If (a, b) is the centroid of the triangle formed by the lines $4x^2-17xy+4y^2=0andx+y-5=0$ and c is the numerical value of the area of the triangle, then a + b + c = 

$\begin{array}{l}GivenEquation4x^2-17xy+4y^2=0\\ \Rightarrow 4x^2-16xy-xy+4y^2=0\\ \Rightarrow 4x(x-4y)-y(x-4y)=0\\ \Rightarrow (x-4y)(4x-y)=0\\ \Rightarrow x-4y=0----\left(1\right)\\ \Rightarrow 4x-y=0----\left(2\right)\\ andx+y-5=0----\left(3\right)\\ P.I.of\left(1\right)\&\left(2\right)is(0,0)\\ Solving\left(1\right)\&\left(3\right)\\ xy1\\ \begin{array}{cccc}-4& 0& 1& -4\\ 1& -5& 1& 1\end{array}\end{array}$

$\begin{array}{l}\frac{x}{20}=\frac{y}{5}=\frac{1}{5}\\ \Rightarrow (x,y)=(4,1)\\ Solving\left(2\right)\&\left(3\right)\\ xy1\\ \begin{array}{cccc}-1& 0& 4& -1\\ 1& -5& 1& 1\end{array}\\ \Rightarrow \frac{x}{5-0}=\frac{y}{0+20}=\frac{1}{4+1}\\ (x,y)=(1,4)\\ O(0,0),A(4,1),B(1,4)\\ Centroid=(\frac{0+4+1}{3},\frac{0+1+4}{3})=(\frac{5}{3},\frac{5}{3})\\ \therefore (a,b)=(\frac{5}{3},\frac{5}{3})\end{array}$

$\begin{array}{l}C=Area=\frac{1}{2}|x_1{y}_2-x_2{y}_1|\\ =\frac{1}{2}|4\left(4\right)-1\left(1\right)|=\frac{15}{2}\\ \therefore a+b+c=\frac{5}{3}+\frac{5}{3}+\frac{15}{2}\\ =\frac{10+10+45}{6}=\frac{65}{6}\end{array}$