Find the equation of a line perpendicular to5, x, plus, y, equals, minus, 75x+y=−7that passes through the point left bracket, 5, comma, minus, 4, right bracket(5,−4).
Question
Find the equation of a line perpendicular to5, x, plus, y, equals, minus, 75x+y=−7that passes through the point left bracket, 5, comma, minus, 4, right bracket(5,−4).
Solution
The first step is to find the slope of the given line. The equation of the line is in the form Ax + By = C, which can be rewritten as y = -Ax/B - C/B. In this case, A=5 and B=1, so the slope of the given line is -A/B = -5/1 = -5.
Lines that are perpendicular to each other have slopes that are negative reciprocals of each other. Therefore, the slope of the line we are trying to find is the negative reciprocal of -5, which is 1/5.
The equation of a line is y = mx + b, where m is the slope and b is the y-intercept. We know that the line passes through the point (5, -4), so we can substitute these values into the equation to solve for b:
-4 = 1/5 * 5 + b -4 = 1 + b b = -4 - 1 b = -5
Therefore, the equation of the line is y = 1/5x - 5.
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