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Q.
The angle between the line $x + y = 3$ and the line joining the points $(1,1)$ and $(-3,4)$ is
Straight Lines
Solution:
Let $P = (1,1)$ and $Q = (-3,4)$
$\therefore $ Slope of line $PQ$, $m_{1}=\frac{1-4}{1+3}=\frac{-3}{4}$
Given line is $x + y = 3$ or $x + y - 3 = 0$
Its slope, $m_{2} = -1$
Now, $tan\,\theta=\left|\frac{m_{1}-m_{2}}{1+m_{1}m_{2}}\right|$
$=\left|\frac{\frac{-3}{4}+1}{1+\frac{3}{4}}\right|=\frac{1}{7}$
$\Rightarrow \theta=tan^{-1}\left(\frac{1}{7}\right)$