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?

Question

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?

Sadie Allen · Accepted Answer

#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 that

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

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

Autumn Armstrong · Answer

undefined To show the distance $ d $ from point $ P(x_1, y_1) $ to the line $ l $ with equation $ ax + by + c = 0 $ is given by $ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} $, we follow these steps:

1. Determine the perpendicular distance from point $ P $ to the line $ l $.
2. Use the formula for the distance from a point to a line.

Given the point $ P(6, 7) $ and the line equation $ 3x + 4y = 11 $, we need to convert the line equation into the standard form $ ax + by + c = 0 $, where $ a $, $ b $, and $ c $ are coefficients.

Converting the line equation:
$$ 3x + 4y = 11 $$
$$ \Rightarrow 3x + 4y - 11 = 0 $$

Comparing with the standard form $ ax + by + c = 0 $, we have:
$$ a = 3, \, b = 4, \, c = -11 $$

Now, plug in the values into the distance formula:
$$ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} $$
$$ = \frac{|3(6) + 4(7) - 11|}{\sqrt{3^2 + 4^2}} $$
$$ = \frac{|18 + 28 - 11|}{\sqrt{9 + 16}} $$
$$ = \frac{|35 - 11|}{\sqrt{25}} $$
$$ = \frac{|24|}{5} $$
$$ = \frac{24}{5} $$

So, the distance $ d $ of the point $ P(6,7) $ from the line $ l $ with equation $ 3x + 4y = 11 $ is $ \frac{24}{5} $.

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-&gt;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?

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?