Given the family of lines, a(3x+4y+6) + b(x+y+2)=0. The line of the family situated at the greatest distance from the point P(2,3) has the equation?

Ans: 4x+3y + 8 = 0

1 Answer
Aug 5, 2018

4x-3y+8=0

Explanation:

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!

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