Two lines $x + 3 y = 0, 3 x + y = 1$ meet at

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Solution

$x + 3 y = 0 x = - 3 y \dots (1) 3 x + y = 1 3 (- 3 y) + y = 1 - 9 y + y = 1 - 8 y = 1 y = \frac{- 1}{8} x + 3 (\frac{- 1}{8}) = 0 x = \frac{3}{8}$
So The point of intersection is $(\frac{3}{8}, \frac{- 1}{8})$

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