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

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)

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Khải Quang 4 years 2021-08-27T13:07:15+00:00 1 Answers 8 views 0

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  1. thienan
    0
    2021-08-27T13:08:18+00:00
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    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.

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