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What is the equation of a line that is perpendicular to y=3/7x and goes through the point (6,−3)(6,−3)?

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Final answer:

To find the equation of a line that is perpendicular to y=3/7x and goes through the point (6,-3), determine the slope of the given line, which is 3/7. Perpendicular lines have slopes that are negative reciprocals, so the slope of the line we are looking for is -7/3. Using the point-slope form of a linear equation, we can find the equation of the line as y + 3 = -7/3(x - 6).

Step-by-step explanation:

To find the equation of a line that is perpendicular to y=3/7x and goes through the point (6,-3), we first need to determine the slope of the given line. The equation y=3/7x is in the form y=mx, where m is the slope. In this case, the slope is 3/7.

Because perpendicular lines have slopes that are negative reciprocals of each other, the slope of the line we are looking for is -7/3.

Now that we have the slope and the point (6,-3), we can use the point-slope form of a linear equation: y - y1 = m(x - x1). Plugging in the values, we get y + 3 = -7/3(x - 6).

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