What is the equation of the line perpendicular to #y=11/16x # that passes through # (3,-7) #?

1 Answer
Dec 29, 2015

Answer:

#y=-16/11x-29/11#.

Explanation:

Any straight line graph as equation #y=mx+c# where #m# is the slope or gradient and #c# is the y-intercept.

Furthermore, perpendicular lines have gradients whose products is #-1#.

So in this case, #y=11/16x # has gradient #m=11/16#.
So another line perpendicular to this one will have gradient #-16/11#.
Now substituting the point #(x,y)=(3,-7)# into its general equation #y=mx+c# yields

#-7=(-16/11)(3)+c#,

from which we obtain the value for #c=-29/11#.

Thus the required line has equation #y=-16/11x-29/11#.