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Find the equation of the line that is perpendicular to y =-3/8x-56 and goes through (12,2)?

User Hkk
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Final answer:

The equation of the line perpendicular to y = -3/8x - 56 and goes through the point (12,2) is y = (8/3)x - 30, determined by finding the negative reciprocal of the given line's slope and then using the point-slope form.

Step-by-step explanation:

To find the equation of a line that is perpendicular to a given line, you first need to determine the slope of the given line. The slope of the line y = -3/8x - 56 is -3/8. A perpendicular line will have a slope that is the negative reciprocal of this slope, which is 8/3.

Next, you use the point-slope form of a line's equation, which is y - y1 = m(x - x1), where m is the slope of the line and (x1, y1) is a point on the line. Using the point given (12, 2), and the perpendicular slope 8/3, we get:

y - 2 = (8/3)(x - 12).

To put this equation into slope-intercept form (y = mx + b), expand and simplify:

y - 2 = (8/3)x - (8/3)×12

y - 2 = (8/3)x - 32

Finally, add 2 to both sides to solve for y:

y = (8/3)x - 30

So, the equation of the line perpendicular to y = -3/8x - 56 and passing through the point (12, 2) is y = (8/3)x - 30.

User Ben Barreth
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