Let #P(x_1, y_1)# be a point and let #l# be the line with equation #ax+ by +c =0#. Show the distance #d# from #P->l# is given by: #d =(ax_1+ by_1 + c)/sqrt( a^2 +b^2)#? Find the distance #d# of the point P(6,7) from the line l with equation 3x +4y =11?

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Answer 1 · 2016-11-16T00:54:55

#d = 7#

Let #l->a x + b y + c=0# and #p_1 = (x_1,y_1)# a point not on #l#.

Supposing that #b ne 0# and calling #d^2=(x-x_1)^2+(y-y_1)^2# after substituting #y=-(a x+c)/b# into #d^2# we have

#d^2=(x - x_1)^2 + ((c + a x)/b + y_1)^2#. The next step is find the #d^2# minimum regarding #x# so we will find #x# such that

#d/(dx)(d^2) = 2 (x - x_1) - (2 a ((c + a x)/b + y_1))/b = 0#. This occours for

#x = (b^2 x_1 - a b y_1-a c)/(a^2 + b^2)# Now, substituting this value into #d^2# we obtain

#d^2=(c + a x_1 + b y_1)^2/(a^2 + b^2)# so

#d = (c + a x_1 + b y_1)/sqrt(a^2 + b^2)#

Now given

#l->3x+4y-11=0# and #p_1=(6,7)# then

#d = (-11+3xx6+4xx7)/sqrt(3^2+4^2)=7#