Question

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Directions: Determine if segments AB and CD are parallel, perpendicular, or neither.
AB formed by (-2, 13) and (0, 3)
CD formed by (-5, 0) and (10, 3)

Answer 1

Given:

AB formed by (-2,13) and (0,3).

CD formed by (-5,0) and (10,3).

To find:

Whether the segments AB and CD are parallel, perpendicular, or neither.

Solution:

Slope formula:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

AB formed by (-2,13) and (0,3). So, the slope of AB is:

$m_1=\dfrac{3-13}{0-(-2)}$

$m_1=\dfrac{-10}{2}$

$m_1=-5$

CD formed by (-5,0) and (10,3). So, slope of CD is:

$m_2=\dfrac{3-0}{10-(-5)}$

$m_2=\dfrac{3}{10+5}$

$m_2=\dfrac{3}{15}$

$m_2=\dfrac{1}{5}$

Since $m_1\neq m_2$, therefore the segments AB and CD are not parallel.

$m_1\times m_2=-5\times \dfrac{1}{5}$

$m_1\times m_2=-1$

Since $m_1\times m_2=-1$, therefore the segments AB and CD are perpendicular because product of slopes of two perpendicular lines is always -1.

Hence, the segments AB and CD are perpendicular.

Answer 2

Answer:

AB is perpendicular to CD.

Step-by-step explanation:

AB formed by (-2, 13) and (0, 3)

CD formed by (-5, 0) and (10, 3)

Slope of a line passing through two points is

$m= \frac{y''-y'}{x''- x'}$

The slope of line AB is

$m= \frac{3- 13}{0+2} = -5$

The slope of line CD is

$m'= \frac{3 -0 }{10+5} = \frac{1}{5}$

As the product of m and m' is -1 so the lines AB and CD are perpendicular to each other.