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

Answer:[tex]y = \frac{1}{2}x+\frac{5}{2}[/tex]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,[tex]Mid-point\ of\ BC = (\frac{6+8}{2},  \frac{8+4}{2})\\= (\frac{14}{2},  \frac{12}{2})\\= (7,6)[/tex]The line will pass through (7,6)Now,[tex]Slope\ of\ BC = m_1 = \frac{y_2-y_1}{x_2-x_1} \\=\frac{4-8}{8-6}\\= \frac{-4}{2}\\= -2[/tex]Letm_2 be the slope of perpendicular bisectorSo,m_1*m_2 = -1-2 * m_2 = -1m_2 = -1/-2 = 1/2The standard equation of line is:y=mx+bWhere m is slopeSo putting the value of slope and point to find the value of b[tex]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}[/tex]..