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## 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 thro

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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)

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Mathematics
5 months
2021-08-27T13:07:15+00:00
2021-08-27T13:07:15+00:00 1 Answers
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## Answers ( )

Answer: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.