Question

either prove or disprove the statement, \the points (-5, -3), (-2, -4), and (5,1) are the vertices of...

○ 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 the slope
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} = -\frac{5}{21} )
( m_{ab} cdot m_{ac} = -\frac{2}{15} )
( m_{bc} cdot m_{ac} = \frac{2}{7} )
(type integers or simplified fractions.)
does the triangle have two perpendicular sides?
○ yes
○ no

Explanation:

Step1: Recall Perpendicular Slopes Rule

Two lines are perpendicular if the product of their slopes is \(-1\).

Step2: Check Each Product

• For \(m_{AB} \cdot m_{BC}\): \(-\frac{1}{3} \cdot \frac{5}{7} = -\frac{5}{21}

eq -1\)

• For \(m_{AB} \cdot m_{AC}\): \(-\frac{1}{3} \cdot \frac{2}{5} = -\frac{2}{15}

eq -1\)

• For \(m_{BC} \cdot m_{AC}\): \(\frac{5}{7} \cdot \frac{2}{5} = \frac{2}{7}

eq -1\)

Since none of the products of the slopes is \(-1\), no two sides are perpendicular.

Answer:

No