SOLUTION: write an equation of the line that contains the indicated point and meets the indicated condition(s). Write the final answer in the standard form Ax + By = C, A ≥ 0.

Algebra ->  Graphs -> SOLUTION: write an equation of the line that contains the indicated point and meets the indicated condition(s). Write the final answer in the standard form Ax + By = C, A ≥ 0.       Log On


   

Question 935263: write an equation of the line that contains the
indicated point and meets the indicated condition(s). Write the
final answer in the standard form Ax + By = C, A ≥ 0.
(0, -4); perpendicular to x + 3y = 9
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Ax+%2B+By+=+C, A+%3E=+0.
(0,-4); perpendicular to x+%2B+3y+=+9
3y+=-x%2B+9
y+=-%281%2F3%29x%2B+3

Solved by pluggable solver: Finding the Equation of a Line Parallel or Perpendicular to a Given Line


Remember, any two perpendicular lines are negative reciprocals of each other. So if you're given the slope of -1%2F3, you can find the perpendicular slope by this formula:

m%5Bp%5D=-1%2Fm where m%5Bp%5D is the perpendicular slope


m%5Bp%5D=-1%2F%28-1%2F3%29 So plug in the given slope to find the perpendicular slope



m%5Bp%5D=%28-1%2F1%29%283%2F-1%29 When you divide fractions, you multiply the first fraction (which is really 1%2F1) by the reciprocal of the second



m%5Bp%5D=3%2F1 Multiply the fractions.


So the perpendicular slope is 3



So now we know the slope of the unknown line is 3 (its the negative reciprocal of -1%2F3 from the line y=%28-1%2F3%29%2Ax%2B3). Also since the unknown line goes through (0,-4), we can find the equation by plugging in this info into the point-slope formula

Point-Slope Formula:

y-y%5B1%5D=m%28x-x%5B1%5D%29 where m is the slope and (x%5B1%5D,y%5B1%5D) is the given point



y%2B4=3%2A%28x-0%29 Plug in m=3, x%5B1%5D=0, and y%5B1%5D=-4



y%2B4=3%2Ax-%283%29%280%29 Distribute 3



y%2B4=3%2Ax-0 Multiply



y=3%2Ax-0-4Subtract -4 from both sides to isolate y

y=3%2Ax-4 Combine like terms

So the equation of the line that is perpendicular to y=%28-1%2F3%29%2Ax%2B3 and goes through (0,-4) is y=3%2Ax-4


So here are the graphs of the equations y=%28-1%2F3%29%2Ax%2B3 and y=3%2Ax-4




graph of the given equation y=%28-1%2F3%29%2Ax%2B3 (red) and graph of the line y=3%2Ax-4(green) that is perpendicular to the given graph and goes through (0,-4)





y+=3x-4in Ax+%2B+By+=+C form: 4=3x-y
or
3x-y+=4 where A=+3=>A+%3E=0