Question

Abscissae and ordinates of $n$ given points are in $A.P.,$ with first term $a$ and common difference $1$ and $2$ respectively. If algebraic sum of perpendiculars drawn from these given points on a variable line which always passes through the point $(\frac{13}{2},11)$ is $0,$ then the value of $a\cdot n$ is

Answer 1

Let the variable line equation be $px+qy+r=0$ ......$\left(1\right)$
and $n$ given points be
$(a,a),(a+1,a+2),(a+2,a+4),...$
$\text{i.e.,}(a+i-1,a+2(i-1));i=1,2,3,4,...,n$
Since, the algebraic sum of perpendiculars drawn from these $n$ points on the variable line $\left(1\right)$ is always $0$,
$\therefore \sum _{i=1}^n\frac{p(a+i-1)+q(a+2i-2)+r}{\sqrt{p^2+q^2}}=0$
$\Rightarrow \sum _{i=1}^n\left[p\right(a+i-1)+q(a+2i-2)+r]=0$
$\Rightarrow p\sum _{i=1}^n(a+i-1)+q\sum _{i=1}^n(a+2i-2)+rn=0$
$\Rightarrow p\sum _{i=1}^n\frac{a+i-1}{n}+q\sum _{i=1}^n\frac{a+2i-2}{n}+r=0$
Hence the line $\left(1\right)$ always passes through a fixed point
$\left(\sum _{i=1}^n\frac{a+i-1}{n},\sum _{i=1}^n\frac{a+2i-2}{n}\right)$
But the fixed point is given that $(\frac{13}{2},11)$
$\therefore \sum _{i=1}^n\frac{a+i-1}{n}=\frac{13}{2}$
and $\sum _{i=1}^n\frac{a+2i-2}{n}=11$
$\Rightarrow a+\frac{1}{n}\cdot \frac{(n-1)n}{2}=\frac{13}{2}\cdots \left(2\right)$
and $a+\frac{2}{n}\cdot \frac{(n-1)n}{2}=11\cdots \left(3\right)$
Solving the above two equations, we get
$a=2$ and $n=10$