The Point which lies on the perpendicular bisector of the line segment joining the points P(-3,2) and B(11,-8) is
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The Point which lies on the perpendicular bisector of the line segment joining the points P(-3,2) and B(11,-8) is
The Point which lies on the perpendicular bisector of the line segment joining the points P(-3,2) and B(11,-8) is
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Step-by-step explanation:
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Answer:
Step-by-step explanation:
Midpoint M of the segment PB is
[tex]M(\frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2}) = M(\frac{-3+11}{2},\frac{2-8}{2})[/tex] = M(4, -3)
Slope "m" of line passing through points P and B is
[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{-8-2}{11+3} = -\frac{5}{7}[/tex]
y + 3 = [tex]-\frac{5}{7}[/tex](x - 4)
[tex]y=-\frac{5}{7}x- \frac{1}{7}[/tex]
Slope of the line perpendicular to line [tex]y=-\frac{5}{7}x- \frac{1}{7}[/tex] is opposite reciprocal to [tex]-\frac{5}{7}[/tex] ; [tex]m = \frac{7}{5}[/tex] and equation [tex]y=\frac{7}{5}x-\frac{43}{5}[/tex]
Any point which lies on the perpendicular bisector satisfies the equation
[tex]y=\frac{7}{5}x-\frac{43}{5}[/tex]