Analyzing Collinearity of Four Points

The question asks us to determine the geometric relationship between four given points: $P_1=(a, b)$, $P_2=(0, 0)$, $P_3=(-a, -b)$, and $P_4=(ab, b^2)$. We are given options that suggest these points might form a parallelogram, rectangle, square, or be collinear.

Let's analyze the points. We have the origin $(0, 0)$ as one of the points. Notice that $P_3=(-a, -b)$ is the negative of $P_1=(a, b)$. This means $P_2$ is the midpoint of the line segment connecting $P_1$ and $P_3$. This strongly suggests that $P_1$, $P_2$, and $P_3$ lie on a straight line passing through the origin.

Checking for Collinearity of P1, P2, and P3

Three points are collinear if the slope between any two pairs of points is the same. Let's calculate the slopes:

• Slope of the line segment connecting $P_1(a, b)$ and $P_2(0, 0)$:

The slope $m_{12}$ is given by $\frac{y_2 - y_1}{x_2 - x_1}$.

$m_{12} = \frac{0 - b}{0 - a} = \frac{-b}{-a} = \frac{b}{a}$, provided $a \neq 0$.

• Slope of the line segment connecting $P_2(0, 0)$ and $P_3(-a, -b)$:

The slope $m_{23}$ is given by $\frac{y_3 - y_2}{x_3 - x_2}$.

$m_{23} = \frac{-b - 0}{-a - 0} = \frac{-b}{-a} = \frac{b}{a}$, provided $a \neq 0$.

Since $m_{12} = m_{23}$, the points $P_1$, $P_2$, and $P_3$ are collinear, provided $a \neq 0$.

Handling Edge Cases (When a=0 or b=0)

• If $a = 0$, the points are $P_1(0, b)$, $P_2(0, 0)$, and $P_3(0, -b)$. These points all lie on the y-axis (the line $x=0$), so they are collinear. The slopes would be undefined (vertical line).
• If $b = 0$, the points are $P_1(a, 0)$, $P_2(0, 0)$, and $P_3(-a, 0)$. These points all lie on the x-axis (the line $y=0$), so they are collinear. The slopes would be 0 (horizontal line).

In all cases, the first three points $P_1$, $P_2$, and $P_3$ are collinear. The line passing through them is $y = \frac{b}{a}x$ if $a \neq 0$ and $b \neq 0$. If $a=0$, it's $x=0$. If $b=0$, it's $y=0$.

Checking if the Fourth Point P4 is Collinear

Now, let's check if the fourth point $P_4(ab, b^2)$ also lies on the same line as $P_1$, $P_2$, and $P_3$.

• Case 1: $a \neq 0$ and $b \neq 0$. The line equation is $y = \frac{b}{a}x$. Let's substitute the coordinates of $P_4(ab, b^2)$:

$b^2 = \frac{b}{a}(ab)$

$b^2 = b^2$

This is true. So, $P_4$ lies on the line.

• Case 2: $a = 0$. The line is $x = 0$. The point $P_4$ is $(0 \cdot b, b^2) = (0, b^2)$. This point $(0, b^2)$ has an x-coordinate of 0, so it lies on the line $x=0$.
• Case 3: $b = 0$. The line is $y = 0$. The point $P_4$ is $(a \cdot 0, 0^2) = (0, 0)$. This point $(0, 0)$ has a y-coordinate of 0, so it lies on the line $y=0$. Note that $(0,0)$ is $P_2$, which is already on the line.

Conclusion on Collinearity

In all possible scenarios for the values of $a$ and $b$, the fourth point $P_4(ab, b^2)$ lies on the line passing through $P_1$, $P_2$, and $P_3$. Therefore, all four points $(a, b)$, $(0, 0)$, $(-a, -b)$, and $(ab, b^2)$ are collinear.

Why other options are incorrect

If four points are collinear, they all lie on a single straight line. A parallelogram, rectangle, and square are two-dimensional shapes formed by four non-collinear points. Since the given points are collinear, they cannot form any of these quadrilaterals.

Thus, the points are collinear.

Revision Table: Key Concepts

Additional Information: Verifying Collinearity with Area

Another way to check if three points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ are collinear is to calculate the area of the triangle formed by them. If the area is zero, the points are collinear.

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

Let's apply this to $P_1(a, b)$, $P_2(0, 0)$, and $P_3(-a, -b)$:

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

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

Area $= \frac{1}{2} |ab + 0 - ab|$

Area $= \frac{1}{2} |0| = 0$

Since the area of the triangle formed by $P_1$, $P_2$, and $P_3$ is zero, these three points are confirmed to be collinear. We then verified that $P_4$ also lies on the same line, proving all four points are collinear.