Newton form of the interpolation polynomial?

Question

Let $ f(x)=\sin(x) $ and let $ p_1 $ be the first degree polynomial that interpolates $ f $ at $0$ and $\pi/2$. Then $ p_1(x)=(2/pi)x. $

How did they get this result; how is $ p_1(x) $ found?

Answer 1

$p_1(x)=ax+b$. We want $p_1(0)=f(0)=0$ and $p_1(\pi/2)=f(\pi/2)=1$. Thus, you get $0=b$, $1=a\pi/2+b=a\pi/2$, so $a=2/\pi$.