Let #L : a(3x+4y+6)+b(x+y+2)=0# be the family of lines.
Observe that, both #a and b# can not be simultaneously zero.
So, if #b!=0#, then dividing the equation of the family by #b#,
# L : x+y+2+p(3x+4y+6)=0, p=a/b, #
# or, L : (1+3p)x+(1+4p)y+(2+6p)=0#.
The #bot"-distance "d" from "P(2,3)# to this line, is given by,
#d=|(1+3p)2+(1+4p)3+(2+6p)|/sqrt{(1+3p)^2+(1+4p)^2}#,
# i,e., d=|7+24p|/sqrt(25p^2+14p+2)#.
Since to maximise #d# is the same as to maximise #d^2#,
we must have the denominator #25p^2+14p+2# minimum.
Now, #25p^2+14p+2=25k^2+14p+49/25+1/25#,
#=(5p+7/5)^2+1/25#.
#AA p in RR, (5p+7/5)^2 ge 0#
#:. d^2" will be minimum" iff" when "(5p+7/5)=0#
# iff" when "p=-7/25#.
For this #p,# the reqd. line from the family #L# is given by,
# l_1 : x+y+2-7/25p(3x+4y+6)=0, i.e., 4x-3y+8=0#.
Recall that the above has been derived on the assumption that,
#b!=0#.
If we work out assuming that #a!=0#, then, the reqd. line
happens to be #l_2 : x-y+2=0#.
To determine which of the lines #l_1 and l_2# fulfills the
given condition , we finally have to work out the #bot-"dist. "d#
from #P(2,3)#.
#"For "l_1, bot-"dist. "d_1=|4(2)-3(3)+8|/sqrt{4^2+(-3)^2}=7/5=1.4#.
#"For "l_2, bot-"dist. "d_2=|2-3+2|/sqrt{1^2+(-1)^2}=sqrt2/2~~0.71#
Consequently, #l_1 : 4x-3y+8=0# is the desired line!
#color(indigo)("Enjoy Maths.!")#
