Question

either prove or disprove the statement, \the points (-5, -3), (-2, -4), and (5,1) are the vertices of a right triangle.\ use a graph only as a guide.
a. b. c. d.
graphs of triangles labeled a, b, c, d
by looking at the graph it seems that the points (-5, -3), (-2, -4), and (5,1) do not form a right triangle.
note that the right triangle has two sides perpendicular to each other. if (-5, -3), (-2, -4), and (5,1) are the vertices of a right triangle, which of the following must be true
a. two sides of the triangle must have the sum of their slopes as 0.
b. two sides of the triangle must have the same slope.
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 (x₁,y₁) and (x₂,y₂) with x₁≠x₂?
a. $\frac{y_2 - y_1}{x_2 - x_1}$
b. $\frac{y_2 - x_2}{y_1 - x_1}$
c. $\frac{x_2 - x_1}{y_2 - y_1}$
d. $\frac{y_1 - x_1}{y_2 - x_2}$

Explanation:

For the first multiple-choice question (about right triangle sides' slopes):

To determine if a triangle is right - angled, we use the property of perpendicular lines. If two lines are perpendicular, the product of their slopes is - 1. So, for a right triangle, two of its sides (which are perpendicular) must have slopes whose product is - 1. Option A is incorrect as the sum of slopes has no relation to perpendicularity. Option B is for parallel lines (same slope), not perpendicular. Option D is incorrect as the product of slopes for perpendicular lines is - 1, not 1.

The slope \(m\) of a line passing through two points \((x_1,y_1)\) and \((x_2,y_2)\) (where \(x_1
eq x_2\)) is given by the formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\). This is derived from the concept of "rise over run", where the "rise" is the change in the \(y\) - values (\(y_2 - y_1\)) and the "run" is the change in the \(x\) - values (\(x_2 - x_1\)).

Step 1: Find the slope of \(AB\)

The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). For points \(A(-5,-3)\) and \(B(-2,-4)\), let \((x_1,y_1)=(-5,-3)\) and \((x_2,y_2)=(-2,-4)\).
\(m_{AB}=\frac{-4 - (-3)}{-2-(-5)}=\frac{-4 + 3}{-2 + 5}=\frac{-1}{3}=-\frac{1}{3}\)

Step 2: Find the slope of \(BC\)

For points \(B(-2,-4)\) and \(C(5,1)\), let \((x_1,y_1)=(-2,-4)\) and \((x_2,y_2)=(5,1)\).
\(m_{BC}=\frac{1-(-4)}{5-(-2)}=\frac{1 + 4}{5 + 2}=\frac{5}{7}\)

Step 3: Find the slope of \(AC\)

For points \(A(-5,-3)\) and \(C(5,1)\), let \((x_1,y_1)=(-5,-3)\) and \((x_2,y_2)=(5,1)\).
\(m_{AC}=\frac{1-(-3)}{5-(-5)}=\frac{1 + 3}{5 + 5}=\frac{4}{10}=\frac{2}{5}\)

Step 4: Check the product of slopes

We check the products of the slopes:

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

eq - 1\)

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

eq - 1\)

• \(m_{BC}\times m_{AC}=\frac{5}{7}\times\frac{2}{5}=\frac{2}{7}

eq - 1\)

Since none of the products of the slopes of the sides is - 1, the two lines (sides) are not perpendicular. So the points do not form a right triangle.

Answer:

C. Two sides of the triangle must have the product of their slopes as - 1.

For the second multiple-choice question (slope formula):