What is the equation of the line that is perpendicular to the line passing through #(3,2)# and #(4,3)# at midpoint of the two points?

1 Answer
Jun 3, 2017

Answer:

#x+y-6+0.#

Explanation:

Observe that the reqd. line is a line through the mid-point of the

sgmt. joining the two given points and perpendicular to the sgmt.

Evidently, it is the Perpendicular Bisector of the sgmt.

Knowing that any point on the per.bisector of the sgmt. is

equidistant from the extremities, we have,

#(x-3)^2+(y-2)^2=(x-4)^2+(y-3)^2,# where #(x,y)# is any pt.

# :. -6x+9-4y+4=-8x+16-6y+9,#

#:. 2x+2y=12, or, x+y-6=0,# as already derived by

Respected Elena C. !

Enjoy Maths.!