Question

Find the point on the line 6x + y = 8 that is closest to the point (−5, 2).(x, y) =​

Answer 1

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Answer:

  (x, y) = (31/37, 2 36/37) ≈ (0.8378, 2.9730)

Step-by-step explanation:

There are several ways we can go at this.

Method 1

Find the point of intersection with the perpendicular line through (-5, 2).

The perpendicular line will have the x- and y-coefficients swapped, one of them negated, and a new constant consistent with the given point.

  x -6y = (-5) -6(2) = -17

So, we're looking for the solution of the system of equations ...

• 6x +y = 8
• x -6y = -17

Using Cramer's Rule, we can find the solution to be ...

  x = ((1)(-17) -(-6)(8))/((1)(1) -(-6)(6)) = 31/37 ≈ 0.837_837*

  y = ((8)(1)-(-17)(6))/37 = 110/37 ≈ 2.972_972

The closest point is (x, y) = (31/37, 2 36/37).

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Method 2

We can write the given line in parametric form, then find the point on it such that the direction along the line is perpendicular to the direction to the given point.

Solving the given equation for y, we have ...

  y = 8 -6x

So, the parametric form of the given line can be written as ...

  (x, y) = (0, 8) +t(1, -6) . . . . for some parameter t

The direction to the given point from (x, y) is then ...

  (-5, 2) -(x, y) = (-5, 2) -((0, 8) +t(1, -6)) = (-5-0-t, 2-8+6t) = (-t-5, 6t-6)

The dot-product of this direction vector with the direction vector of the original line will be zero for the value of the parameter at the closest point.

  (1, -6) · (-t-5, 6t-6) = 0 = (1)(-t -5) +(-6)(6t -6)

  0 = -t -5 -36t +36 = -37t +31

  31/37 = t

So, filling this value into the parametric equation gives the closest point as ...

  (0, 8) +(31/37)(1, -6) = (31/37, 2 36/37) = (x, y)

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Method 3

Find the point of tangency of the given line with a circle centered at (-5, 2).

The equation of a circle centered at (-5, 2) is ...

  (x +5)^2 +(y -2)^2 = r^2

The given line will meet the circle at a point that satisfies this equation and the equation of the circle. Writing an expression for y using the given line, we can substitute into the circle equation to get ...

  (x +5)^ +((8 -6x) -2)^2 = r^2

  x^2 +10x +25 +36 -72x +36x^2 = r^2

  37x^2 -62x + (some constant) = 0

This will have one solution when the constant is chosen to make the left-side expression a perfect square. That perfect square will be ...

  37(x -31/37)^2 = 0

which is to say the tangent point of the line and the circle will have x-coordinate x = 31/37. The y-coordinate is 8 -6(31/37) = 2 36/37.

The closest point is (x, y) = (31/37, 2 36/37).

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* repeated decimal fractions are signified in typeset text with an overbar over the repeating digits. In plain text, we sometimes use a preceding underscore to identify the repeating digits.