Family of Straight line : Consider a family of straight lines $(x+y) +\lambda (2x-y +1) =0$. Find the equation of the straight ....

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 that is farthest from $(1,-3)$.

Solution:

Let the point of intersection of the family of lines be P. If solve :

$$\left\{\begin{matrix} x+y=0 & \\ 2x-y+1=0 & \end{matrix}\right.$$ We get the point of intersection which is $P \left(-\dfrac{1}{3}, \dfrac{1}{3} \right)$

Now let us denote the point $(1,-3)$ as $Q$. So, now how to find $\lambda$ so that this will be fartheset from $Q$.

If we see the slope of $PQ = m_{PQ} = -\dfrac{5}{2}$

Any line perpendicular to $PQ$ will have slope $\dfrac{2}{5}$ Please suggest further.. thanks.

lab bhattacharjee · Answer 1 · 2018-04-04T09:52:51.890

2

HINT:

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

If $d$ is the perpendicular distance from $(1,-3)$

$$d^2=\frac{\{1(1+2\lambda)+(-3)(1-\lambda)+\lambda\}^2}{(1+2\lambda)^2+(1-\lambda)^2}$$

We need to maximize this which can be done using the pattern described here or here

edited Apr 04 '18 at 09:52

answered Nov 25 '13 at 03:10

lab bhattacharjee

274,582

I think you have to place square on the numerator of $d^2$ too – Archer Sep 29 '17 at 05:02
@Abcd, Thanks. It remained unnoticed for such a long time:) – lab bhattacharjee Sep 29 '17 at 05:12
I found the maximum of d= $\sqrt{89}/3$. What to do after that? – Archer Sep 29 '17 at 05:25
@Abcd, u have the $\lambda$ and hence the required equation – lab bhattacharjee Sep 29 '17 at 05:50
How can $d$ be the $\lambda$? – Archer Sep 29 '17 at 06:10
@Abcd, the maximum value of $d$ corresponds to one value of $\lambda$, right? – lab bhattacharjee Sep 29 '17 at 08:36
Hey, I think it's just supposed to be lambda for the constant, not (1+lambda). Is there something I'm missing? – Reya Apr 04 '18 at 09:48
@Reya, Thanks for your obsrevation – lab bhattacharjee Apr 04 '18 at 09:53

score 1 · Answer 2 · edited Nov 09 '15 at 08:05

1

The farthest line of the family of lines through $(-1/3, 1/3)$ from the point $(1, -3)$ cannot be farther than the distance $\sqrt{116}/3$ (this line segment will act as hypotenuse) that line must have the slope $2/5.$
You can find the value of $\lambda$ to match this slope.

edited Nov 09 '15 at 08:05

Max Payne

3,447

answered Jan 02 '15 at 03:04

abel

29,170

How did you find the maximum distance and the slope? Please elaborate. – Archer Sep 29 '17 at 06:27

score 1 · Answer 3 · answered Nov 09 '15 at 08:25

1

HINT : The line can be either perpendicular to it or it could also be a parallel line to the line passing through (1,-3). Thus you can find out required lambda and find the equation of line.

answered Nov 09 '15 at 08:25

Archis Welankar

15,910

score 1 · Answer 4 · answered Sep 29 '17 at 06:25

1

The point of intersection of the family of lines is $(-1/3, 1/3)$.

Thus, the required straight line is the line that is perpendicular to the join of $(-1/3,1/3)$ and $(1,-3)$ which is $15y-6x-7=0$.

answered Sep 29 '17 at 06:25

Archer

6,051

aditya gupta · Answer 5 · 2019-12-15T18:08:01.787

Note that the given family of lines always passes through the point A(-1/3, 1/3). So any line belonging to this family will always be at a distance less than or equal to AP (P is (1,-3)). For distance to be maximum, D= AP. This can obviously happen only when the line is perpendicular to AP. So we need to find the equation of line through P and having a slope -1/m, where m is the slope of the join of A and P. Finally we obtain 15y= 6x+7 as the reqd eqn

Family of Straight line : Consider a family of straight lines $(x+y) +\lambda (2x-y +1) =0$. Find the equation of the straight ....

5 Answers5