Let the algebraic sum of the perpendicular distance from the points $(2,0),(0,2)$ and $(1,1)$ to a v

Question

Let the algebraic sum of the perpendicular distance from the points $(2,0),(0,2)$ and $(1,1)$ to a variable straight line be zero; then the line passes through a fixed point whose coordinates are

Professor PC Thomas Classes · Answer

Let the given line be $a x+b y-p=0 \quad . .$. (i) Now the distance of perpendiculars from a point $(x, y)$ to this line is given by $\frac{a x+b y-p}{\sqrt{a^{2}+b^{2}}}$ And we are given that the sum of the perpendiculars from the given three points to this line is zero. So we get\ $\frac{2 a-p}{\sqrt{a^{2}+b^{2}}}+\frac{2 b-p}{\sqrt{a^{2}+b^{2}}}+\frac{a+b-p}{\sqrt{a^{2}+b^{2}}}=0$\ or $2 a-p+2 b-p+a+b-p=0 \Rightarrow a+b=p$ Putting this in equation (i), we get $a x+b y-(a+b)=0 \Rightarrow a x+b y-a-b=0$ or $a$ $(x-1)+b(y-1)=0$\ It can be seen that line passes through the point ( 1,1 ).\