The equation of the perpendicular bisector of the line segment joining A(- 2, 3) and B(6, - 5) is

Question

The equation of the perpendicular bisector of the line segment joining A(- 2, 3) and B(6, - 5) is (A) x - y = - 1 (B) x - y = 3 (C) x + y = 3 (D) x + y = 1

Tardigrade Admin · Accepted Answer

Mid point of given line segment AB is (2, -1). Now, slope of the line segment AB is - 8/8 = -1 Slope of the required line segment is 1. So equation of the required line is y + 1 = 1(x - 2) ⇒ x - y = 3.

Q. The equation of the perpendicular bisector of the line segment joining $A (- 2, 3)$ and $B (6, - 5)$ is

$x - y = - 1$

$x - y = 3$

$x + y = 3$

$x + y = 1$

Mid point of given line segment $A B$ is $(2, - 1)$ .
Now, slope of the line segment $A B$ is $- 8/8 = - 1$
Slope of the required line segment is $1$ .
So equation of the required line is
$y + 1 = 1 (x - 2)$
$\Rightarrow x - y = 3$ .

Q. The equation of the perpendicular bisector of the line segment joining A(−2,3) and B(6,−5) is

x−y=−1

x−y=3

x+y=3

x+y=1