Let C( , ) be the circumcenter of the triangle formed by the lines 4x+3y=69 4y-3x=17, and …

Question

Let $C(\alpha ,\beta )$ be the circumcenter of the triangle formed by the lines

$4x+3y=69$

$4y-3x=17$, and

$x+7y=61$.

Then $(\alpha -\beta {)}^{2}+\alpha +\beta $ is equal to :

Accepted Answer

We have,

$\begin{gathered} 4x+3y=69.......(i) \ 4y-3x=17.......(ii) \ x+7y=61.......(iii) \end{gathered}$

On solving (i) and (iii), we get,

$\begin{gathered} x=12,y=7 \ A\equiv (12,7) \end{gathered}$

On solving (ii) and (iii), we get,

$\begin{gathered} x=5y=8 \ \equiv (5,8) \end{gathered}$

$\begin{gathered} \equiv (\frac{12+5}{2},\frac{7+8}{2}) \ \equiv (\frac{17}{2},\frac{15}{2}) \end{gathered}$

$\begin{gathered} 	herefore \alpha =\frac{17}{2},\beta =\frac{15}{2} \ 	herefore (\alpha -\beta {)}^{2}+(\alpha +\beta )={(\frac{17}{2}-\frac{15}{2})}^{2}+(\frac{17}{2}+\frac{15}{2}) \ =(1{)}^{2}+(16)=17 \end{gathered}$

Solutions · 1