Find the point on the line 4 x + 5 y + 6 =0 which is closest to the point ( 4, 3 ).

can anyone solve this so i can see how it is done correctly. not really sure

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- Oct 5th 2007, 07:44 PMpseizure2000Finding a point on a line closest to a given point
Find the point on the line 4 x + 5 y + 6 =0 which is closest to the point ( 4, 3 ).

can anyone solve this so i can see how it is done correctly. not really sure - Oct 5th 2007, 09:52 PMTKHunny
You can calculate the distance directly.

$\displaystyle \frac{|4(4)+5(3)+6|}{\sqrt{4^{2}+5^{2}}}$

...but that doesn't quite find the point. However, given that distance, you can construct a circle on on the point, using the distance as radius. Find the intersection of the line and the circle.

A little easier would be just to use lines.

1) Find the slope of the line.

2) Realize that the slope of the perpendicular is the negative reciprocal of that slope.

3) Realize that the closest point on the line is at the perpendicular through the remote point.

4) Construct the equation of the line through your point and perpendicular to the line.

6) Solve the two-equation system for x and y. - Oct 5th 2007, 09:55 PMticbol
Here is one way.

The point on the line closest to (4,3) is the point of intersection of the normal line from (4,3) and the said line.

So, the slopes of the said line and the normal line are negative reciprocals.

Slope of the said line:

4x +5y +6 = 0 -----------------(i)

[y = mx +b]

5y = -4x -6

y = -(4/5)x -6/5 -------------(ii)

So, m1 = -4/5

Hence, m2 = 5/4 <------the slope of the normal line.

The equation of the normal line, using the point-slope form, is:

(y-3) = (5/4)(x-4)

y -3 = (5/4)x -5

y = (5/4)x -2 ----------------(2)

At the intersection of lines (ii) and (2), their y's are the same.

So,

-(4/5)x -6/5 = (5/4)x -2

Clear the fractions, multiply both sides by 4*5,

-16x -24 = 25x -40

-16x -25x = -40 +24

-41x = -16

x = 16/41 --------***

Hence,

y = (5/4)x -2 ----------------(2)

y = (5/4)(16/41) -2

y = 20/41 -2

y = (20 -82)/41

y = -62/41 -------------***

Check,

4x +5y +6 = 0 ------------------------(i)

4(16/41) +5(-62/41) +6 =? 0

64/41 -310/41 +6 =? 0

-246/41 +6 =? 0

-6 +6 =? 0

0 =? 0

Yes, so, OK.

Therefore, (16/41,-62/41) is the point on the line 4x +5y +6 = 0 that is closest to (4,3). ----------------------answer. - Oct 6th 2007, 12:06 AMearboth
Hi,

here is another approach:

All points on the given line have the coordinates $\displaystyle P\left(x, -\frac45 x - \frac65\right)$

Use the distance formula to calculate the distance between Q(4, 3) and P:

$\displaystyle d^2 = (x-4)^2 + \left(-\frac45 x - \frac65 - 3\right)^2 = \frac{41}{25} x^2 - \frac{32}{25} x + \frac{841}{25}$

You'll get the extreme distance (minimum or maximum) if the first derivation of (dē) equals zero:

$\displaystyle (d^2)' = \frac{82}{25} x - \frac{32}{25}$

$\displaystyle (d^2)' = 0~\implies~\frac{82}{25} x - \frac{32}{25}=0~\implies~x = \frac{32}{82}=\frac{16}{41}$

Plug in this value into the y-coordinate of P:$\displaystyle P\left(\frac{16}{41}, -\frac45 \cdot \left(\frac{16}{41} - \frac65\right)\right) = \left( \frac{16}{41}, -\frac{62}{41}\right)$ - Oct 6th 2007, 12:21 AMred_dog
Another solution:

Let $\displaystyle \displaystyle M\left(x,-\frac{4}{5}x-\frac{6}{5}\right)$ be an arbitrary point on the line.

The distance from $\displaystyle M$ to $\displaystyle P(4,3)$ is

$\displaystyle \displaystyle MP=\sqrt{(x-4)^2+\left(-\frac{4}{5}x-\frac{6}{5}-3\right)}=\sqrt{\frac{41}{25}x^2-\frac{32}{25}x+\frac{841}{25}}$

Then, $\displaystyle x_{min}=-\frac{b}{2a}=\frac{16}{41}$

Now, plug x in the equation of the line and solve for y.