Jane drew a point A located at (-3,-0) and point Clocated at (12,-6). What is the x-coordinate of point B that bisects AC?

Question

Jane drew a point A located at (-3,-0) and point Clocated at (12,-6). What is the x-coordinate of point B that bisects AC?

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Helga 3 years 2021-07-19T15:51:54+00:00 1 Answers 6 views 0

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    2021-07-19T15:53:23+00:00

    Answer:

    The x-coordinate of point B that bisects AC is \frac{9}{2}.

    Step-by-step explanation:

    The statement is not correct, the correct form is:

    Jane drew a point A located at (-3,0) and point Clocated at (12,-6). What is the x-coordinate of point B that bisects AC?

    Where a point bisects a segment, it means that segment is divided into two equal parts. If we know that \vec A = (-3, 0) and \vec C = (12, -6) are endpoints of the segment, the location of the endpoint can be found by the following vectorial formula:

    \vec B = \vec A + \frac{1}{2}\cdot \overrightarrow {AC}

    \vec B = \vec A + \frac{1}{2} \cdot (\vec C-\vec A)

    \vec B = \frac{1}{2}\cdot \vec A + \frac{1}{2}\cdot \vec C

    \vec B = \frac{1}{2}\cdot (-3,0)+\frac{1}{2}\cdot (12,-6)

    \vec B = \left(-\frac{3}{2}, 0 \right)+\left(6,-3\right)

    \vec B = \left(-\frac{3}{2}+6, 0-3\right)

    \vec B = \left(\frac{9}{2},-3\right)

    The x-coordinate of point B corresponds to the first component of the ordered pair found above, that is, x_{B} = \frac{9}{2}.

    The x-coordinate of point B that bisects AC is \frac{9}{2}.

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