Question

On the coordinate plane, point P is located at (3, y) and point Q is located at (1, -4). The distance between

points P and Q is 29 units.
What are the two possible values of y?

Answer 1

The two possible values of y are -20 and 12.

Step-by-step explanation:

The distance between two points on a coordinate plane can be found using the distance formula:

d = sqrt((x2 - x1)^2 + (y2 - y1)^2)

In this case, the coordinates of point P are (3, y) and the coordinates of point Q are (1, -4).

We are given that the distance between P and Q is 29 units.

Plugging in these values into the distance formula, we get:

29 = sqrt((1 - 3)^2 + (-4 - y)^2)

Simplifying, we get:

841 = 4 + y^2 + 8y

Rearranging and solving the quadratic equation, we find two possible values of y: -20 and 12.