Question

segment jk is a diameter of a circle with center ( pleft(-\frac{1}{2}, \frac{7}{3}
ight) ). if point j has coordinates ( left(3, -\frac{1}{3}
ight) ), what are the coordinates of point k? enter the correct answer in the boxes.

Explanation:

Step1: Recall midpoint formula

The midpoint \( M(x_m, y_m) \) of a segment with endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by \( x_m=\frac{x_1 + x_2}{2} \), \( y_m=\frac{y_1 + y_2}{2} \). Here, the center \( P \) is the midpoint of \( JK \), so let \( J=(x_J,y_J)=(3,-\frac{1}{3}) \), \( K=(x_K,y_K) \), and \( P=(x_P,y_P)=(-\frac{1}{2},\frac{7}{3}) \).

Step2: Solve for \( x_K \)

Using the midpoint formula for \( x \)-coordinate: \( x_P=\frac{x_J + x_K}{2} \). Substitute the known values: \( -\frac{1}{2}=\frac{3 + x_K}{2} \). Multiply both sides by 2: \( -1 = 3 + x_K \). Subtract 3 from both sides: \( x_K=-1 - 3=-4 \).

Step3: Solve for \( y_K \)

Using the midpoint formula for \( y \)-coordinate: \( y_P=\frac{y_J + y_K}{2} \). Substitute the known values: \( \frac{7}{3}=\frac{-\frac{1}{3}+y_K}{2} \). Multiply both sides by 2: \( \frac{14}{3}=-\frac{1}{3}+y_K \). Add \( \frac{1}{3} \) to both sides: \( y_K=\frac{14}{3}+\frac{1}{3}=\frac{15}{3}=5 \).

Answer:

\( (-4, 5) \)