The equation of the perpendicular bisector of the segment with endpoints Y(10,−7) and Z(−4,1) is 3y - 7x = -30
The equation of a line in point-slope form [tex]y-y_1 = m(x-x_1)[/tex]
m is the slope
(x0, y0) is the point on the line
For this question, the point on the line will be the midpoint between the given endpoints Y(10,−7) and Z(−4,1)
x0 = 10-4/2
x0 = 6/2 = 3
Similarly,
y0 = -7+1/2
y0 = -6/2 = -3
The required point will be (3, -3)
Get the slope of the line passing through the endpoints Y(10,−7) and Z(−4,1)
[tex]m=\frac{1+7}{-4-10}\\m=\frac{-8}{14} \\m = \frac{-4}{7}[/tex]
The slope of the line perpendicular will be [tex]\frac{7}{4}[/tex]
Get the required equation:
Recall that [tex]y-y_1 = m(x-x_1)[/tex]
[tex]y-(-3)= \frac{7}{4} (x-3)\\y+3=\frac{7}{3}(x-3)\\3(y+3)=7(x-3)\\3y+9=7x-21\\3y-7x = -30[/tex]
Hence the equation of the perpendicular bisector of the segment with endpoints Y(10,−7) and Z(−4,1) is 3y - 7x = -30
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