I'm confused with this:

Question

Let A = (−2,4) and B = (7,−2). Find the point Q on the line y = 2 that makes the total distance AQ + BQ as small as possible.

Answer 1

If you draw a sketch:

you can see that the way to get AQ + BQ as small as possible is to have point Q be on the line AB.

First, find the equation of the line between A and B:

Find the slope (m): m = (y₂ - y₁)/(x₂ - x₁) = (-2 - 4)/(7 - -2) = -6/9 = -2/3

Then use the point-slope form of a line, y - y₁ = m(x - x₁), to find the equation of the line:

y - 4 = -2/3(x - -2)

y -4 = -2/3x - 4/3

y = -2/3x - 4/3 + 4

y = -2/3x + 8/3

Now, since you know that the point on the line y = 2 will have a y value of 2, you plug in y = 2 into your equation:

2 = -2/3x + 8/3

2 - 8/3 = -2/3x

-2/3 = -2/3x

x = 1

So Point Q is the point is (1, 2)

Answer 2

The shortest distance between 2 points is a straight line, SO

First you need to find the line containinf both A and B. Start by finding the slope.

m = (4 - -2) / (-2 - 7) = 6 / (-9) = -(2/3) use this and one of the points to write the equation of the line.

y - 2 = - (2/3)(x - 4)

If the line goes through y = 2, substitute 2 for y

2 - 2 = -(2/3)(x - 4) multiply both sides by -(3/2)

0 = x - 4

x = 4

So the point is (4, 2)