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The slope of CD is 4/3. Which segments are perpendicular to CD? Select each correct answer.

O GF, where G is at (6, 5) and F is at (10, 2)
O LM, where L is at (-2, 3) and M is at (-10, 9)
O JK, where J is at (1,-2) and K is at (10, -14)
O EF, where E is at (4, 1) and F is at (7, 5)

Respuesta :

devishri1977

Answer:

Line CD and GF are perpendicular to each other.

Step-by-step explanation:

The product of the slope of perpendicular lines is -1.

[tex]\sf \text{Slope of line CD, $m_1$ = } \dfrac{4}{3}\\\\\text{So, the line perpendicular to CD should have a slope of } \ \dfrac{-1}{m_1} \ which \ is \ \dfrac{-3}{4}[/tex]

Slope of line GF:

(6,5) ; (10 ,2)

[tex]\boxed{\bf Slope= \dfrac{y_2-y_1}{x_2-x_1}}[/tex]

          [tex]\sf =\dfrac{2-5}{10-6}\\\\\\=\dfrac{-3}{4}\\[/tex]

Line CD and GF are perpendicular to each other.

Slope of line LM:

(-2,3) ; (-10 , 9)

[tex]\sf Slope = \dfrac{9-3}{-10-[-2]}\\\\\\[/tex]

         [tex]\sf =\dfrac{6}{-10+2}\\\\=\dfrac{6}{-7}\\\\\\=\dfrac{-6}{7}[/tex]

LM and CD are not perpendicular to each other.

Slope of JK:

J(1 ,-2) & K(10, -14)

[tex]\sf Slope = \dfrac{-14-[-2]}{10-1}[/tex]

          [tex]\sf = \dfrac{-14+2}{9}\\\\\\=\dfrac{-12}{9}\\\\\\=\dfrac{-4}{3}\\[/tex]

JK and CD are not perpendicular to each other.

Slope of EF:

E(4,1); F(7,5)

[tex]\sf Slope=\dfrac{5-1}{7-4}\\\\[/tex]

         [tex]\sf =\dfrac{4}{3}[/tex]

EF and CD are not perpendicular to each other.

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