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

Question

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

$4 x+3 y=69$

$4 y-3 x=17$, and

$x+7 y=61$.

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

Accepted Answer

We have,

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

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

$$
\begin{aligned}
& x=12, \text { and } y=7 \\
& \text { So, } A \equiv(12,7)
\end{aligned}
$$

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

$$
\begin{aligned}
& x=5 \text { and } y=8 \\
& \text { So, B } \equiv(5,8)
\end{aligned}
$$

$$
\begin{aligned}
& \text { Hence, circumcentre } \equiv\left(\frac{12+5}{2}, \frac{7+8}{2}\right) \\
& \equiv\left(\frac{17}{2}, \frac{15}{2}\right)
\end{aligned}
$$

$$
\begin{aligned}
& \therefore \alpha=\frac{17}{2}, \beta=\frac{15}{2} \\
& \therefore(\alpha-\beta)^2+(\alpha+\beta)=\left(\frac{17}{2}-\frac{15}{2}\right)^2+\left(\frac{17}{2}+\frac{15}{2}\right) \\
& =(1)^2+(16)=17
\end{aligned}
$$

Solutions · 1