Question

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)

Answer 1

The equation of a line perpendicular to line segment JK and passing through (3, -12) is y = -3x - 3, where we have used the negative reciprocal of the slope of JK to find the required slope of the perpendicular line.

Step-by-step explanation:

To find the equation of a line that is perpendicular to line segment JK with endpoints J(-3,4) and K(3,6), and passes through the point (3,-12), we first need to determine the slope of JK. The slope of a line passing through two points (x1, y1) and (x2, y2) is given by (y2 - y1) / (x2 - x1). For JK, the slope is (6 - 4) / (3 - (-3)) = 2/6 = 1/3. A line perpendicular to this must have a slope that is the negative reciprocal of 1/3, which is -3.

The equation of a line with slope m and passing through (x0, y0) is given by y - y0 = m(x - x0). Substituting the given point (3, -12) and the slope -3, we get y - (-12) = -3(x - 3), which simplifies to y + 12 = -3x + 9 or y = -3x - 3 as the equation of the desired perpendicular line.

Answer 2

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.