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What is the equation of the line that is perpendicular to the line y=3/5x+10 and passes through the point (15,-5)?

Oy=3/5x-20
Oy=-3/5x+20
O y=5/3x-20
y=-5/3x+20

User Ypx
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1 Answer

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Answer:

(d) y = -5/3x+20

Explanation:

You want the equation of the line that is perpendicular to the line y=3/5x+10 and passes through the point (15,-5).

Slope

The slopes of perpendicular lines are opposite reciprocals of one another. The slope (m') of the perpendicular line is ...

m' = -1/m . . . . . where m is the slope of the given line

Application

The given line's equation is in slope-intercept form. The slope of it is the coefficient of x: 3/5. Then the slope of the perpendicular line is ...

m' = -1/(3/5) = -5/3

Only one answer choice is an equation of a line with a slope of -5/3.

y = -5/3x +20

User Shuwei
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