Understanding the Problem: Collinearity of Points

The question asks us to determine the geometric relationship between four given points: \((-a, -b)\), \((0, 0)\), \((a, b)\), and \((a^2, ab)\). We are given four options describing possible relationships: lying on the same circle, being vertices of a square, being vertices of a parallelogram (not a square), or being collinear.

What Does Collinear Mean?

Points are said to be collinear if they all lie on the same straight line. If a set of points is collinear, they cannot form a polygon like a square or a parallelogram (unless there are fewer than 3 distinct points, in which case they trivially lie on a line but don't form a polygon with area). Points lying on a circle generally form a curve, not a straight line, unless only two distinct points are considered.

Checking for Collinearity using Slopes

A common way to check if three or more points are collinear is to calculate the slope between different pairs of points. If the slope between any two pairs of distinct points is the same, and they share a common point, then all three points are collinear. If we extend this to four points, we check the slopes between multiple pairs. If all pairs of distinct points have the same slope, they lie on the same line and are collinear.

The formula for the slope (\(m\)) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\(\qquad m = \frac{y_2 - y_1}{x_2 - x_1}\)

Let's label the given points:

• \(P_1 = (-a, -b)\)
• \(P_2 = (0, 0)\)
• \(P_3 = (a, b)\)
• \(P_4 = (a^2, ab)\)

Step-by-Step Collinearity Check

We will calculate the slopes between various pairs of these points.

Case 1: General Case (assuming \(a \neq 0\) and \(b \neq 0\))

Slope of \(P_1P_2\):

\(\qquad m_{P_1P_2} = \frac{0 - (-b)}{0 - (-a)} = \frac{b}{a}\)

Slope of \(P_2P_3\):

\(\qquad m_{P_2P_3} = \frac{b - 0}{a - 0} = \frac{b}{a}\)

Since \(m_{P_1P_2} = m_{P_2P3}\) and they share point \(P_2\), points \(P_1, P_2, P_3\) are collinear. They all lie on the line passing through the origin with slope \(\frac{b}{a}\).

Slope of \(P_2P_4\):

\(\qquad m_{P_2P4} = \frac{ab - 0}{a^2 - 0} = \frac{ab}{a^2}\)

If \(a \neq 0\), this simplifies to \(\frac{b}{a}\).

So, \(m_{P_2P4} = \frac{b}{a}\). This means points \(P_2, P_3, P_4\) are also collinear with slope \(\frac{b}{a}\).

Slope of \(P_1P_4\):

\(\qquad m_{P_1P4} = \frac{ab - (-b)}{a^2 - (-a)} = \frac{ab + b}{a^2 + a} = \frac{b(a + 1)}{a(a + 1)}\)

If \(a \neq 0\) and \(a \neq -1\), this simplifies to \(\frac{b}{a}\).

Summary of Slopes (assuming \(a \neq 0, b \neq 0, a \neq -1\)):

• \(m_{P_1P_2} = \frac{b}{a}\)
• \(m_{P_2P_3} = \frac{b}{a}\)
• \(m_{P_2P4} = \frac{b}{a}\)
• \(m_{P_1P4} = \frac{b}{a}\)

Since the slopes between various pairs of points are equal, the points \(P_1, P_2, P_3, P_4\) are collinear in the general case.

Case 2: Special Cases

What happens if \(a=0\) or \(b=0\) or \(a=1\) or \(a=-1\)?

• If \(a = 0\): The points become \((-0, -b), (0, 0), (0, b), (0^2, 0b)\), which are \((0, -b), (0, 0), (0, b), (0, 0)\). The distinct points are \((0, -b), (0, 0), (0, b)\). These points all lie on the y-axis (where \(x=0\)). Points on a vertical line are collinear.

• If \(b = 0\): The points become \((-a, -0), (0, 0), (a, 0), (a^2, a0)\), which are \((-a, 0), (0, 0), (a, 0), (a^2, 0)\). These points all lie on the x-axis (where \(y=0\)). Points on a horizontal line are collinear.

