SOLUTION: Lines l and l′ with equations y=(11k+17)x+1 x=(61k−23)y−2 are perpendicular, where k is a constant. Find k. k = (Do NOT USE DECIMALS if k is a fraction or an ir

Algebra ->  Test -> SOLUTION: Lines l and l′ with equations y=(11k+17)x+1 x=(61k−23)y−2 are perpendicular, where k is a constant. Find k. k = (Do NOT USE DECIMALS if k is a fraction or an ir      Log On


   

Question 1170509: Lines l and l′ with equations
y=(11k+17)x+1
x=(61k−23)y−2
are perpendicular, where k is a constant. Find k.

k =
(Do NOT USE DECIMALS if k is a fraction or an irrational number)
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

Lines l and l′ with equations
y=%2811k%2B17%29x%2B1....slope: %2811k%2B17%29

x=%2861k-23%29y-2.........solve for y
y=x%2F%2861k-23%29%2B2
y=%281%2F%2861k-23%29%29x%2B2....slope: 1%2F%2861k-23%29

since lines are perpendicular, slopes are negative+reciprocal to each other

%2811k%2B17%29+=-1%2F%281%2F%2861k-23%29%29.....solve for+k
%2811k%2B17%29+=-%2861k-23%29
11k%2B17+=-61k%2B23
11k%2B61k+=-17%2B23
72k+=6
k+=6%2F72
k+=1%2F12

your lines are:
line l
y=%2811k%2B17%29x%2B1
y=%2811%281%2F12%29%2B17%29x%2B1
y=%28215%2F12%29x%2B1

line l’
y=%281%2F%2861k-23%29%29x%2B2
y=%281%2F%2861%281%2F12%29-23%29%29x%2B2
y=-%2812%2F215%29x%2B2