Question

c. two sides of the triangle must have the product of their slopes as -1.
d. two sides of the triangle must have the product of their slopes as 1.
what is the formula for the slope of a line through ((x_1,y_1)) and ((x_2,y_2)) with (x_1
eq x_2)?

a. (\frac{y_2 - y_1}{x_2 - x_1}) (checked)

b. (\frac{y_2 - x_2}{y_1 - x_1})

c. (\frac{x_2 - x_1}{y_2 - y_1})

d. (\frac{y_1 - x_1}{y_2 - x_2})

recall that the three points ((-5, -3)), ((-2, -4)), and ((5,1)) are labeled as (a), (b), and (c) respectively. find th
slope (m_{ab}) of the side (ab = -\frac{1}{3}) (type an integer or a simplified fraction.)
slope (m_{bc}) of the side (bc = \frac{5}{7}) (type an integer or a simplified fraction.)
slope (m_{ac}) of the side (ac = \frac{2}{5}) (type an integer or a simplified fraction.)
find (m_{ab} cdot m_{bc}), (m_{ab} cdot m_{ac}), and (m_{bc} cdot m_{ac}).
(m_{ab} cdot m_{bc} = square)
(m_{ab} cdot m_{ac} = square)
(m_{bc} cdot m_{ac} = square)
(type integers or simplified fractions.)
view an example get more help

Explanation:

Step1: Calculate \( m_{AB} \cdot m_{BC} \)

We know \( m_{AB} = -\frac{1}{3} \) and \( m_{BC} = \frac{5}{7} \). Multiply them: \( -\frac{1}{3} \times \frac{5}{7} = -\frac{5}{21} \)

Step2: Calculate \( m_{AB} \cdot m_{AC} \)

We know \( m_{AB} = -\frac{1}{3} \) and \( m_{AC} = \frac{2}{5} \). Multiply them: \( -\frac{1}{3} \times \frac{2}{5} = -\frac{2}{15} \)

Step3: Calculate \( m_{BC} \cdot m_{AC} \)

We know \( m_{BC} = \frac{5}{7} \) and \( m_{AC} = \frac{2}{5} \). Multiply them: \( \frac{5}{7} \times \frac{2}{5} = \frac{2}{7} \)

Answer:

\( m_{AB} \cdot m_{BC} = -\frac{5}{21} \)
\( m_{AB} \cdot m_{AC} = -\frac{2}{15} \)
\( m_{BC} \cdot m_{AC} = \frac{2}{7} \)