Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10th
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Find the equation of the perpendicular bisector of the line joining the points A(− 4, 2) and B(6, − 4) - Mathematics

Sum

Find the equation of the perpendicular bisector of the line joining the points A(− 4, 2) and B(6, − 4)

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Solution

“C” is the midpoint of AB also CD ⊥ AB.

Slope of AB = `(y_2 - y_1)/(x_2 - x_1)`

= `(-4 - 2)/(6 + 4)`

= `(-6)/10`

= `-3/5`

Slope of the ⊥r AB is `5/3`

Mid point of AB = `((x_1 + x_2)/2, (y_1 + y_2)/2)`

= `((-4 + 6)/2, (2 - 4)/2)`

= `(2/2, (-2)/2)`

= (1, −1)

Equation of the perpendicular bisector of CD is

y – y1 = m(x – x1)

y + 1 = `5/3(x - 1)`

5(x – 1) = 3(y + 1)

5x – 5 = 3y + 3

5x – 3y – 5 – 3 = 0

5x – 3y – 8 = 0

Equation of the perpendicular bisector is 5x – 3y – 8 = 0

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APPEARS IN

Samacheer Kalvi Mathematics Class 10 SSLC Tamil Nadu State Board
Chapter 5 Coordinate Geometry
Exercise 5.4 | Q 8 | Page 235
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