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What is an equation of the line that passes through the point (5,-8) and is perpendicular to the line 5x-4y=16.

User Galactus
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1 Answer

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To find the equation of the line that passes through the point (5,-8) and is perpendicular to the line 5x-4y=16, we need to follow these steps:

1: Rewrite the given equation in slope-intercept form:

5x - 4y = 16

-4y = -5x + 16

y = (5/4)x - 4

So the slope of the given line is 5/4.

2: The slope of a line perpendicular to this line will be the negative reciprocal of 5/4, which is -4/5.

3: Use the point-slope form of the equation of a line, y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line, to find the equation of the line:

y - (-8) = (-4/5)(x - 5)

y + 8 = (-4/5)x + (4/5)*5

y + 8 = (-4/5)x + 4

y = (-4/5)x - 4

Therefore, the equation of the line that passes through the point (5,-8) and is perpendicular to the line 5x-4y=16 is y = (-4/5)x - 4.

User UsmanJ
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