Question

If a straight line in XY plane is passes through \((-a,-b),(a,b),(k,k),(a^2,a^3)\) for some real number \(a,b\) and \(k\),where \(a≠0\),then which of the following option is correct?

• A. k≠0 when a≠b
• B. k is necessarily a positive real number when a=b
• C. k is any positive real number when a≠b
• D. k=a or k=b necessarily
• E. k=0 when a≠b

Answer 1

Answer: (E)

Given that:
Points are, \((-a, -b), (a, b), (k, k), and (a^2, a^3)\)

First find the slope of the line between the points \((-a,-b)\) and \((a,b)\).

\(m_1 = \dfrac{(b - (-b))}{(a - (-a))} =\dfrac{b}{a}\)

Now, the equation of the line passing through the point \((k, k)\)with slope \(m_1\) is:

\(y - k = \dfrac{b}{a}(x - k)\)

Now, check if another point \((a^2, a^3)\) lies on this line:

\(a^3 - k = \dfrac{b}{a}(a^2 - k)\)

\(⇒a^3 - k= \dfrac{b}{a}(a^2 - k)\)

\(⇒a^3 - k= \dfrac{b}{a}(a^2 - k)\)

\(⇒a^3 - k = \dfrac{b}{a}(a^2 - k)\)

\(⇒a^3 - k= \dfrac{b}{a}(a^2 - k)\)

\(a^3 - k = ba - \dfrac{bk}{a}\)

Now analyzing the options we can state that: 

\(k ≠ 0\) when \(a ≠ b\) Since we have shown that\(k = 0\) when \(a = b\), this statement is true. k cannot be 0 when \(a ≠ b\) because in that case \(k = a^3\). 

Answer 2

Step 1: Understand the problem and given points:

• The line passes through the points \((-a, -b)\), \((a, b)\), \((k, k)\), and \((a^2, a^3)\).
• We need to determine the relationship between \(k\), \(a\), and \(b\) based on the given conditions.

Step 2: Equation of the straight line:

• A straight line can be represented as \(y = mx + c\), where \(m\) is the slope and \(c\) is the y-intercept.
• The slope \(m\) of the line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
  • \[ m = \frac{y_2 - y_1}{x_2 - x_1}. \]

Step 3: Calculate the slope using two points:

• Using the points \((-a, -b)\) and \((a, b)\):
  • \[ m = \frac{b - (-b)}{a - (-a)} = \frac{2b}{2a} = \frac{b}{a}. \]
• Thus, the equation of the line becomes:
  • \[ y = \frac{b}{a}x + c. \]

Step 4: Find the y-intercept \(c\):

• Substitute one of the points, say \((a, b)\), into the line equation \(y = \frac{b}{a}x + c\):
  • \[ b = \frac{b}{a}(a) + c. \]
  • \[ b = b + c. \]
  • \[ c = 0. \]
• Thus, the equation of the line simplifies to:
  • \[ y = \frac{b}{a}x. \]

Step 5: Verify the point \((k, k)\):

• If \((k, k)\) lies on the line, it must satisfy the equation \(y = \frac{b}{a}x\):
  • \[ k = \frac{b}{a}k. \]
  • Rearrange:
  • \[ k \left(1 - \frac{b}{a}\right) = 0. \]
  • This implies:
  • \[ k = 0 \quad \text{or} \quad \frac{b}{a} = 1. \]
• If \(\frac{b}{a} = 1\), then \(b = a\).

Step 6: Analyze the options:

• (A) \(k = 0\) when \(a \neq b\):
  • If \(a \neq b\), then \(\frac{b}{a} \neq 1\), so \(k = 0\) is the only solution. This is correct.
• (B) \(k\) is necessarily a positive real number when \(a = b\):
  • If \(a = b\), then \(\frac{b}{a} = 1\), so \(k\) can be any real number (not necessarily positive). This is incorrect.
• (C) \(k\) is any positive real number when \(a \neq b\):
  • If \(a \neq b\), then \(k = 0\) is the only solution. This is incorrect.
• (D) \(k = a\) or \(k = b\) necessarily:
  • There is no requirement that \(k = a\) or \(k = b\). This is incorrect.
• (E) \(k \neq 0\) when \(a \neq b\):
  • If \(a \neq b\), then \(k = 0\) is the only solution. This is incorrect.

Final Answer:

The correct option is (E) \(k = 0\) when \(a \neq b\).