Question

J(-4,2) is the center of a circle. K(-1,5) is any point on the circle. A ling y=x+9 is a chord of the circle. Prove that the line segment JK is Parallel to the Chord.

Answer 1

By calculating the slope of the chord from the equation y=x+9 and the slope of line segment JK using their coordinates and comparing them, we are able to prove that segment JK is parallel to the chord, as both have a slope of 1.

Step-by-step explanation:

We are asked to prove that the line segment JK is parallel to the chord of a circle where J(-4,2) is the center of the circle and K(-1,5) is a point on the circle with the chord described by the equation y = x + 9. To show that segments are parallel, we need to demonstrate that they have the same slope.

The slope of the chord comes from its equation, which is in the form y = mx + b, with m representing the slope. In this case, m = 1 because the coefficient of x in the equation y = x + 9 is 1, indicating a slope of 1.

Now, we calculate the slope of the segment JK using the coordinates of J and K:

• Let J be (x1, y1) = (-4, 2)
• Let K be (x2, y2) = (-1, 5)

We use the slope formula m = (y2 - y1) / (x2 - x1) to find the slope of JK:

m = (5 - 2) / (-1 + 4)
m = 3 / 3
m = 1

Since the slope of JK is also 1, the slopes are equal and therefore line segment JK is parallel to the chord.