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What is an equation of the line that passes through the point (-6, -7) and is perpendicular to the line 6x - 5y = 30?

 Jun 8, 2021
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Key idea:

1) Slope of the perpendicular line is oposite recipocal slope to the other

line equation  y  = mx + b  ,

perpendicular line y = (- 1/m)x  + b.

or

[Formula:  oposite recipocal    m1 * m2 = -1].

 

2) The given line equation  y = (6/5)x -30 .....(1).

slope of the given line = 6/5

Then oposite recipocal to (6/5)  =  negative of *(5/6) = - (5/6)

3) Get the  perpendicuar line equation by using the given point ( x= -6,  y= -7)

4)  y = mx + b    ,  -7 = (-5/6)x +b .....(2)

5) Find the  b value :  where x = 0   in (2)   .  Then  b =  -7.

The perpendicular line y = (oposite recipocal slope) *x + b

 Jun 8, 2021

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