(4,0), x-3y=0?

Question

Write an equation of the line containing the speified point and perpindicular to the indicated line.


slope = -1/3x
y-0=-3(x-4)
y-0=-3x+12
y=-3x+12

Is this right?

Answer 1

Find the slope of the line you have
-3y = -x
y = 1/3 x

So the slope you need is -3.

I think you got it right but you were lucky. Where you say
"slope = -1/3x", it should have been positive because when the given equation is solved for y, you'll be dividing two negatives.

Since the rule for perpendiculars is that the slopes are reciprocals and have opposite signs, -3 was the correct slope to use.

Answer 2

You're close! The answer is y = 3x-12. It looks like you forgot that the slope of perpendicular lines is are the opposites AND reciprocals of each other.

For example:
Slope of line A is 4
Slope of a line perpendicular to line A is going to be the opposite (-4) reciprocal (-1/4).

Answer 3

equation id right...error in the method
x - 3y = 0 in y=mx+b form gives you y = 1/3x...so slope is 1/3...slope of line perp is neg reciprocal, which is -3...point slope gives you (y-0)= -3(x-4) so equation is y = -3x + 12

check by plugging in the point (0,4) ...does 0=-3(4) + 12 YES

Answer 4

No. The slope should be 3, not -3. Therefore y = 3x-12.