Let G be the circumcentre of the triangle formed by the lines 4 x + 3 y = 69 , 4 y - 3 x = 17 and x + 7 y = 61 . Then 2 is equal to [2023]

Question

Let $G (α, β)$ be the circumcentre of the triangle formed by the lines $4 x + 3 y = 69, 4 y - 3 x = 17$ and $x + 7 y = 61$ . Then ${(α - β)}^{2} + α + β$ is equal to [2023]

Smart Achievers · Accepted Answer

(2)

$Let A B C be a triangle with$

$A B : 4 x + 3 y = 69 \dots (i)$

$B C : 4 y - 3 x = 17 \dots (i i)$

$A C : x + 7 y = 61 \dots (i i i)$

Solving (i) and (ii), we get $B \equiv (9, 11)$

Solving (ii) and (iii), we get $C \equiv (5, 8)$

Solving (i) and (iii), we get $A \equiv (12, 7)$

Now, $A G^{2} = G B^{2}$

$\Rightarrow {(12 - α)}^{2} + {(7 - β)}^{2} = {(9 - α)}^{2} + {(11 - β)}^{2}$

$\Rightarrow 8 β - 6 α = 9 ...(iv)$

$Similarly, G C^{2} = G B^{2}$

$\Rightarrow {(5 - α)}^{2} + {(8 - β)}^{2} = {(9 - α)}^{2} + {(11 - β)}^{2}$

$\Rightarrow 8 α + 6 β = 113 \dots (v)$

Now solving (iv) and (v), we get

$α = \frac{17}{2}, β = \frac{15}{2}$

$∴$ ${(α - β)}^{2} + α + β = 17$

Solution

Q.
Let $G (α, β)$ be the circumcentre of the triangle formed by the lines $4 x + 3 y = 69, 4 y - 3 x = 17$ and $x + 7 y = 61$ . Then ${(α - β)}^{2} + α + β$ is equal to [2023]

1 16

2 17

3 18

4 15

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