Plotting the Points

To begin, let's plot the given points P(1, 0), Q(4, 0), and S(1, 3) on a coordinate plane. This will help us visualize the problem and find the coordinates of the point R.

Calculating the Distance

To determine if PQRS is a square, we need to find the distances between the points. We can use the distance formula:

Distance = √[(x2 - x1)^2 + (y2 - y1)^2]

Let's calculate the distances PQ, QR, RS, and SP.

- PQ: Distance between P(1, 0) and Q(4, 0)
PQ = √[(4 - 1)^2 + (0 - 0)^2] = √[3^2 + 0] = √9 = 3

- QR: Distance between Q(4, 0) and S(1, 3)
QR = √[(1 - 4)^2 + (3 - 0)^2] = √[(-3)^2 + 3^2] = √[9 + 9] = √18

- RS: Distance between S(1, 3) and R(x, y)
RS = √[(x - 1)^2 + (y - 3)^2]

- SP: Distance between R(x, y) and P(1, 0)
SP = √[(1 - x)^2 + (0 - y)^2]

Properties of a Square

To form a square, PQRS must satisfy the following conditions:

1. PQ = QR = RS = SP (all sides are equal)
2. The diagonals of PQRS are equal and perpendicular to each other.

Finding the Coordinates of R

Since PQ = QR = RS, we can equate the distances obtained:

√18 = √[(x - 1)^2 + (y - 3)^2]

Squaring both sides, we get:

18 = (x - 1)^2 + (y - 3)^2

Expanding the equation, we have:

18 = x^2 - 2x + 1 + y^2 - 6y + 9

Combining like terms, we get:

x^2 + y^2 - 2x - 6y - 8 = 0

This equation represents a circle centered at (1, 3) with a radius of √18.

Plotting the Circle

Now, let's plot the circle on the coordinate plane using the equation x^2 + y^2 - 2x - 6y - 8 = 0.

Intersection of Circle and Line PQ

To find the coordinates of R, we need to find the intersection point of the circle and the line PQ. The equation of line PQ is y = 0.

Substituting y = 0 into the equation of the circle, we get:

x^2 - 2x - 8 = 0

Solving this quadratic equation, we find two possible x-values: x = -2 and x = 4.

Calculating the y-Coordinate of R