Line segment JK has endpoints at J(-3,4) and K(3,6). Which is an equation of a line that is perpendicular to line segment JK and passes through the point (3,-12)

y = -3x - 3.

Step-by-step explanation:

Find the slope of the line JK:

Slope = (6-4)/(3- (-3))

= 2 / 6

= 1/3.

Therefore the slope of the line perpendicular to it is - 1 / 1/3

= -3.

Now we use the point-slope form of a line to get the required equation:

y - y1 = m(x - x1)

Here m = -3 and (x1, y1) = (3, -12) so we have

y - (-12) = -3(x - 3)

y + 12 = -3x + 9

y = -3x + 9 - 12

y = -3x - 3 is the answer.

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step-by-step explanation:

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