Is P (3,5) on y=2x-1 closet to A (−2, 7)?

Question

After drawing the line y = 2x − 1 and marking the point A = (−2, 7), Kendall is trying to decide which point on the line is closest to A. The point P = (3, 5) looks promising. To check that P really is the point on y = 2x − 1 that is closest to A, what should Kendall do? Is P closest to A?

Answer 1

The point closest to A shall be on the line perpendicular to y = 2x - 1.

y = (-x/2) + b, point A is on this line

7 = [-(-2)/2] + b

7 = 1 + b

6 = b

y = (-x/2) + 6

If P (3, 5) is the closet point it shall be on that line.

Can you test and determine?

If P (

Answer 2

Yefim has explicitly shown that the closest point on the line is not P.

An alternative: show the point P is on the line \, and show that the line from P to A is orthogonal to the line y = 2x-1; thus, as the slope of the given line is 2, the slope of the orthogonal line is -1/2.

(3, 5) is on the line y = 2x - 1 as 2*3 - 1 = 5

The slope of the line from (3, 5) to (-2, 7) = (7 - 5)/(-2 - 3) = 2/-5 = -2/5. This is not -1/2. Therefore this is not the closest point.

Answer 3

Let f(x) = d² = (x + 2)² + (2x - 1 - 7)² = (x + 2)² + (2x - 8)² = x² + 4x + 4 + 4x² - 32x + 64 =

5x² - 28x + 68 = 5(x²- 28/5x + 196/25) - 196/5 + 68 = 5(x - 14/5)² + 144/5.

So, minimum distance happened if x = 14/5 y = 28/5 - 1 = 23/5.

Answer: (14/5, 23/5)