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Determine if the lines that pass through the given points are parallel, perpendicular or neither.
Line A: (-8, 8) and (-5, 10) Line B: (-7, 12) and (-5. 15)

Respuesta :

frika

The equation of the line passing through the points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] is

[tex]\dfrac{x-x_1}{x_2-x_1}=\dfrac{y-y_1}{y_2-y_1}.[/tex]

Then

1) the equation of line A is

[tex]\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.[/tex]

2) the equation of line B is

[tex]\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.[/tex]

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

Since [tex]\dfrac{2}{3}\neq \dfrac{-3}{-2},[/tex] then lines are not parallel.

Answer: neither parallel, nor perpendicular


alinakincsem

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:

[tex]mA= \frac{(10-8)}{(-5-(-8))} = \frac{2}{3}[/tex]

[tex]mB = \frac{15-12}{-5-(-7)} = \frac{3}{2}[/tex]

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

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