$\large A = \left(5,3 \right) ; B = \left(1 , 9 \right)$

$A$ and $B$ are two points on $xy$ -plane. Let $'M'$ be a point on the line $L \equiv 2x+3y - 6 =0$ such that $AM +BM$ is minimized. The coordinates of $M$ are $\left(x, y \right)$ . Find the value of $x+y$ .

The answer is 2.333333333.

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So in this problem we're given the points A and B which are (5,3) and (1,9) respectively. We are asked to find the sum of the x and y coordinates of a point M on the line given by $L\quad \equiv \quad 2x\quad +\quad 3y\quad -\quad 6$ such that $MA+MB$ is minimized.

We can first start by rewriting the equation for L and express $MA+MB$ as $M(A+B)$

In terms of y, the line L is

$y\quad =\quad 2\quad -\quad \frac { 2 }{ 3 } x$

This is all we need to solve this problem. We find that the y-intercept is (0,2) and the x-intercept is (3,0). The values of A and B are irrelevant to the solution of this problem. We can think of A, B, and M as vectors drawn from the origin. In order to minimize $M(A+B)$ we simply have to minimize M.

So if $\vec{M}$ is a vector with its tail at the origin then the smallest value of $\vec{M}$ will be where this vector is normal to the line L.

Here's a visual of the line L in blue and a red line representing the direction of $\vec{M}$ . We're only interested in the red line from the origin to its point of intersection with L.

As you can see $\vec{M}$ will be at a minimum where it intersects the hypotenuse at a right angle in the triangle formed by the y-intercept and x-intercept and x and y axes.

So now we just do a little bit of trigonometry to find the x and y components of this vector $\vec{M}$

The angle formed between the x-axis and L is simply going to be $\tan ^{ -1 }{ \frac { 2 }{ 3 } }$ which we'll call ${ \theta }_{ 1 }$ .

Since $\vec{M}$ is perpendicular to L, a second triangle is formed inside the larger one with $\vec{M}$ , line L and the x-axis and since we know two angles we can find the angle between $\vec{M}$ and the x-axis which is simply going to be $180-({ \theta }_{ 1 }+90)$ which we'll call ${ \theta }_{ 2 }$ .

As you can see the hypotenuse of this second triangle is 3. So the length of $\vec{M}$ is $3\sin { { \theta }_{ 1 } }$ and substituting $\tan ^{ -1 }{ \frac { 2 }{ 3 } }$ in for ${ \theta }_{ 1 }$

$\Longrightarrow \quad 3\sin { (\tan ^{ -1 }{ \frac { 2 }{ 3 } } ) }$

$\cong \quad 1.664$

We can now find the x and y components of $\vec{M}$ .

By drawing a vertical line perpendicular to the x-axis and intersecting at point M we have a similar triangle with the same angles with $\vec{M}$ being the hypotenuse of this smaller triangle. We already know that the angle between $\vec{M}$ and the x-axis is ${ \theta }_{ 2 }$ so we can use this to find the x and y components.

$x\quad =\quad \cos { { \theta }_{ 2 } }$

$y\quad =\quad \sin { { \theta }_{ 2 } }$

substituting for ${ \theta }_{ 2 }$

$x\quad =\quad 1.664\cos { (180-(\tan ^{ -1 }{ (\frac { 2 }{ 3 } )+90))\quad \cong \quad .923 } }$

$y\quad =\quad 1.664\sin { (180-(\tan ^{ -1 }{ (\frac { 2 }{ 3 } )+90))\quad \cong \quad 1.3845 } }$

$x\quad +\quad y\quad \cong \quad 2.3079$