In the xy coordinate plane the slope of line p is 1/2 and its x-intercept is -3. How do you find the equation of a perpendicular to p and intersects p at is x-intercept.?

1 Answer
Shwetank Mauria
Dec 29, 2016

#2x+y+6=0#

Explanation:

The equation of a line having a slope #m# and passing through point #(x_1,y_1)# is #(y-y_1)=m(x-x_1)#. As the slope of line is #1/2# and its #x#-intercept is #-3# (i.e. it passes through #(-3,0)#, its equation is

#(y-0)=1/2(x-(-3))# or #y=1/2x+3/2# i.e. #x-2y+3=0#.

As the slope of line #p# is #1/2#, slope of line perpendicular to it is #(-1)-:1/2=-1xx2=-2# and as it intersects #p# at its #x#-intercept, it tpp passes through #(-3,0)# and its equation will be

#(y-0)=-2(x-(-3))# or #y=-2x-6# i.e. #2x+y+6=0#
graph{(x-2y+3)(2x+y+6)=0 [-13.96, 6.04, -4.16, 5.84]}

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