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A line passes through the point (12,–4) and is perpendicular to the line with the equation y = 6x+3. what's the equation of the line? options: a) y = –6x+68 b) y = –1∕6x – 2 c) y = –1∕6x 2 d) y = 6x – 76

User Ketsia
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Final answer:

The equation of the line that is perpendicular to y = 6x+3 and passes through (12, -4) is y = –1/6x + 2. This equation is found by using the concept that perpendicular lines have slopes that are negative reciprocals of each other and the point-slope formula.

Step-by-step explanation:

The subject of this problem falls within Mathematics, specifically Geometry and Algebra. The problem is seeking the equation of a line that passes through the point (12,-4) and is perpendicular to the line with the equation y = 6x+3.

To find the equation of the line that is perpendicular to another, you need to know that perpendicular lines have slopes that are negative reciprocals. The slope in the equation y = 6x+3 is 6. So, the slope of the line perpendicular to it would be -1/6.

Now you have the slope and a point (12, -4) through which the line passes. Use the point-slope formula which is y - y1 = m(x - x1). Here m is the slope and (x1, y1) is the point. So, the equation becomes y - (-4) = -1/6*(x - 12). Simplify this equation to get the answer. You will end up with option a) y = –1/6x + 2.

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User Longneck
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