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What is the equation of a line that is perpendicular to y = 3x-2 and passes
through the point (6, 8)?

What Is The Equation Of A Line That Is Perpendicular To Y 3x2 And Passes Through The Point 6 8 class=

Sagot :

Answer:

8 = (-1/3)(6) + b

8 = -2 + b

b = 10

y = (-1/3)x + 10

The correct answer is A.

Answer:

Step-by-step explanation:

To find the equation of a line that is perpendicular to y = 3x - 2, we first need to determine the slope of the given line. In this case, the slope of y = 3x - 2 is 3.

The slope of a line perpendicular to another line is the negative reciprocal of the original slope. So, the slope of the perpendicular line will be -1/3.

Next, we use the point-slope form of a linear equation to find the equation of the perpendicular line passing through point (6,8). The point-slope form is given by y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.

Substitute m = -1/3 and (x1, y1) = (6,8) into the point-slope form:

y - 8 = (-1/3)(x - 6)

y - 8 = (-1/3)x + 2

y = (-1/3)x + 10

Therefore, the equation of the line that is perpendicular to y = 3x - 2 and passes through point (6,8) is y = (-1/3)x + 10.

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