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ANSWER WITH SOLUTION:

Find the value of K so that Kx-3y=5 and 5x-6y=8 will be parallel and perpendicular

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Find the value of K so that Kx-3y=5 and 5x-6y=8 will be parallel and perpendicular

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1)

- 6y = - 5x + 8

- 6y/-6 = -5x/-6 + 8/-6

y = 5x/6 - 4/3

kx - 3y = 5

- 3y = -kx + 5

- 3y/-3 = -kx/-3 + 5/-3

y = kx/3 - 5/3

m = 5/6

k/3 = 5/6

6k = (3) (5)

6k = 15

6k/6 = 15/6

Therefore, Equation 2 is:

5x/2 - 3y = 5 LCD: 2

2 (5x/2 - 3y = 5)2

-6y = -5x + 10

-6y/-6 = -5x/-6 + 10/-6

y = 5x/6 - 5/3

2) For perpendicular equations, the product of the slopes of the two equations is -1.

k/3 = -6/5

5k = (3) (-6)

5k = -18

5k/5 = -18/5

Therefore, to make the equations perpendicular, Equation 2 is:

kx - 3y = 5

-18x/5 - 3y = 5 LCD: 5

5 (-18x/5 - 3y = 5) 5

Slope-intercept form:

-15y = 18x + 25

-15y/-15 = 18x/-15 + 25/-15

y = -6x/5 - 5/3