Which of the following lines are perpendicular to the line $3x + 2y = 7$?

Question

The following are a list of 6 equations of lines. I am trying to determine which are perpendicular to the line $3x + 2y = 7$. This is not homework but a practice exercise for the GRE.

(a) $y = \dfrac{2x}{3} + 8$

(b) $y = \dfrac{-2x}{3} - 6$

(c) $y = \dfrac{3x}{2} + 5$

(d) $y = \dfrac{-3x}{2} - \dfrac{4}{7}$.

(e) $y = \dfrac{19 - 2x}{3}$

(f) $y = \dfrac{2}{3} + \dfrac{2x}{3}$.

More importantly, what is the general method of solving for whether a line is perpendicular to a given line? I know the definition using dot products $-$ i.e. that if the dot product of two vectors is equal to zero then they are perpendicular $-$ however I'm not sure how to reason about these situations.

Answer 1

Huge Hint:

If $y=ax+b$ is a linear function, then

$y=-\frac{x}{a}+c$ is perpendicular.

Where $c$ can be any real number

Answer 2

Hint #1: Write $3x + 2y = 7$ in the slope-intercept form $y = mx + b$, where $m$ is the slope of the line.

Hint #2: If two lines are perpendicular, then their slopes $m_1$ and $m_2$ satisfy ${m_1}{m_2} = -1$.

Answer 3

Slope of given line is $-3/2$ . We know product os slopes of perpendicular lines is -1 so the other perpendicular line must have slope $2/3$. Here all options are in form y=mx+c so by checking we come to know that options a and f are perpendicular to line 3x+2y=7.

Answer 4

Proof:

$m_1 = \frac{\Delta y}{\Delta x}=\tan A$ where $\theta$ is the angle made with the x axis.

If $m_2$ is perpendicular then $m_2=\tan B$ where $B = A+90^\circ$

So we have that $$m_1\times m_2 = \tan A \times \tan (90^\circ+A)$$ $$=tan A \times -\cot A =-1$$