# Find the magnitude of the acute angle between the lines $2y+3x=4$ and $x+y=5$.

Find the magnitude of the acute angle between the lines $2y+3x=4$ and $x+y=5$.

I have no idea how to start the above equation. I try to draw the graph of $2y+3x=4$ and $x+y=5$ in the calculator but nothing show in the calculator.

The formula provided in the text book is
gradient of $l_{1}=m_{1}=\tan \theta_{1}$
gradient of $l_{2}=m_{2}=\tan \theta_{2}$

Help me out! thanks.

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The equations can be rewritten as:

$$y = -\frac{3}{2}x + 2$$ $$y = -x + 5$$ So, the slopes are: $\tan(\theta_1) =m_1 = -1.5$ and $\tan(\theta_2) = m_2 = -1$.

Therefore, the acute angle($\phi$) between the two lines is:

$$\tan(\phi) = \tan(|\theta_2 - \theta_1|) = \left|\frac{\tan(\theta_2) - \tan(\theta_1)}{1 + \tan(\theta_1)\tan(\theta_2)}\right|$$ or, $$\tan(\phi) = \left|\frac{m_2 - m_1}{1 + m_1m_2}\right|$$ $$= \left|\frac{(-1.5) - (-1)}{1 + (-1.5)(-1)}\right|$$ $$= \left|-\frac{0.5}{2.5}\right| = \frac{1}{5}$$

Hence, the angle will be:

$$\phi = \tan^{-1}\left(\frac{1}{5}\right) = 11.309932^\circ$$ or $\phi = 11.31^\circ$

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The acute angle between the lines is the same as that between their normals. The normals have directions $v=(2,3)$ and $w=(1,1)$
Just use inner products i.e, $$\theta=\cos^{-1}\frac{\left<v,w\right>}{\|v\|\|w\|}$$ subtract from $\pi$ radians ($180^o$) if you get an obtuse angle.
In case you do not have exposure to inner products, it's worth noting that at school level, one normally learns that the angle between the lines $ax+by=c$ and $a'x+b'y=c'$ is $$\cos^{-1}\frac{aa'+bb'}{\sqrt{(a^2+b^2)(a'^2+b'^2)}}=\cos^{-1}\frac{2+3}{\sqrt{13}\sqrt2}=\cos^{-1}\frac5{\sqrt{26}}=\tan^{-1}\frac15$$