Answer
(a) see explanations.
(b) $\langle \frac{8}{5}, -\frac{4}{5} \rangle$
(c) see sketch and explanations.
(d) see procedure below.
Work Step by Step
(a) To show the points are on the line, plug-in the coordinates, we have $2\times0+4\times2=8$ and $2\times2+4\times1=8$, indicating that the coordinates of both points satisfy the line equation.
(b) We can fine the vectors $\vec u=\langle 3-0, 4-2 \rangle=\langle 3, 2 \rangle$, and $\vec v=\langle 2-0, 1-2 \rangle=\langle 2, -1\rangle$. Use the projection formula, we get $\vec w=proj_v\vec u=\frac{3\times2-2\times1}{2^2+1^2}\langle 1, 3 \rangle=\frac{4}{5}\langle 2, -1 \rangle=\langle \frac{8}{5}, -\frac{4}{5} \rangle$
(c) As shown in the sketch (not to scale), $\vec w$ is represented by the solid black arrow on the line, and vector $\vec u-\vec w$ is represented by the dashed arrow. As $\vec w$ is the projection of $\vec u$ on the line, we know the dashed arrow is perpendicular to the line, and hence the modulus of the dashed arrow $|\vec u-\vec w|$ represents the distance from point P to the line.
(d) Procedures to find the distance from a given point P to a given line: 1. choose two points Q, R on the line, 2. establish two vectors $\vec {QP}$ and $\vec {QR}$, 3. Find the projection of $\vec {QP}$ on $\vec {QR}$ denoted by $proj_{\vec{QR}}\vec {QP}$, 4. Find the modulus of the vector $|\vec {QP}-proj_{\vec{QR}}\vec {QP}|$ which would be the distance from point P to the line.
