2 years ago

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

1 answer:

3 0

<span>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.</span>

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