# find the point on the line 6x+4y-6=0 which is the closest to the point (5,2)?

Relevance

the closest point will be on a line that is perpendicular to 6x + 4y - 6 = 0 and goes through point (5, 2)

6x + 4y - 6 = 0

4y = -6x + 6

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

slope of given line

m = -3/2

slope of perpendicular line

m = 2/3

y - 2 = (2/3)(x - 5)

y - 2 = (2/3)x - 10/3

y = (2/3)x - 4/3

3y = 2x - 4

-2x + 3y = -4

2x - 3y = 4

where the two lines intersect is your point

6x + 4y = 6

2x - 3y = 4

18x + 12y = 18

8x - 12y = 16

26x = 34

x = 17/13

2(17/13) - 3y = 4

-3y = 4 - (34/13)

-3y = 18/13

y = -18/39

y = -6/13

(17/13, -6/13)

• the non-calculus way:

the point on the line closest to (5 , 2) will be on the line through this point perpendicular to the given line

6x + 4y - 6 = 0

3x + 2y - 3 = 0

2y = -3x + 3

y = -3/2 x + 3/2 ==> slope of given line is -3/2

slope of perpendicular line will be 2/3

this will also be the slope between (5 , 2) and (x , -3/2 x + 3/2)

(-3/2 x + 3/2 - 2) / (x - 5) = 2/3

(-3/2 x - 1/2) / (x - 5) = 2/3

cross-multiply to give:

2(x - 5) = 3(-3/2 x - 1/2)

2x - 10 = -9/2 x - 3/2

2x + 9/2 x = 10 - 3/2

4/2 x + 9/2 x = 20/2 - 3/2

13/2 x = 17/2

13x = 17

x = 17 / 13

and when x = 17/13, then y = -3/2 (17/13) + 3/2

y = -51/26 +39 /26 = -12 / 26 = -6 / 13

the point closest to (5 , 2) on the given line is (17 / 13 , -6 / 13)

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another non-calculus way:

you're trying to minimize the distance between (5 , 2) and (x , -3/2 x + 3/2)

d = sqrt[(x - 5)^2 + (-3/2 x + 3/2 - 2)^2]

plot this and find the minimum point (somewhere in the neighborhood of 1.5 to set your windows...)

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or, the calculus way, but not as direct as the perpendicular distance technique

d = sqrt[(x - 5)^2 + (-3/2 x + 3/2 - 2)^2]

d^2 = (x - 5)^2 + (-3/2 x + 3/2 - 2)^2

d^2 = x^2 - 10x + 25 + 9/4 x^2 + 3/2 x + 1/4

d^2 = 13/4 x^2 - 17/2 x + 101/4

(d^2) ' = 13/2 x - 17/2

(d^2) " = 13/2, which is positive, confirming that this will be a min

(d^2) ' = 13/2x - 17/2

setting this to 0, 13/2 x - 17/2 = 0

13/2 x = 17/2

13x = 17

x = 17/13 (wow, the same answer, go figure!)

and plug in x = 17/13 to solve for y, which will again be -6/13

point closest to the given line is (17/13 , -6 / 13)

more than one way to skin a math cat

• You have got to be kidding. (5,2) definitely is not near the line.

The fastest way to get a line is plotting the points of x=0 & y=0

6x+4y-6=0 substitute y=0 6x+ (0) -6=0 6x =6 x=1 (1,0)

" x=0 (0)+4y -6=0 4y=6 y=1.5 (0,1.5)

Ploy these 2 points on a graph & you're a ways from (5,2)

• Line 6x + 4y - 6 = 0 -----> 3x + 2y - 3 = 0

Let Q(x,y) be point on line 3x + 2y - 3 = 0 that is closest to point P(5,2)

Shortest distance from point to line is perpendicular distance.

Therefore line PQ is perpendicular to line 3x + 2y - 3 = 0

Line PQ also passes through point P(5,2), so it has equation:

2(x-5) - 3(y-2) = 0

2x - 10 - 3y + 6 = 0

2x - 3y - 4 = 0

Point Q is intersection of the 2 lines:

3x + 2y - 3 = 0

2x - 3y - 4 = 0

Solving, we get:

x = 17/13, y = -6/13

Q(17/13, -6/13) is point on given line that is closest to P(5,2)

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