• If \(a = 1\): The points become \((-1, -b), (0, 0), (1, b), (1^2, 1b)\), which are \((-1, -b), (0, 0), (1, b), (1, b)\). Points \(P_3\) and \(P_4\) are the same. The distinct points are \((-1, -b), (0, 0), (1, b)\). The slope between \((-1, -b)\) and \((0, 0)\) is \(\frac{0 - (-b)}{0 - (-1)} = \frac{b}{1} = b\). The slope between \((0, 0)\) and \((1, b)\) is \(\frac{b - 0}{1 - 0} = \frac{b}{1} = b\). Since slopes are equal, the points are collinear.

• If \(a = -1\): The points become \((-(-1), -b), (0, 0), (-1, b), ((-1)^2, (-1)b)\), which are \((1, -b), (0, 0), (-1, b), (1, -b)\). Points \(P_1\) and \(P_4\) are the same. The distinct points are \((1, -b), (0, 0), (-1, b)\). The slope between \((1, -b)\) and \((0, 0)\) is \(\frac{0 - (-b)}{0 - 1} = \frac{b}{-1} = -b\). The slope between \((0, 0)\) and \((-1, b)\) is \(\frac{b - 0}{-1 - 0} = \frac{b}{-1} = -b\). Since slopes are equal, the points are collinear.

In all cases, including the special cases where \(a\) or \(b\) are zero, or \(a\) is 1 or -1 leading to repeated points, the distinct points always lie on the same line.

Evaluating Other Options

• Lying on the same circle: Collinear points (more than two distinct points) cannot lie on the same circle. A circle is a curve.

• Vertices of a square or parallelogram: A square or a parallelogram requires four distinct vertices that do not lie on a single straight line. Since the given points are collinear, they cannot form a quadrilateral with area, and thus cannot be the vertices of a square or parallelogram.

Therefore, the only valid description for the relationship between the four points is that they are collinear.

Conclusion

By calculating the slopes between pairs of points and considering special cases, we consistently find that the points \((-a, -b)\), \((0, 0)\), \((a, b)\), and \((a^2, ab)\) lie on the same straight line, provided they are distinct. Thus, the points are collinear.

Revision Table: Key Concepts for Collinearity

Additional Information: Alternative Collinearity Check

Besides the slope method, another way to check for collinearity of three points \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is to calculate the area of the triangle formed by them. If the area is zero, the points are collinear.

The area of a triangle can be calculated using the determinant formula:

\(\qquad \text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\)

For points \(P_1(-a, -b)\), \(P_2(0, 0)\), \(P_3(a, b)\), let's check the area:

\(\qquad \text{Area}(P_1P_2P_3) = \frac{1}{2} |(-a)(0 - b) + 0(b - (-b)) + a(-b - 0)|\)

\(\qquad \text{Area}(P_1P_2P_3) = \frac{1}{2} |(-a)(-b) + 0(2b) + a(-b)|\)

\(\qquad \text{Area}(P_1P_2P_3) = \frac{1}{2} |ab + 0 - ab| = \frac{1}{2} |0| = 0\)

Since the area of triangle \(P_1P_2P_3\) is zero, points \(P_1, P_2, P_3\) are collinear. You could similarly check other combinations like \(P_1P_2P_4\), \(P_1P_3P_4\), etc., but showing that \(P_1, P_2, P_3\) are collinear and \(P_2, P_3, P_4\) are collinear (using slopes or area) is sufficient to show all four are collinear, provided \(P_2\) and \(P_3\) are distinct (which they are unless \(a=0, b=0\)). If \(P_2=P_3\), they are the origin \((0,0)\) and the points become \((-a, -b), (0,0), (a^2, ab)\). We already covered the \(a=0, b=0\) case where all points are the origin and thus collinear.