How do you write the slope-intercept equation of the line perpendicular to y = 5/2 x - 2, which passes though the point (0, 2)?

1 Answer
Sep 20, 2016

Answer:

#y=-2/5x+2#

Explanation:

Thinks to remember:
[1]#color(white)("XXX")color(green)(y=color(blue)(m)x+b)# is the slope-intercept form
#color(white)("XXXX")#with a slope of #color(blue)(m)# and a y-intercept of #color(red)(b)#

[2]#color(white)("XXX")#If a line has a slope of #color(blue)(m)#
#color(white)("XXXX")#all lines perpendicular to it have a slope of #color(blue)(""(-1/m))#

Based on this we know that:
given a line #y=color(blue)(5/2)x-2#
any line perpendicular to this will have a slope of #color(blue)(-2/5)#

and

any such perpendicular line will have the form:
#color(white)("XXX")y=color(blue)(-2/5)x+color(red)(b)#
where #color(red)(b)# is the y-intercept;
that is #color(red)(b)# is the value of #y# when #color(darkmagenta)(x=0)#

and

given #(color(darkmagenta)(x),color(red)y)=(color(darkmagenta)(0),color(red)(2))# is a point on the required perpendicular line
#color(white)("XXX")rArr color(red)(b)=color(red)(2)#

So the equation of the required perpendicular line is:
#color(white)("XXXXXXXX")color(green)(y=color(blue)(-2/5)x+color(red)(2)#