What is the equation of the line that is perpendicular to the line passing through #(3,18)# and #(6,14)# at midpoint of the two points?

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Answer 1 · 2018-07-28T04:17:10

#6x-8y+121=0#

The slope #m# of line perpendicular to the line joining the points #(x_1, y_1)\equiv(3, 18)# & #(x_2, y_2)\equiv(6, 14)# is given as

#m=-1/{\frac{y_2-y_1}{x_2-x_1}}#

#=-1/{\frac{14-18}{6-3}}#

#=3/4#

Now, the mid-point of line joining the points #(x_1, y_1)\equiv(3, 18)# & #(x_2, y_2)\equiv(6, 14)# are given as

#(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})#

#\equiv(\frac{3+6}{2}, \frac{18+14}{2})#

#\equiv(\frac{9}{2}, 16)#

Now, the equation of line passing through the point #(x_0, y_0)\equiv(9/2, 16)# & having slope #m=3/4# is given as

#y-y_0=m(x-x_0)#

#y-16=3/4(x-9/2)#

#6x-8y+121=0#