QUESTION

# If the algebraic sum of the perpendicular distances from the points (2,0), (0,2) & (1,1) to a variable straight line be zero, then the line passes through the pointA. (-1, 1)B. (1, 1)C. (1, -1)D. (-1, -1)

Hint: As mentioned in this question, we have perpendicular distances from the points (2,0), (0,2) & (1,1) to a variable straight line. First, we consider a general equation of a variable straight-line ax+by+c=0, Which represents a family of straight lines passing through a fixed point. And then after we use a formula for calculating the perpendicular distance between a straight line and a point.
The perpendicular distance of P (x1, y1) from the line ax + by + c = 0 is $\dfrac{{{\text{a}}{{\text{x}}_{\text{1}}}{\text{ + b}}{{\text{y}}_{\text{1}}}{\text{ + c}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ + }}{{\text{b}}^{\text{2}}}} }}$

Let the variable line be ax+by+c=0.
Given, the algebraic sum of the perpendicular from the points (2,0), (0,2) and (1,1) to this line is zero.
the perpendicular distance of P (x1, y1) from the line ax + by + c = 0 is $\dfrac{{{\text{a}}{{\text{x}}_{\text{1}}}{\text{ + b}}{{\text{y}}_{\text{1}}}{\text{ + c}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ + }}{{\text{b}}^{\text{2}}}} }}$
So, as given in question, the algebraic sum of the perpendicular distances from the points (2,0), (0,2) & (1,1) to a variable straight line be zero.
$\Rightarrow$ $\dfrac{{{\text{2a + 0 + c}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ + }}{{\text{b}}^{\text{2}}}} }} + \dfrac{{{\text{0 + 2b + c}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ + }}{{\text{b}}^{\text{2}}}} }} + \dfrac{{{\text{a + b + c}}}}{{\sqrt {{{\text{a}}^{\text{2}}}{\text{ + }}{{\text{b}}^{\text{2}}}} }}$ = 0
⇒2a+c+2b+c+a+b+c=0
⇒3a+3b+3c=0
⇒a+b+c=0
This is a linear relation between a, b and c.
So, the equation ax+by+c=0 represents a family of straight lines passing through a fixed point.
Comparing,
ax+by+c=0 and a+b+c=0.
We obtain, x=1, y=1 i.e.
The coordinates of fixed points are (1,1).
So, option (B) is the correct answer.

Note- When we say distance, we mean the shortest possible distance from the point to the line, which happens to be when the distance line through the point is also perpendicular to the line. But why is the shortest line segment perpendicular? This is because the longest side in a right triangle is the hypotenuse.