SOLUTION: Consider the line y=7x+8. Find the equation of the line that is parallel to this line and passes through the point(5,9). Find the equation of the line that is perpendicul

Algebra ->  Graphs -> SOLUTION: Consider the line y=7x+8. Find the equation of the line that is parallel to this line and passes through the point(5,9). Find the equation of the line that is perpendicul      Log On


   

Question 98369: Consider the line y=7x+8.

Find the equation of the line that is parallel to this line and passes through the point(5,9).
Find the equation of the line that is perpendicular to this line and passes through the point (5,9).

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
"Find the equation of the line that is parallel to this line and passes through the point(5,9)."


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


Since any two parallel lines have the same slope we know the slope of the unknown line is 7 (its from the slope of y=7%2Ax%2B8 which is also 7). Also since the unknown line goes through (5,9), 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-9=7%2A%28x-5%29 Plug in m=7, x%5B1%5D=5, and y%5B1%5D=9



y-9=7%2Ax-%287%29%285%29 Distribute 7



y-9=7%2Ax-35 Multiply



y=7%2Ax-35%2B9Add 9 to both sides to isolate y

y=7%2Ax-26 Combine like terms

So the equation of the line that is parallel to y=7%2Ax%2B8 and goes through (5,9) is y=7%2Ax-26


So here are the graphs of the equations y=7%2Ax%2B8 and y=7%2Ax-26



graph of the given equation y=7%2Ax%2B8 (red) and graph of the line y=7%2Ax-26(green) that is parallel to the given graph and goes through (5,9)






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"Find the equation of the line that is perpendicular to this line and passes through the point(5,9)."


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 7, 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%287%2F1%29 So plug in the given slope to find the perpendicular slope



m%5Bp%5D=%28-1%2F1%29%281%2F7%29 When you divide fractions, you multiply the first fraction (which is really 1%2F1) by the reciprocal of the second



m%5Bp%5D=-1%2F7 Multiply the fractions.


So the perpendicular slope is -1%2F7



So now we know the slope of the unknown line is -1%2F7 (its the negative reciprocal of 7 from the line y=7%2Ax%2B8). Also since the unknown line goes through (5,9), 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-9=%28-1%2F7%29%2A%28x-5%29 Plug in m=-1%2F7, x%5B1%5D=5, and y%5B1%5D=9



y-9=%28-1%2F7%29%2Ax%2B%281%2F7%29%285%29 Distribute -1%2F7



y-9=%28-1%2F7%29%2Ax%2B5%2F7 Multiply



y=%28-1%2F7%29%2Ax%2B5%2F7%2B9Add 9 to both sides to isolate y

y=%28-1%2F7%29%2Ax%2B5%2F7%2B63%2F7 Make into equivalent fractions with equal denominators



y=%28-1%2F7%29%2Ax%2B68%2F7 Combine the fractions



y=%28-1%2F7%29%2Ax%2B68%2F7 Reduce any fractions

So the equation of the line that is perpendicular to y=7%2Ax%2B8 and goes through (5,9) is y=%28-1%2F7%29%2Ax%2B68%2F7


So here are the graphs of the equations y=7%2Ax%2B8 and y=%28-1%2F7%29%2Ax%2B68%2F7




graph of the given equation y=7%2Ax%2B8 (red) and graph of the line y=%28-1%2F7%29%2Ax%2B68%2F7(green) that is perpendicular to the given graph and goes through (5,9)