Line JK passes through points J(-3, 11) and K(1, -3). What is the equation of line JK in standard form? The answers are 7x + 2y = -1
7x + 2y = 1
14x + 4y = -1
14x + 4y = 1

To find the standard form of line JK, we have to find the slope, so we can put these information to y-y1 = m (x-x1)

To find the slope, we can use the equation of m = y2 - y1 / x2 - x1

So the slow is -3 - 11 / 1 - (-3) = -14 / 4 = -7/2

To put this on that equation,

y - 11 = -7/2 ( x + 3)

y - 11 = -7/2 x - 21/2

y + 7/2 x = 11 - 21/2

To multiply this result and 2,

2y + 7x = 22 - 21 = 1

So the answer is 7x + 2y = 1

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