Find the equation of a line which passes through point (6, 8) and is perpendicular to the line formed by the equation y = 6x + 1. Express it in slope-intercept form.

Question

amistre64 · Answer

perp lines are 90 degrees from each other; and they have the inherent property that when you multiply their slopes together you get -1

anonymous · Answer

ok here he goes.

anonymous · Answer

haha

amistre64 · Answer

one simple way to get a perp line is to swith out the numbers on your x and y variables; and negate a sign

amistre64 · Answer

y = 6x + 1

6y = x + n; negate one of them like this

6y = -x + n ; we can solve for y by dividing out the 6 now

y = (-x+n)/6

amistre64 · Answer

the "n" is determined by the point being used; fill in the points and solve for n

amistre64 · Answer

y = (-x + n)/6;   (6,8)

8 = (-6 + n)/6

48 = -6 + n

42 = n

y = (-x+42)/6 whould do it

amistre64 · Answer

42 is wrong, I forgot how to add lol

amistre64 · Answer

y = (-x+54)/6

anonymous · Answer

or you can try this. we  know (as sensei said) that the slope if 
$$-\frac{1}{6}$$now use "point-slope" formula 
$$y-y_1=m(x-x_1)$$ with 
$$m=-\frac{1}{6}, x_1=6, y_1=8$$

amistre64 · Answer

another way to write it is:

y = (-1/6)x + 54/6
   = (-1/6)x + 9

anonymous · Answer

get 
$$y-8=-\frac{1}{6}(x-6)$$
$$y-8=-\frac{1}{6}x+1$$
$$y=-\frac{1}{6}x+9$$