Perpendicular Positive Slopes?

Method 1 : Directly apply the formula for perpendicular lines.

Let $m_1$ and $m_2$ denote the gradient/slopes of the two lines, then $m_1 m_2 > 0$ and so they can't satisfy the formula for the perpendicular lines, $m_1 m_2 = -1$ , hence the claim is false.

Method 2 : Apply the general formula, $\tan \theta = \dfrac{m_1 - m_2}{1 + m_1 m_2}$ .

Let $\theta$ denote the angle formed between these two straight lines, then $\theta$ satisfy the condition, $\tan \theta = \dfrac{m_1 - m_2}{1 + m_1 m_2}$ , where $m_1$ and $m_2$ denote the gradient/slopes of these straight lines, and because $m_1$ and $m_2$ are strictly positive, we have $1 + m_1 m_2 > 0$ , so $\tan\theta$ is a finite number, and this tells us that $\theta \ne 90^\circ$ .

This tells us that the angle formed between these two straight lines is not a right angle, so they are not perpendicular, hence the claim is false.

they both have to be 0 "slope" but that would not work because 0 is not a positive number