# 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: $4 x + 3 y + 8 = 0$

Aug 5, 2018

$4 x - 3 y + 8 = 0$

#### Explanation:

Let $L : a \left(3 x + 4 y + 6\right) + b \left(x + y + 2\right) = 0$ be the family of lines.

Observe that, both $a \mathmr{and} b$ can not be simultaneously zero.

So, if $b \ne 0$, then dividing the equation of the family by $b$,

$L : x + y + 2 + p \left(3 x + 4 y + 6\right) = 0 , p = \frac{a}{b} ,$

$\mathmr{and} , L : \left(1 + 3 p\right) x + \left(1 + 4 p\right) y + \left(2 + 6 p\right) = 0$.

The $\bot \text{-distance "d" from } P \left(2 , 3\right)$ to this line, is given by,

$d = | \left(1 + 3 p\right) 2 + \left(1 + 4 p\right) 3 + \left(2 + 6 p\right) \frac{|}{\sqrt{{\left(1 + 3 p\right)}^{2} + {\left(1 + 4 p\right)}^{2}}}$,

$i , e . , d = | 7 + 24 p \frac{|}{\sqrt{25 {p}^{2} + 14 p + 2}}$.

Since to maximise $d$ is the same as to maximise ${d}^{2}$,

we must have the denominator $25 {p}^{2} + 14 p + 2$ minimum.

Now, $25 {p}^{2} + 14 p + 2 = 25 {k}^{2} + 14 p + \frac{49}{25} + \frac{1}{25}$,

$= {\left(5 p + \frac{7}{5}\right)}^{2} + \frac{1}{25}$.

$\forall p \in \mathbb{R} , {\left(5 p + \frac{7}{5}\right)}^{2} \ge 0$

$\therefore {d}^{2} \text{ will be minimum" iff" when } \left(5 p + \frac{7}{5}\right) = 0$

$\iff \text{ when } p = - \frac{7}{25}$.

For this $p ,$ the reqd. line from the family $L$ is given by,

${l}_{1} : x + y + 2 - \frac{7}{25} p \left(3 x + 4 y + 6\right) = 0 , i . e . , 4 x - 3 y + 8 = 0$.

Recall that the above has been derived on the assumption that,

$b \ne 0$.

If we work out assuming that $a \ne 0$, then, the reqd. line

happens to be ${l}_{2} : x - y + 2 = 0$.

To determine which of the lines ${l}_{1} \mathmr{and} {l}_{2}$ fulfills the

given condition , we finally have to work out the $\bot - \text{dist. } d$

from $P \left(2 , 3\right)$.

$\text{For "l_1, bot-"dist. } {d}_{1} = | 4 \left(2\right) - 3 \left(3\right) + 8 \frac{|}{\sqrt{{4}^{2} + {\left(- 3\right)}^{2}}} = \frac{7}{5} = 1.4$.

$\text{For "l_2, bot-"dist. } {d}_{2} = | 2 - 3 + 2 \frac{|}{\sqrt{{1}^{2} + {\left(- 1\right)}^{2}}} = \frac{\sqrt{2}}{2} \approx 0.71$

Consequently, ${l}_{1} : 4 x - 3 y + 8 = 0$ is the desired line!

$\textcolor{\in \mathrm{di} g o}{\text{Enjoy Maths.!}}$