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I'll state the question from my textbook, with the hint given, below.

A variable line passes through a fixed point $P$. The algebraic sum of the perpendiculars drawn from the points $(2, 0)$, $(0, 2)$ and $(1, 1)$ on the line is zero. Find the coordinates of the point $P$. $[$Hint: Let the slope of the line be $m$. Then the equation of the line passing through the fixed point $P(h,k)$ is $y−k=m(x−h)$. Taking the algebraic sum of perpendicular distances equal to zero, we get $y−1=m(x−1)$. Thus $(h,k)$ is $(1,1)$.$]$

I tried solving it as given below:

Let the fixed point be $P(h,k)$ and the variable line be $Ax+By+C=0$.

Since $P(h,k)$ lies on the variable line $Ax+By+C=0$,

$Ah+Bk+C=0 \tag 1$

Using the formula for the distance of a point from a line we have:

$\begin{align} \frac {A(2)+B(0)+C}{\sqrt {A^2+B^2}} + \frac {A(0)+B(2)+C}{\sqrt {A^2+B^2}} \frac {A(1)+B(1)+C}{\sqrt {A^2+B^2}} & = 0 &&\because\text{Algebraic sum of distances is zero}\\ \implies A + B + C & = 0 &&\tag2 \end{align}$

Now by comparing equations $(1)$ and $(2)$ I can get $(h,k)=(1,1)$. This is also the answer given in my textbook. But is it the only answer?

When I subtract equation $(2)$ from $(1)$ I get:

$A(h-1) + B(k-1) = 0\\ \implies \frac {k-1}{h-1} = - \frac AB$

So, aren't there other possible points $(h,k)$ where $h$ and $k$ satisfy the above relation? Also, is just comparing the two equations and concluding $(h,k)=(1,1)$ correct?

Any help would be appreciated.

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    $\begingroup$ but the Hessian Normalform is given by $$\frac{Ax+By+C}{\pm\sqrt{A^2+B^2}}=0$$ $\endgroup$ – Dr. Sonnhard Graubner Dec 31 '17 at 16:56
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    $\begingroup$ and for the three summands are the possible signs are $$+++,---,+--,$$ and so on $\endgroup$ – Dr. Sonnhard Graubner Dec 31 '17 at 16:58
  • $\begingroup$ @Dr.SonnhardGraubner I see. I thought an algebraic sum of distances meant the adding the values, not the absolute values. In that case I get more such points, right? $\endgroup$ – SamInuyasha ANMF Jan 1 '18 at 14:32
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    $\begingroup$ yes this is right, the distances have signes $\endgroup$ – Dr. Sonnhard Graubner Jan 1 '18 at 14:34
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The trouble is, the problem is VERY poorly stated. Perhaps this text is translated from some other language? In that case, the issues may be in the translation, not the original problem.

However, I believe what it is trying to say is that all the lines for which the sum of the signed distances (or "directed distances") from the line to $(2, 0), (1,1), (0,2)$ is $0$ intersect in a point $P$. What is $P$?

Your work has shown that $(1,1)$ is a point on all such lines. You are correct that for any given line $Ax + By = A+B$, there are other points through which the line runs. But it is only $(1,1)$ that lies on all such lines, regardless of the values of $A, B$.

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