SOLUTION: find an equation of the line containing the point (5,-1) and perpendicular to the line 5x-2y=10.

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Question 127416: find an equation of the line containing the point (5,-1) and perpendicular to the line 5x-2y=10.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!



First convert the standard equation 5x-2y=10 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)


5x-2y=10 Start with the given equation


5x-2y-5x=10-5x Subtract 5x from both sides


-2y=-5x%2B10 Simplify


%28-2y%29%2F%28-2%29=%28-5x%2B10%29%2F%28-2%29 Divide both sides by -2 to isolate y


y+=+%28-5x%29%2F%28-2%29%2B%2810%29%2F%28-2%29 Break up the fraction on the right hand side


y+=+%285%2F2%29x-5 Reduce and simplify


The original equation 5x-2y=10 (standard form) is equivalent to y+=+%285%2F2%29x-5 (slope-intercept form)


The equation y+=+%285%2F2%29x-5 is in the form y=mx%2Bb where m=5%2F2 is the slope and b=-5 is the y intercept.







Now let's find the equation of the line that is perpendicular to y=%285%2F2%29x-5 which goes through (5,-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 5%2F2, 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%285%2F2%29 So plug in the given slope to find the perpendicular slope



m%5Bp%5D=%28-1%2F1%29%282%2F5%29 When you divide fractions, you multiply the first fraction (which is really 1%2F1) by the reciprocal of the second



m%5Bp%5D=-2%2F5 Multiply the fractions.


So the perpendicular slope is -2%2F5



So now we know the slope of the unknown line is -2%2F5 (its the negative reciprocal of 5%2F2 from the line y=%285%2F2%29%2Ax-5). Also since the unknown line goes through (5,-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=%28-2%2F5%29%2A%28x-5%29 Plug in m=-2%2F5, x%5B1%5D=5, and y%5B1%5D=-1



y%2B1=%28-2%2F5%29%2Ax%2B%282%2F5%29%285%29 Distribute -2%2F5



y%2B1=%28-2%2F5%29%2Ax%2B10%2F5 Multiply



y=%28-2%2F5%29%2Ax%2B10%2F5-1Subtract -1 from both sides to isolate y

y=%28-2%2F5%29%2Ax%2B10%2F5-5%2F5 Make into equivalent fractions with equal denominators



y=%28-2%2F5%29%2Ax%2B5%2F5 Combine the fractions



y=%28-2%2F5%29%2Ax%2B1 Reduce any fractions

So the equation of the line that is perpendicular to y=%285%2F2%29%2Ax-5 and goes through (5,-1) is y=%28-2%2F5%29%2Ax%2B1


So here are the graphs of the equations y=%285%2F2%29%2Ax-5 and y=%28-2%2F5%29%2Ax%2B1




graph of the given equation y=%285%2F2%29%2Ax-5 (red) and graph of the line y=%28-2%2F5%29%2Ax%2B1(green) that is perpendicular to the given graph and goes through (5,-1)