Question

If m and b are real numbers and mb > 0, then the line whose equation is y = mx + b cannot contain the point
A) $\left( 0, 2009 \right)$
B) $\left( 2009, 0 \right)$
C) $\left( 0, -2009 \right)$
D) $\left( 20, -100 \right)$

Answer 1

In this problem, we just have to substitute the options to check for the point which is not contained. We can start by choosing the second option for checking the possibility, which does not contain the point for the condition mb > 0 for the equation y = mx + b.

Complete step-by-step solution:
We have the given equation as
y = mx + b …….. (1)
To check the possibility for the given option, which does not contain the point, we will consider each option one by one.
We can take option A) $\left( 0,2009 \right)$
Substituting this option in equation (1), we will get
$2009=m\left( 0 \right)+b$
Here mb > 0, the condition is satisfied and hence it is not the correct option as the question is the point which is not contained for the condition.
Now we can take the point B) $\left( 2009,0 \right)$, to check the possibility for the condition.
For $\left( 2009,0 \right)$
Substituting in the given equation (1), we get
$2900m+b=0$
This is possible only if mb < 0, which contradicts the given condition mb > 0. Therefore, the correct option is B) $\left( 2009, 0 \right)$.

Note: Check for other options too, for the possibilities. So, for option C), we will get b = -2009. This also doesn’t satisfy mb > 0 condition. Similarly, option D), we will get -100 = 20m + b. Here, both m and b must be positive for mb > 0, but then it won’t satisfy -100. So, this is also not possible. Students may make mistakes in writing incorrect symbols for the relational operations. Students may also get confused in the question part, read it twice and understand what has been asked for. In this question, they asked for the point which does not satisfy the given condition, that cannot contain the point.