JEE MAIN - Mathematics (2023 - 8th April Morning Shift - No. 10)
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 :
15
17
16
18
Explanation
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,
$$ \begin{aligned} & x=12, \text { and } y=7 \\\\ & \text { So, } A \equiv(12,7) \end{aligned} $$
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On solving (ii) and (iii), we get,
$$ \begin{aligned} & x=5 \text { and } y=8 \\\\ & \text { So, B } \equiv(5,8) \end{aligned} $$
$$ \begin{aligned} & \text { Hence, circumcentre } \equiv\left(\frac{12+5}{2}, \frac{7+8}{2}\right) \\\\ & \equiv\left(\frac{17}{2}, \frac{15}{2}\right) \end{aligned} $$
$$ \begin{aligned} & \therefore \alpha=\frac{17}{2}, \beta=\frac{15}{2} \\\\ & \therefore(\alpha-\beta)^2+(\alpha+\beta)=\left(\frac{17}{2}-\frac{15}{2}\right)^2+\left(\frac{17}{2}+\frac{15}{2}\right) \\\\ & =(1)^2+(16)=17 \end{aligned} $$
$$ \begin{aligned} & 4 x+3 y=69 .......(i) \\\\ & 4 y-3 x=17 .......(ii) \\\\ & x+7 y=61 .......(iii) \end{aligned} $$
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,
$$ \begin{aligned} & x=5 \text { and } y=8 \\\\ & \text { So, B } \equiv(5,8) \end{aligned} $$
$$ \begin{aligned} & \text { Hence, circumcentre } \equiv\left(\frac{12+5}{2}, \frac{7+8}{2}\right) \\\\ & \equiv\left(\frac{17}{2}, \frac{15}{2}\right) \end{aligned} $$
$$ \begin{aligned} & \therefore \alpha=\frac{17}{2}, \beta=\frac{15}{2} \\\\ & \therefore(\alpha-\beta)^2+(\alpha+\beta)=\left(\frac{17}{2}-\frac{15}{2}\right)^2+\left(\frac{17}{2}+\frac{15}{2}\right) \\\\ & =(1)^2+(16)=17 \end{aligned} $$
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