Answer:
[tex]y=\frac{1}{3}x[/tex]
Step-by-step explanation:
The given line segment has a midpoint at (3, 1) and goes through (2, 4), (3, 1), and (4, -2). We can use any two of the three points to calculate the equation of the line. Let us use the points (2, 4) and (4, -2)
Therefore the line goes through (2, 4) and (4, -2). The equation of a line passing through [tex](x_1,y_1)\ and\ (x_2,y_2)[/tex] is:
[tex]\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}[/tex].
Therefore the line passing through (2, 4) and (4, -2) has an equation:
[tex]\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}\\\frac{y-4}{x-2}=\frac{-2-4}{4-2}\\\frac{y-4}{x-2}=\frac{-6}{2}\\y-4=x-2(-3)\\y-4=-3x+6\\y=-3x+10[/tex]
Comparing with the general equation of line: y = mx + c, the slope (m) = -3 and the intercept on the y axis (c) = 10
Two lines are said to be perpendicular if the product of their slope is -1. If the slope of line one is m1 and the slope of line 2 = m2, then the two lines are perpendicular if:
[tex]m_1m_2=-1[/tex].
Therefore The slope (m2) of the perpendicular bisector of y = -3x + 10 is:
[tex]m_1m_2=-1\\-3m_2=-1\\m_2=\frac{1}{3}[/tex]
Since it is the perpendicular bisector of the given line segment, it passes through the midpoint (3, 1). The equation of the perpendicular bisector is:
[tex]\frac{y-y_1}{x-x_1}=m\\\frac{y-1}{x-3}=\frac{1}{3}\\ y-1= \frac{1}{3}(x-3)\\ y-1=\frac{1}{3}x-1\\y=\frac{1}{3}x[/tex]
the equation, in slope-intercept form, of the perpendicular bisector of the given line segment is [tex]y=\frac{1}{3}x[/tex]
