SOLUTION: 1) how do you write in slope-intercept form of the line that passes through (6,-13) and is perpendicular to the graph 2x-9y=5? 2) (-3,1), y=1/3x+2 3) (6,-1), 3y+x=3

Algebra ->  Equations -> SOLUTION: 1) how do you write in slope-intercept form of the line that passes through (6,-13) and is perpendicular to the graph 2x-9y=5? 2) (-3,1), y=1/3x+2 3) (6,-1), 3y+x=3      Log On


   

Question 119821: 1) how do you write in slope-intercept form of the line that passes through (6,-13) and is perpendicular to the graph 2x-9y=5?

2) (-3,1), y=1/3x+2


3) (6,-1), 3y+x=3
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
#1






First convert the standard equation 2x-9y=5 into slope intercept form

Solved by pluggable solver: Converting Linear Equations in Standard form to Slope-Intercept Form (and vice versa)
Convert from standard form (Ax+By = C) to slope-intercept form (y = mx+b)


2x-9y=5 Start with the given equation


2x-9y-2x=5-2x Subtract 2x from both sides


-9y=-2x%2B5 Simplify


%28-9y%29%2F%28-9%29=%28-2x%2B5%29%2F%28-9%29 Divide both sides by -9 to isolate y


y+=+%28-2x%29%2F%28-9%29%2B%285%29%2F%28-9%29 Break up the fraction on the right hand side


y+=+%282%2F9%29x-5%2F9 Reduce and simplify


The original equation 2x-9y=5 (standard form) is equivalent to y+=+%282%2F9%29x-5%2F9 (slope-intercept form)


The equation y+=+%282%2F9%29x-5%2F9 is in the form y=mx%2Bb where m=2%2F9 is the slope and b=-5%2F9 is the y intercept.







Now let's find the equation of the line that is perpendicular to y=%282%2F9%29x-5%2F9 which goes through (6,-13)

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 2%2F9, 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%282%2F9%29 So plug in the given slope to find the perpendicular slope



m%5Bp%5D=%28-1%2F1%29%289%2F2%29 When you divide fractions, you multiply the first fraction (which is really 1%2F1) by the reciprocal of the second



m%5Bp%5D=-9%2F2 Multiply the fractions.


So the perpendicular slope is -9%2F2



So now we know the slope of the unknown line is -9%2F2 (its the negative reciprocal of 2%2F9 from the line y=%282%2F9%29%2Ax-5%2F9). Also since the unknown line goes through (6,-13), 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%2B13=%28-9%2F2%29%2A%28x-6%29 Plug in m=-9%2F2, x%5B1%5D=6, and y%5B1%5D=-13



y%2B13=%28-9%2F2%29%2Ax%2B%289%2F2%29%286%29 Distribute -9%2F2



y%2B13=%28-9%2F2%29%2Ax%2B54%2F2 Multiply



y=%28-9%2F2%29%2Ax%2B54%2F2-13Subtract -13 from both sides to isolate y

y=%28-9%2F2%29%2Ax%2B54%2F2-26%2F2 Make into equivalent fractions with equal denominators



y=%28-9%2F2%29%2Ax%2B28%2F2 Combine the fractions



y=%28-9%2F2%29%2Ax%2B14 Reduce any fractions

So the equation of the line that is perpendicular to y=%282%2F9%29%2Ax-5%2F9 and goes through (6,-13) is y=%28-9%2F2%29%2Ax%2B14


So here are the graphs of the equations y=%282%2F9%29%2Ax-5%2F9 and y=%28-9%2F2%29%2Ax%2B14




graph of the given equation y=%282%2F9%29%2Ax-5%2F9 (red) and graph of the line y=%28-9%2F2%29%2Ax%2B14(green) that is perpendicular to the given graph and goes through (6,-13)











#2


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%281%2F3%29 So plug in the given slope to find the perpendicular slope



m%5Bp%5D=%28-1%2F1%29%283%2F1%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=%281%2F3%29%2Ax%2B2). Also since the unknown line goes through (-3,1), 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-1=-3%2A%28x%2B3%29 Plug in m=-3, x%5B1%5D=-3, and y%5B1%5D=1



y-1=-3%2Ax%2B%283%29%28-3%29 Distribute -3



y-1=-3%2Ax-9 Multiply



y=-3%2Ax-9%2B1Add 1 to both sides to isolate y

y=-3%2Ax-8 Combine like terms

So the equation of the line that is perpendicular to y=%281%2F3%29%2Ax%2B2 and goes through (-3,1) is y=-3%2Ax-8


So here are the graphs of the equations y=%281%2F3%29%2Ax%2B2 and y=-3%2Ax-8




graph of the given equation y=%281%2F3%29%2Ax%2B2 (red) and graph of the line y=-3%2Ax-8(green) that is perpendicular to the given graph and goes through (-3,1)













#3








First convert the standard equation x%2B3y=3 into slope intercept form

Solved by pluggable solver: Converting Linear Equations in Standard form to Slope-Intercept Form (and vice versa)
Convert from standard form (Ax+By = C) to slope-intercept form (y = mx+b)


1x%2B3y=3 Start with the given equation


1x%2B3y-1x=3-1x Subtract 1x from both sides


3y=-1x%2B3 Simplify


%283y%29%2F%283%29=%28-1x%2B3%29%2F%283%29 Divide both sides by 3 to isolate y


y+=+%28-1x%29%2F%283%29%2B%283%29%2F%283%29 Break up the fraction on the right hand side


y+=+%28-1%2F3%29x%2B1 Reduce and simplify


The original equation 1x%2B3y=3 (standard form) is equivalent to y+=+%28-1%2F3%29x%2B1 (slope-intercept form)


The equation y+=+%28-1%2F3%29x%2B1 is in the form y=mx%2Bb where m=-1%2F3 is the slope and b=1 is the y intercept.







Now let's find the equation of the line that is perpendicular to y=%28-1%2F3%29x%2B1 which goes through (6,-1)

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%2B1). Also since the unknown line goes through (6,-1), 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%2B1=3%2A%28x-6%29 Plug in m=3, x%5B1%5D=6, and y%5B1%5D=-1



y%2B1=3%2Ax-%283%29%286%29 Distribute 3



y%2B1=3%2Ax-18 Multiply



y=3%2Ax-18-1Subtract -1 from both sides to isolate y

y=3%2Ax-19 Combine like terms

So the equation of the line that is perpendicular to y=%28-1%2F3%29%2Ax%2B1 and goes through (6,-1) is y=3%2Ax-19


So here are the graphs of the equations y=%28-1%2F3%29%2Ax%2B1 and y=3%2Ax-19




graph of the given equation y=%28-1%2F3%29%2Ax%2B1 (red) and graph of the line y=3%2Ax-19(green) that is perpendicular to the given graph and goes through (6,-1)