Family of Lines

Question

Problem:

Consider a family of straight lines $(x+y)+\lambda(2x-y+1)=0$. Find the equation of the straight line belonging to this family which is farthest from $(1,-3)$.

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Any help with this problem would be really appreciated!

Answer 1

Think of it geometrically. Since the family of lines all pass through $I\left(-\frac{1}{3}, \frac{1}{3}\right)$, the largest distance from any point $A$ to a line passing through $I$ will be the length $IA$. Therefore, the problem becomes:

Find the equation of the line containing $I\left(-\frac{1}{3}, \frac{1}{3}\right)$ and perpendicular to $IA$ where $A(1,-3)$.

Solution

The slope of the line passing through $IA$ is

$$ m_{IA} = \frac{-3 - \frac{1}{3}}{1 - (-\frac{1}{3})} = -\frac{5}{2} $$

You want the slope of the line perpendicular to this, which is

$$m_{\lambda} = -\frac{1}{m_{IA}}=\frac{2}{5}$$

Combined with the known point, the equation is

$$y = \frac{2}{5}x + \frac{7}{15}$$

Edit: Reasoning for perpendicularity

Let $H$ be the intersection of a perpendicular line drawn from $A$. We have $AH$ is the distance from $A$ to the line.

Suppose $IA$ makes an angle $\theta$ with the line in question, it follows that $AH = IA\sin\theta \le IA$. Thus $AH = IA$ when $\theta = 90^\circ$ or $H \equiv I$

Answer 2

We can write the equation as $\displaystyle (2\lambda+1)x+(1-\lambda)y+\lambda+1 =0$

Now Perpendicular Distance of Line from Point $\bf{P(1,-3)}$ is $ = PQ=\displaystyle \left|\frac{2\lambda+1+3\lambda-3+\lambda+1}{\sqrt{(2\lambda+1)^2+(1-\lambda)^2}}\right|$

Now $\displaystyle PQ = \left|\frac{6\lambda-1}{\sqrt{5\lambda^2+2\lambda+2}}\right|\;,$ Then we have to $\bf{Maximize}$ the value of $$\displaystyle \frac{6\lambda-1}{\sqrt{5\lambda^2+2\lambda+2}}$$

Now Let $$\displaystyle f(\lambda) = \frac{(6\lambda-1)^2}{5\lambda^2+2\lambda+2}$$.(Bcz If $f(\lambda)$ is $\bf{Max.}\;,$ Then $\left[f(\lambda)\right]^2$ is $\bf{Max.}$)

Now Let $$\displaystyle z = \frac{36\lambda^2+1-12\lambda}{5\lambda^2+2\lambda+2}\Rightarrow 5\lambda^2\cdot z+2\lambda \cdot z+2z = 36\lambda^2-12\lambda+1$$

So we get $$\displaystyle \left(5z-36\right)\lambda^2+2(z+6)\lambda+(2z-1) =0\;,$$ For real roots, Then $\bf{Discriminant\geq 0}$

So $$\displaystyle 4\left(z+6\right)^2-4\cdot (5z-36)\cdot (2z-1)\geq 0$$

So $(z+6)^2-(5z-36)\cdot (2z-1)\geq 0$

So $$\displaystyle z^2+36+12z-10z^2+77z-36\geq 0$$

So $$\displaystyle -9z^2+89z\geq 0\Rightarrow 9z^2-89z\leq 0$$

So we get $$\displaystyle z\leq \frac{89}{9}$$

So $$\displaystyle \bf{Max.(z) = \frac{89}{9}}$$

Now Put $\displaystyle z = \frac{89}{9}$ in $$\displaystyle z = \frac{(6\lambda-1)^2}{5\lambda^2+2\lambda+2}\;,$$ We Get $\displaystyle \lambda = -\frac{13}{11}$

Answer 3

Well, if the distance from a point $(x_1,y_1)$ to a line $ax+by+c=0$ is given by:$\dfrac{ax_1+by_1+c}{\sqrt{a^2+b^2}}$, then maybe you could rewrite that formula for the point $\left(1,-3\right)$ and the line $(x+y)+\lambda(2x-y+1)=0$ ?

After that you will just need to find the $\lambda$ for which this formula is maximal. Then the farthest line is described by that value of $\lambda$.

Answer 4

Notice, the equation of straight lines $$x+y+\lambda(2x-y+1)=0$$ $$(2\lambda +1)x+(1-\lambda)y+\lambda=0$$

Now, the distance say $D$ of the above line from the given point $(1, -3)$ is given as follows $$D=\left|\frac{(2\lambda +1)(1)+(1-\lambda)(-3)+\lambda}{\sqrt{(2\lambda +1)^2+(1-\lambda )^2}}\right|$$

$$=\left|\frac{(2\lambda +1)(1)+(1-\lambda)(-3)+\lambda}{\sqrt{(2\lambda +1)^2+(1-\lambda )^2}}\right|$$ $$=\left|\frac{6\lambda-2}{5\lambda^2+2\lambda+2}\right|$$ Now, considering the following function: $f(\lambda)=\frac{6\lambda-2}{5\lambda^2+2\lambda+2}$

$$\frac{df(\lambda)}{d\lambda}=\frac{d}{d\lambda}\left(\frac{6\lambda-2}{5\lambda^2+2\lambda+2}\right)$$ $$=\frac{(5\lambda^2+2\lambda+2)(6)-(6\lambda-2)(10\lambda+2)}{(5\lambda^2+2\lambda+2)^2}$$ $$=\frac{-30\lambda^2+20\lambda+16}{(5\lambda^2+2\lambda+2)^2}$$

Now, for maximum value of $D$ we have $$\frac{df(\lambda)}{d\lambda}=0$$

$$\frac{-30\lambda^2+20\lambda+16}{(5\lambda^2+2\lambda+2)^2}=0$$

$$15\lambda^2-10\lambda-8=0$$ $$\lambda=\frac{5\pm\sqrt{145}}{15}$$$$\iff \lambda =\frac{5+\sqrt{145}}{15}\ \vee\ \lambda=\frac{5-\sqrt{145}}{15}$$

Now, find $\frac{d^2f(\lambda)}{d\lambda^2}$ & substitute these values of $\lambda $ individually in $\frac{d^2f(\lambda)}{d\lambda^2}$ then you will find that for $\lambda=\frac{5+\sqrt{145}}{15}$, $\frac{d^2f(\lambda)}{d\lambda^2}<0$

Thus, the function $f(\lambda)$ will be maximum & hence the distance $D$ of the given point from the line will be maximum at $\lambda=\frac{5+\sqrt{145}}{15}$ Now, substituting this value of $\lambda $ in the given equation of family of straight lines we get the equation of the straight line farthest from the given point $(1, -3)$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{x+y+\left(\frac{5+\sqrt{145}}{15}\right)(2x-y-1)=0}}$$