Question

Determine if the lines that pass through the given points are parallel, perpendicular or neither.

Answer 1

The equation of the line passing through the points $(x_1,y_1)$ and $(x_2,y_2)$ is

$\dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}.$

Then

1) the equation of line A is

$\dfrac{x-(-8)}{-5-(-8)}=\dfrac{y-8}{10-8},\\ \\\dfrac{x+8}{3}=\dfrac{y-8}{2},\\ \\2(x+8)=3(y-8),\\ \\2x+16=3y-24,\\ \\2x-3y+40=0.$

2) the equation of line B is

$\dfrac{x-(-7)}{-5-(-7)}=\dfrac{y-12}{15-12},\\ \\\dfrac{x+7}{2}=\dfrac{y-12}{3},\\ \\3(x+7)=2(y-12),\\ \\3x+21=2y-24,\\ \\3x-2y+45=0.$

Since $2\cdot 3+(-3)\cdot (-2)=12\neq 0,$ then lines are not perpendicular.

Since $\dfrac{2}{3}\neq \dfrac{-3}{-2},$ then lines are not parallel.

Answer: neither parallel, nor perpendicular

Answer 2

Answer:

Lines A and B are neither parallel nor perpendicular lines.

Step-by-step explanation:

For two lines,

if their slopes are equal then the lines are parallel; and

if the slope of one line is the negative reciprocal of the slope of the other line, then the two lines are equal.

Finding the slopes of line A and B:

$mA= \frac{(10-8)}{(-5-(-8))} = \frac{2}{3}$

$mB = \frac{15-12}{-5-(-7)} = \frac{3}{2}$

$\frac{2}{3}$ $\neq \frac{3}{2}$ and neither they are the negative reciprocal of each other so the lines A and B are neither parallel nor perpendicular lines.