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)

k≠0 when a≠b

Answer 2

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 $k = a^3$.