What is an equation of the line that passes through the point (-6, -7) and is perpendicular to the line 6x - 5y = 30?
+287 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
