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MakcuM [25]
3 years ago
12

Line segment NY has endpoints N(-11, 5) and Y(3,-3).

Mathematics
1 answer:
777dan777 [17]3 years ago
6 0

Given:

Line segment NY has endpoints N(-11, 5) and Y(3,-3).

To find:

The equation of the perpendicular bisector of NY.

Solution:

Midpoint point of NY is

Midpoint=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)

Midpoint=\left(\dfrac{-11+3}{2},\dfrac{5-3}{2}\right)

Midpoint=\left(\dfrac{-8}{2},\dfrac{2}{2}\right)

Midpoint=\left(-4,1\right)

Slope of lines NY is

m=\dfrac{y_2-y_1}{x_2-x_1}

m=\dfrac{-3-5}{3-(-11)}

m=\dfrac{-8}{14}

m=\dfrac{-4}{7}

Product of slopes of two perpendicular lines is -1. So,

m_1\times \dfrac{-4}{7}=-1

m_1=\dfrac{7}{4}

The perpendicular bisector of NY passes through (-4,1) with slope \dfrac{7}{4}. So, the equation of perpendicular bisector of NY is

y-y_1=m_1(x-x_1)

y-1=\dfrac{7}{4}(x-(-4))

y-1=\dfrac{7}{4}(x+4)

y-1=\dfrac{7}{4}x+7

Add 1 on both sides.

y=\dfrac{7}{4}x+8

Therefore, the equation of perpendicular bisector of NY is y=\dfrac{7}{4}x+8.

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