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.!")