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

1 Answer
Trevor Ryan.
Dec 29, 2015

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.

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