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Find the equation of the line that passes through the point (3,−5) and is perpendicular to the line y=1/5x-2

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To prove that the line y = -5x + 10 passes through the point (3,-5) and is perpendicular to the line y = 1/5x - 2, we need to show two things:

1. The point (3,-5) lies on the line y = -5x + 10.
2. The slope of the line y = -5x + 10 is -5, which is the negative reciprocal of the slope of the line y = 1/5x - 2.

1. To show that the point (3,-5) lies on the line y = -5x + 10, we can substitute x = 3 and y = -5 into the equation y = -5x + 10:

-5 = -5(3) + 10

-5 = -15 + 10

-5 = -5

Since the equation is true, the point (3,-5) lies on the line y = -5x + 10.

2. To show that the slope of the line y = -5x + 10 is -5, we can rewrite the equation in slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept:

y = -5x + 10

Comparing this to the general form of the slope-intercept equation, we see that the slope is -5. Therefore, the slope of the line y = -5x + 10 is -5.

Since we have shown that the point (3,-5) lies on the line y = -5x + 10 and the slope of the line is -5, which is the negative reciprocal of the slope of the line y = 1/5x - 2, we have proven that the line y = -5x + 10 passes through the point (3,-5) and is perpendicular to the line y = 1/5x - 2.
User Rifky Niyas
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