Anonymous
Anonymous asked in Science & MathematicsMathematics · 9 years ago

# Find the point on the line -2x+3y+4 =0 which is closest to the point ( 4, -5 )?

Find the point on the line -2x+3y+4 =0 which is closest to the point ( 4, -5 ).

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• 9 years ago

The shortest distance between a point and a line is the distance that is coincident with the line passing through the point and perpendicular to the line. So in this problem, our strategy will be to (1) find the equation of the perpendicular line passing through (4,-5), and then (2) find the point of intersection between it and the given line.

(1) To find the equation of the line perpendicular to L: -2x+3y+4=0, passing through (4, -5), we note we already have a point on that line, namely (4,-5), and so only need a slope of that line in order to find the equation. Using L's slope, we can find the slope of the line perpendicular by recognizing that perpendicular lines slopes are negative reciprocals. Thus, rewriting L in slope-intercept form, we obtain y = (2/3)x - (4/3), and the slope of L is 2/3. Thus the slope of the line perpendicular to L is -3/2.

Now, to find the equation of the line perpendicular to L passing through the point (4,-5), we use that point and the slope of the line, -3/2, to find that the equation of the line perpendicular to L is y = (-3/2) x + 1.

(2) Finally, we can find the the point of intersection of L and the line perpendicular. This can be found using substitution (or any other method of solving linear systems of two equations in two unknowns) to find x = 14/13. Thus, y = -8/13 and the point on L closest to (4,-5) is (14/13,-8/13).

To help reinforce your understanding of these concepts, I've searched and found a webpage and a video tutorial that address problems similar to this one, and I thought they might be helpful to you. I've listed them below.

As always, if you need more help, please clarify where you are in the process and what's giving you trouble. I'd be more than happy to continue to assist.

• 9 years ago

Any point to line -2x+3y+4 =0 has coordinates:

(x, (2x-4)/3 )

Let be a mobile point M (x, (2x-4)/3 ) to this line

Distance D between point (4, -5) and M is:

D^2 = (x-4)^2 + ((2x-4)/3 + 5)^2 =

x^2 - 8x + 16 + (2x+ 11)^2 / 9 =

(1+4/9)x^2 - x(8 - 44/9) + 16 + 121/9 =

(13/9)x^2 - (28/9)x + 265/9

For x = (28/9)/(2*13/9) = (28/9)*(9/26) = 14/13 you have a local minimum for this parabola

For x = 14/13 you get y = (2*(14/13)-4)/3 = (28/13- 52/13)/3 = -8/13

Closest point is

(x,y) = (14/13, -8/13)