Question

Triangle ABC has vertices A(0,0) B(6,8) and C(8,4). Which equation represents the perpendicular bisected of BC?

Answer 1

Answer:

$y = \frac{1}{2}x+\frac{5}{2}$

Step-by-step explanation:

The perpendicular bisector of a line passes through the mid-point of the line and the product of slopes of the line and perpendicular bisector will be -1.

So,

$Mid-point\ of\ BC = (\frac{6+8}{2}, \frac{8+4}{2})\\= (\frac{14}{2}, \frac{12}{2})\\= (7,6)$

The line will pass through (7,6)

Now,

$Slope\ of\ BC = m_1 = \frac{y_2-y_1}{x_2-x_1} \\=\frac{4-8}{8-6}\\= \frac{-4}{2}\\= -2$

Let

m_2 be the slope of perpendicular bisector

So,

m_1*m_2 = -1

-2 * m_2 = -1

m_2 = -1/-2 = 1/2

The standard equation of line is:

y=mx+b

Where m is slope

So putting the value of slope and point to find the value of b

$6 = \frac{1}{2}*7 +b\\ 6 = \frac{7}{2} + b\\b = 6 - \frac{7}{2}\\ b = \frac{12-7}{2}\\b = \frac{5}{2}\\So,\ the\ equation\ of\ perpendcular\ bisector\ of\ BC\ is:\\y = \frac{1}{2}x+\frac{5}{2}$

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