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Given that equation of line passing through points (2 , 2) and (6 , 3)
\Use determinants
\
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x (2(1) - 3(1)) - y (2(1) - 6(1)) + 1 (2(3) - 6(2)) = 0
\x (2 - 3) - y (2 - 6) + 1 (6 - 12) = 0
\x (-1) - y (-4) + 1 (-6) = 0
\-x + 4y - 6 = 0
\Take negitive symbol common
\- (x - 4y + 6) = 0
\x - 4y + 6 = 0
\x - 4y = -6
\Therefore the required equation of line is x - 4y = -6
\option (d)
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