Question 9. Write the equation of a line that is PERPENDICULAR to 6x+3y=15 and passes through the coordinate (12,-5).
Answer: y = 1/2x – 11 Step-by-step explanation: Step one convert to y = mx+b form 6x + 3y =15 Isolate Y 3y = -6x + 15 Divde the 3 among the other numbers y = -2x + 5 That is the equation perpindiculr to that is y = 1/2x + 5 but solve for the points and plug in the points are (12,-5) So -5 = 1/2 (12) +b -5 = 6 + b Solve for B -5 – 6 = b b = -11 FINAL EQUATION y = 1/2x – 11 Log in to Reply
Answer:
y = 1/2x – 11
Step-by-step explanation:
Step one convert to y = mx+b form
6x + 3y =15
Isolate Y
3y = -6x + 15
Divde the 3 among the other numbers
y = -2x + 5
That is the equation
perpindiculr to that is y = 1/2x + 5 but solve for the points and plug in
the points are (12,-5)
So -5 = 1/2 (12) +b
-5 = 6 + b
Solve for B
-5 – 6 = b
b = -11
FINAL EQUATION
y = 1/2x – 11