Find the Point Through which the Variable Line Passes

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.

• but the Hessian Normalform is given by $$\frac{Ax+By+C}{\pm\sqrt{A^2+B^2}}=0$$ Dec 31 '17 at 16:56
• and for the three summands are the possible signs are $$+++,---,+--,$$ and so on Dec 31 '17 at 16:58
• @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? Jan 1 '18 at 14:32
• yes this is right, the distances have signes Jan 1 '18 at 14:34

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 $A, B$.