250 Followers Problem - Dedicated to my friend

Geometry Level 5

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

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


The answer is 2.333333333.

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2 solutions

Nick Smith
Nov 2, 2015

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 2 x + 3 y 6 L\quad \equiv \quad 2x\quad +\quad 3y\quad -\quad 6 such that M A + M B MA+MB is minimized.

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

In terms of y, the line L is

y = 2 2 3 x 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 ) M(A+B) we simply have to minimize M.

So if M \vec{M} is a vector with its tail at the origin then the smallest value of M \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 M \vec{M} . We're only interested in the red line from the origin to its point of intersection with L.

As you can see M \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 M \vec{M}

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

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

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

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

1.664 \cong \quad 1.664

We can now find the x and y components of M \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 M \vec{M} being the hypotenuse of this smaller triangle. We already know that the angle between M \vec{M} and the x-axis is θ 2 { \theta }_{ 2 } so we can use this to find the x and y components.

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

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

substituting for θ 2 { \theta }_{ 2 }

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

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

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

AM +BM will be minimum when A,M,b or a,M,B will be collinear. a and b are mirror images of A and B respectively about given line . We take second choice and find 'a' is (1,-3. Then using condition for collinearity we get point M as (1,4/3)

Yep Did It The Same Way

Prakhar Bindal - 5 years, 7 months ago

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