Question

Find the equation of the right bisector of the line segment joining the points (3, 4) and ( – 1, 2).

Answer 1

The line segment is formed by joining the points ( 3,4 ) and ( −1,2 ) .

The formula for the slope of a line passes through points ( x 1 , y 1 ) and ( x 2 , y 2 ) is given by,

m= y 2 − y 1 x 2 − x 1 (1)

Let m 1 be the slope of the line segment which passes through the points ( 3,4 ) and ( −1,2 ) .

Substitute the value for ( x 1 , y 1 ) and ( x 2 , y 2 ) as ( 3,4 ) and ( −1,2 ) . respectively in equation (3).

m 1 = 2−4 −1−3 = −2 −4 = 1 2 (2)

The right bisector intersect the line segment at 90° .

The product of the slope of two lines perpendicular to each other is −1 .

m 1 ⋅ m 2 =−1 (3)

Let m 2 be the slope of the right bisector.

Substitute the value of m 1 from equation (2) to equation (3).

1 2 ⋅ m 2 =−1 m 2 =− 1 1 2 =−2

The right bisector intersects at the mid-point of line segment.

Let the coordinates of the mid-point are ( x m , y m ) .

The formula for the mid-point of a line segment passing through the points ( x 1 , y 1 ) and ( x 2 , y 2 ) is given by

( x m , y m )=( x 1 + x 2 2 , y 1 + y 2 2 ) (4)

Substitute the values of ( x 1 , y 1 ) and ( x 2 , y 2 ) as ( 3,4 ) and ( −1,2 ) in equation (4).

( x m , y m )=( 3−1 2 , 4+2 2 ) =( 2 2 , 6 2 ) =( 1,3 )

The formula for the equation of the line having slope m passes through the point ( x 1 , y 1 ) is given by,

( y− y 1 )=m( x− x 1 ) (5)

Substitute the values of m as −2 and ( x 1 , y 1 ) as ( 1,3 ) in equation (5).

( y−3 )=−2⋅( x−1 ) y−3=−2x+2 2x+y−2−3=0 2x+y−5=0

Thus, the equation of right bisector of the line segment joining the points ( 3,4 ) and ( −1,2 ) is 2x+y−5=0 .