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 \(k = a^3\).