Aug 23, 2019
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Praneeth Peiris
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