cjboboriq2n
contestada

The coordinates of the vertices of △JKL are J(0, 2) , K(3, 1) , and L(1, −5) .

Drag and drop the choices into each box to correctly complete the sentences.

The slope of JK¯¯¯¯¯ is , the slope of KL¯¯¯¯¯ is , and the slope of JL¯¯¯¯¯ is . △JKL a right triangle because .

The coordinates of the vertices of JKL are J0 2 K3 1 and L1 5 Drag and drop the choices into each box to correctly complete the sentences The slope of JK is the class=

Respuesta :

calculista

Answer:

Part 1) The slope of JK is [tex]m_J_K=-\frac{1}{3}[/tex]

Part 2) The slope of KL is [tex]m_K_L=3[/tex]

Part 3) The slope of JL is [tex]m_J_L=-7[/tex]

Part 4) Triangle JKL is a right triangle because two of these slopes have a product of -1

Step-by-step explanation:

we know that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

The formula to calculate the slope between two points is equal to

[tex]m=\frac{y2-y1}{x2-x1}[/tex]

we have

[tex]J(0, 2), K(3, 1),L(1, -5)[/tex]

Part 1) Find the slope JK

we have

[tex]J(0, 2), K(3, 1)[/tex]

substitute in the formula

[tex]m=\frac{1-2}{3-0}[/tex]

[tex]m_J_K=-\frac{1}{3}[/tex]

Part 2) Find the slope KL

we have

[tex]K(3, 1),L(1, -5)[/tex]

substitute in the formula

[tex]m=\frac{-5-1}{1-3}[/tex]

[tex]m_K_L=\frac{-6}{-2}[/tex]

[tex]m_K_L=3[/tex]

Part 3) Find the slope JL

we have

[tex]J(0, 2),L(1, -5)[/tex]

substitute in the formula

[tex]m=\frac{-5-2}{1-0}[/tex]

[tex]m_J_L=\frac{-7}{1}[/tex]

[tex]m_J_L=-7[/tex]

Part 4) Compare the slopes

[tex]m_J_K=-\frac{1}{3}[/tex]

[tex]m_K_L=3[/tex]

[tex]m_J_L=-7[/tex]

we have that

JK and KL are perpendicular because their slopes are opposite reciprocal

The product of their slopes is equal to -1

therefore

Triangle JKL is a right triangle because two of these slopes have a product of -1

Otras preguntas