Question

One possible condition for the three points \((a,b), (b,a)\) and \((a^2, -b^2)\) to be collinear, is

• A. \(a - b = 2\)
• B. \(a + b = 2\)
• C. \(a = 1 + b\)
• D. \(a = 1 - b\)

Hint:

Collinearity condition: slope between any two pairs is equal.

Answer 1

The Correct Option is C

To determine the condition under which the three points \((a,b)\), \((b,a)\), and \((a^2, -b^2)\) are collinear, we need to ensure that the slope between any two pairs of these points is the same. Specifically, we will ensure that the slope between points \((a,b)\) and \((b,a)\) is equal to the slope between points \((b,a)\) and \((a^2, -b^2)\).

The slope between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

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

First, we calculate the slope between points \((a,b)\) and \((b,a)\):

\(m_1 = \frac{a-b}{b-a} = -1\)

Next, we calculate the slope between points \((b,a)\) and \((a^2, -b^2)\):

\(m_2 = \frac{-b^2 - a}{a^2 - b}\)

For the points to be collinear, these slopes must be equal:

\(m_1 = m_2\)

This gives us the equation:

\(-1 = \frac{-b^2 - a}{a^2 - b}\)

Cross-multiplying, we have:

\(a^2 - b = -(-b^2 - a)\)

Simplifying, we get:

\(a^2 - b = b^2 + a\)

Rearrange the terms to form a quadratic equation:

\(a^2 - a - b^2 - b = 0\)

To find specific solutions or conditions, we look at possible relationships between \(a\) and \(b\). Let's test the given options:

1. \(a - b = 2\): This doesn't simplify or fit directly into the quadratic.
2. \(a + b = 2\): This would require substitution into the quadratic and again doesn't directly simplify systematically.
3. \(a = 1 + b\): Substituting gives \(a = b + 1\).
   • Upon substitution into the equation \(a^2 - a - b^2 - b = 0\):
   • Left: \((b+1)^2 - (b+1) - b^2 - b = 0\)
   • Expanding: \((b^2 + 2b + 1) - (b + 1) - b^2 - b = 0\)
   • Simplifies to: \(0\), so this is a valid condition.
4. \(a = 1 - b\): This substitution doesn't zero out the equation, suggesting non-collinearity.

Thus, the correct condition is:

\(a = 1 + b\)