Find the distance of the points A(3,6) B(-7,8) from the line 4x+6y-8=0?

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Answer 1 · 2018-02-27T16:45:44

#p_1=20/sqrt13#

#p_2=6/sqrt13#

distance p between the point (z,y) to the line ax+by+c=0 is given by

#p=|(ax+by+c)/(sqrt(a^2+b^2))|#
#A-=(3,6)#
#4x+6y-8#
#p=p_1,x=3, y=6#
#a=4, b=6, c=-8#
Substituting
#p_1=|(4xx3+6xx6-8)/(sqrt(4^2+6^2))|#

#p_1=|(12+36-8)/(sqrt(16+36))|#

#p_1=|(48-8)/(sqrt52)|#

#p_1=40/(2sqrt13)#

#p_1=20/sqrt13#

#B-=(-7,8)#
#4x+6y-8#
#p=p_2,x=-7, y=8#
#a=4, b=6, c=-8#
Substituting
#p_2=|(4xx-7+6xx8-8)/(sqrt(4^2+6^2))|#

#p_2=|(-28+48-8)/(sqrt(16+36))|#

#p_2=|(20-8)/(sqrt52)|#

#p_2=12/(2sqrt13)#

#p_2=6/sqrt13#