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A diagram shown has point A( -2 , 4 ) , B ( 6, 2 ), C (-4,-4) find the equation of the line perpendicular to BC and passing through the midpoint of BC (M). Give answers in general form.

Gerry Myerson
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sid
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    Author unresponsive to comments. Question appears to be abandoned. Voting to close. – Gerry Myerson Oct 31 '13 at 12:06

1 Answers1

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To find the equation of the line perpendicular to the line $BC$ formed by the points $B = ( 6, 2 ), \;C = (-4,-4)$, we need to

  • first, find the slope of line $BC$: $$\text{slope of BC}:\;\; m = \dfrac{y_2-y_1}{x_2 - x_1}$$

  • Then, the slope of the line perpendicular to $BC$ is given by $m_\perp = \left( -\dfrac 1m\right)$.

  • Next, you need to determine the midpoint $(x_0, y_0)$ of line $BC$ using the given points: $B = ( 6, 2 ), \;C = (-4,-4)$.

  • Now you'll have the slope $m_\perp$ and the point $(x_0,y_0)$ on the perpendicular line from which you can create the equation of the desired line: $$y - y_0 = m(x - x_0)$$

amWhy
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  • I like answers like this, nicely steps them through, but they have to do work. +1 – Amzoti Oct 31 '13 at 13:10
  • @Amzoti Thanks. Yes, with questions like this, I prefer to outline the process. When I get feedback from the OP, if I get feedback showing effort, I'm willing to elaborate, clarify, confirm. – amWhy Oct 31 '13 at 13:21