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# Find the point on the line -6 x + 5 y - 6 =0 which is closest to the point ( 4, 5 ).

Hello seth,
When I plugged x back into the equation I got a reducible fraction which gave y = 330/61
Seth,

Also note that

330/61 ≅ 5.409836,

which is consistent with the Answer that my method gives.  So your point is (214/61, 330/61), in fractional form.  But also note Michael's comment about whether the on-line instructional tool that you use accepts fractions, or just decimal answers.

### 3 Answers by Expert Tutors

Michael J. | Effective High School STEM Tutor & CUNY Math Peer LeaderEffective High School STEM Tutor & CUNY ...
5.0 5.0 (5 lesson ratings) (5)
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5y = 6x + 6

y = (6/5)x + (6/5)

The point (x , (6/5)x + (6/5)) is closer to (4, 5).

d = √[(x - 4)2 + ((6/5)x + (6/5) - 5)2

d = √[(x2 - 8x + 16) + ((6/5)x - (19/5))2]

d = √[(x2 - 8x + 15) + (1.44x2 - 9.12x + 14.44)]

d = √(2.44x2 - 17.12x - 29.44)

Take the derivative of this and set it equal to zero.

4.88x - 17.12 = 0

Solving for x,

4.88x = 17.12

x = 214/68

This confirms that Christopher's work is correct.   If the program you are using says it is wrong, then try using entering decimal numbers rather than the fractional answer.
Arturo O. | Experienced Physics Teacher for Physics TutoringExperienced Physics Teacher for Physics ...
5.0 5.0 (66 lesson ratings) (66)
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Another way to solve this, without calculus:

Find the equation of the line perpendicular to the given line, and passing through (4,5).  Its slope must be

m = -1/(6/5) = -5/6

y = mx + b = (-5/6)x + b

Find b from the given point.

5 = -(5/6)(4) + b
b = ?

You will have the equation of a line perpendicular to the given line and passing through (4,5).  The intersection of the given line and the perpendicular line is the point on the given line that is closest to (4,5).  Just solve the system of 2 linear equations in x, y to get the intersection.
Christopher S. | Professionally Trained Math and Physics TutorProfessionally Trained Math and Physics ...
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So here you want to minimize the distance between the point (4,5) and some point (x,y) that exists on the line. We know that the distance formula is D = sqrt((x-4)^2 + (y-5)^2), from this relationship and the equation of the line above we can get d to be a function of either x alone or y alone. from there we take the derivative and set it equal to zero to find the critical points. This is done easiest using implicit differentiation:
2Dd(D)/dx = 2(x-4) + 2(y-5)dy/dx, solve for d(D)/dx
d(D)/dx = ((x-4) + (y-5)dy/dx)/D, and set this to zero
(x-4) + (y-5)dy/dx = 0, note that I didn't divide by D since I can just multiply it out once we've set the equation to zero
Now plug in y and dy/dx and solve for x
x-4 + (36/25)x + (36/25) - 6 = 0 , here dy/dx = 6/5, and y = (6/5)(x + 1)
solving this for x we get: x = 214/61
Then plug this back into the y equation and you will get the point on the line which is closest

when i plugged it back in i got the point to be (214/61,918/305). It says it is still wrong

Seth,

I got the same value for x as Christopher.  In decimal form, I got

x ≅ 3.508197

Christopher got

214/61 ≅ 3.508197

Plug this into

y = (6/5)x + (6/5)

and get

y ≅ 5.409836

I got

y = 5.409836

by the different method I described in my Answer above.  Try these numbers.  It looks like you already have the correct value for x, but I think your y value is incorrect.