What is the equation of a line that passes through the point (10, 3) and is perpendicular to the line represented by the equation y = 5x + 7?

To find the equation of a line that passes through the point (10, 3) and is perpendicular to the line described by the equation y = 5x + 7, we first need to determine the slope of the original line.

The given line has a slope of 5. Two lines are said to be perpendicular if the product of their slopes is -1.

To find the slope of the perpendicular line, we can use the formula:

m1 × m2 = -1

Here, m1 is the slope of the original line (5), and m2 is the slope of the line we are trying to find. By rearranging the formula, we find:

m2 = -1/m1 = -1/5

Now that we have the slope of the new line, which is -1/5, we can use the point-slope form of the equation of a line to find the equation:

y – y1 = m(x – x1)

Here, (x1, y1) is the point (10, 3) through which the line passes, and m = -1/5.

Substituting the known values into the point-slope formula gives us:

y – 3 = -1/5(x – 10)

To express this equation in slope-intercept form (y = mx + b), we simplify:

First, distribute the slope:

y – 3 = -1/5x + 2

Then, add 3 to both sides:

y = -1/5x + 5

Therefore, the equation of the line that passes through the point (10, 3) and is perpendicular to the line y = 5x + 7 is:

y = -1/5x + 5

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