How do you write the slope-intercept equation for the line that is perpendicular to the line 3x-5y=9 and that contains the point (2,-5)?

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Answer 1 · 2016-06-06T21:23:21

# y = -9/5x - 1 2/5, " or " y= -9/5 -1.4#

First change the given equation into standard form, #y = mx + c#

#5y = 3x -9 , "which gives " y= 3/5x - 9/5#

The slope is therefore #3/5#
The slope perpendicular to this is #-5/3#

Now, we have the slope of the required line as well as a point.

The easiest way to find the equation is to substitute these into the formula: #y - y_1 = m(x - x_1)#

#y -(-5) = -9/5(x - 2)#

#y + 5 = -9/5x +18/5#

# y = -9/5x - 1 2/5, " or " y= -9/5 -1.4#