Question

sort the equations into a pair of parallel lines, a pair of perpendicular lines, and a pair of oblique lines.
these lines are parallel
these lines are perpendicular
these lines are oblique
: y - 6 = 2(x + 7)
: y - 4 = \frac{2}{3}(x - 3)
: y - 1 = -\frac{2}{3}(x + 4)
: y = \frac{3}{2}x + 5
: y = \frac{2}{3}x + 7
: y = - 2x + 3

Explanation:

Step1: Convert to slope - intercept form

The slope - intercept form is $y = mx + b$, where $m$ is the slope.

1. For $y - 6=2(x + 7)$, expand: $y-6 = 2x+14$, then $y=2x + 20$, slope $m_1 = 2$.
2. For $y - 4=\frac{2}{3}(x - 3)$, expand: $y-4=\frac{2}{3}x - 2$, then $y=\frac{2}{3}x+2$, slope $m_2=\frac{2}{3}$.
3. For $y - 1=-\frac{3}{2}(x + 4)$, expand: $y-1=-\frac{3}{2}x-6$, then $y=-\frac{3}{2}x - 5$, slope $m_3=-\frac{3}{2}$.
4. For $y=\frac{3}{2}x + 5$, slope $m_4=\frac{3}{2}$.
5. For $y=\frac{2}{3}x + 7$, slope $m_5=\frac{2}{3}$.
6. For $y=-2x + 3$, slope $m_6=-2$.

Step2: Determine parallel lines

Parallel lines have equal slopes. So, $y - 4=\frac{2}{3}(x - 3)$ and $y=\frac{2}{3}x + 7$ are parallel since $m_2=m_5=\frac{2}{3}$.

Step3: Determine perpendicular lines

Perpendicular lines have slopes such that $m_1\times m_2=-1$. The product of the slopes of $y - 6=2(x + 7)$ (with $m_1 = 2$) and $y - 1=-\frac{3}{2}(x + 4)$ (with $m_3=-\frac{3}{2}$) is $2\times(-\frac{3}{2})=- 3
eq - 1$. But the product of the slopes of $y - 6=2(x + 7)$ (slope $m = 2$) and $y=-2x + 3$ (slope $m=-2$) is $2\times(-2)=-4
eq - 1$. The product of the slopes of $y=\frac{3}{2}x + 5$ and $y - 1=-\frac{3}{2}(x + 4)$ is $\frac{3}{2}\times(-\frac{3}{2})=-\frac{9}{4}
eq - 1$. The product of the slopes of $y=\frac{3}{2}x + 5$ (slope $m=\frac{3}{2}$) and $y - 4=\frac{2}{3}(x - 3)$ (slope $m=\frac{2}{3}$) is $\frac{3}{2}\times\frac{2}{3}=1
eq - 1$. However, the product of the slopes of $y=\frac{3}{2}x + 5$ and $y - 1=-\frac{2}{3}(x + 4)$ is $\frac{3}{2}\times(-\frac{2}{3})=-1$. So, $y=\frac{3}{2}x + 5$ and $y - 1=-\frac{2}{3}(x + 4)$ are perpendicular.

Step4: Determine oblique lines

Oblique lines are neither parallel nor perpendicular. So, $y - 6=2(x + 7)$ and $y=\frac{2}{3}x + 7$ are oblique.

Answer:

These lines are Parallel: $y - 4=\frac{2}{3}(x - 3)$, $y=\frac{2}{3}x + 7$
These lines are Perpendicular: $y=\frac{3}{2}x + 5$, $y - 1=-\frac{2}{3}(x + 4)$
These lines are Oblique: $y - 6=2(x + 7)$, $y=\frac{2}{3}x + 7$