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Write an equation of the given line passing through point P that is perpendicular to the given line. P(-8,0) , 3x-5y = 6

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Final answer:

The equation of the line perpendicular to 3x - 5y = 6 and passing through P(-8,0) is y = (-5/3)x + 40/3.

Step-by-step explanation:

To write an equation of a line that passes through point P(-8,0) and is perpendicular to the given line 3x - 5y = 6, we need to find the slope of the given line and then use the negative reciprocal of that slope for our perpendicular line. Start by solving the given equation for y to get it into slope-intercept form (y = mx + b), where m represents the slope. The given equation can be rewritten as:

y = (3/5)x - 6/5

The slope of the given line is 3/5. The slope (m) of the line perpendicular to this would be the negative reciprocal, so m = -5/3. Now, using the point-slope form of a line equation, which is y - y1 = m(x - x1), where (x1, y1) are the coordinates of the given point, we can plug in our values:

y - 0 = (-5/3)(x - (-8))

Simplifying the equation, we get:

y = (-5/3)x - (-5/3)(-8)

y = (-5/3)x + 40/3

Therefore, the equation of the line perpendicular to 3x - 5y = 6 and passing through point P(-8,0) is:

y = (-5/3)x + 40/3

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