Question

1. here are 3 points in the plane. explain how to determine whether point c is closer to point a or point b. (from unit 1, lesson 9.)

Explanation:

Step1: Recall distance formula

If the coordinates of point $A=(x_1,y_1)$, point $B=(x_2,y_2)$ and point $C=(x_0,y_0)$ in the plane, the distance between two points $(x_1,y_1)$ and $(x_0,y_0)$ is given by $d(A,C)=\sqrt{(x_0 - x_1)^2+(y_0 - y_1)^2}$, and the distance between $(x_2,y_2)$ and $(x_0,y_0)$ is $d(B,C)=\sqrt{(x_0 - x_2)^2+(y_0 - y_2)^2}$.

Step2: Compare distances

Calculate $d(A,C)$ and $d(B,C)$. If $d(A,C)d(B,C)$, then point $C$ is closer to point $B$. If $d(A,C) = d(B,C)$, then point $C$ is equidistant from point $A$ and point $B$.

Answer:

Calculate the distance between point $C$ and point $A$ using the distance - formula $d(A,C)=\sqrt{(x_0 - x_1)^2+(y_0 - y_1)^2}$ and the distance between point $C$ and point $B$ using $d(B,C)=\sqrt{(x_0 - x_2)^2+(y_0 - y_2)^2}$. Then compare $d(A,C)$ and $d(B,C)$. If $d(A,C)d(B,C)$, $C$ is closer to $B$; if $d(A,C)=d(B,C)$, $C$ is equidistant from $A$ and $B$.