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

The distance $d$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ in a plane is given by $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Let the coordinates of point $A$ be $(x_A,y_A)$, point $B$ be $(x_B,y_B)$ and point $C$ be $(x_C,y_C)$.

Step2: Calculate distance $AC$

Using the distance formula, $d_{AC}=\sqrt{(x_C - x_A)^2+(y_C - y_A)^2}$.

Step3: Calculate distance $BC$

Using the distance formula, $d_{BC}=\sqrt{(x_C - x_B)^2+(y_C - y_B)^2}$.

Step4: Compare distances

If $d_{AC}d_{BC}$, then point $C$ is closer to point $B$. If $d_{AC}=d_{BC}$, then point $C$ is equidistant from points $A$ and $B$.

Answer:

Calculate the distance between $C$ and $A$ and the distance between $C$ and $B$ using the distance formula $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. Compare the two - calculated distances. If the distance between $C$ and $A$ is smaller, $C$ is closer to $A$. If the distance between $C$ and $B$ is smaller, $C$ is closer to $B$. If the two distances are equal, $C$ is equidistant from $A$ and $B$.