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Find the acute angle between the lines 3x - y = 4 and 2x + y = 3
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For 3x-y = 4,

$slope m_{1}=-\frac{a}{b}=-\frac{3}{-1}=3$

$For 2 x+y=3$

$slope m_{2}=-\frac{a}{b}=-\frac{2}{1}=-2$

$\therefore \tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$

$=\left|\frac{3-(-2)}{1+3 \times(-2)}\right|$

$=1$

$\therefore \theta=\tan ^{-1}(1)$

$\therefore \theta=\frac{\pi}{4} or 45^{0}$

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