Coordinate geometry question

Given 2 ponts P(1,3) and Q(3,2) find a point A on the x-axis so that AP + AQ is minimum.

Hi,

What is the length of the line AP? Or of the line AQ?

Draw a rough sketch and you will see that you can apply the Pythagorean theorem:

Consider A to be A(Ax,0), because it is on the x-axis.

By Pythagoras:

AQ^2 = (Qx - Ax)^2 + (Qy - Ay)^2 = (3 - Ax)^2 + (2 - 0)^2 = (3 - Ax)^2 + 4.

AP^2 = (Px - Ax)^2 + (Py - Ay)^2 = (1 - Ax)^2 + (3 - 0)^2 = (1 - Ax)^2 + 9.

So, taking square roots:

AP + AQ = root[(3 - Ax)^2 + 4] + root[(1 - Ax)^2 + 9].

Differentiate with respect to Ax:

d/d(Ax) [AP + AQ] = (Ax - 3) / root[(3 - Ax)^2 + 4] + (Ax - 1) / root[(1 - Ax)^2 + 9].

Set this equation equal to zero and solve:

Ax = 11 / 5.

Solution:

A = (Ax,0) = (11/5, 0).

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Minimum distance:

AP + AQ [min] = root[(3 - 11 / 5)^2 + 4] + root[(1 - 11 / 5)^2 + 9]

= root[(4 / 5)^2 + 4] + root[(-6 / 5)^2 + 9] = root [116 / 25] + root [261 / 25]

= (1 / 5) (root[116] + root[261]) = (1 / 5) (root[29 * 4] + root[29 * 9])

= (1 / 5) (2 root[29] + 3 root[29]) = (1 / 5) (5 root[29])

= root[29].

AP + AQ = sqrt[(x-1)^2 + 9] + sqrt[(3-x)^2 + 4]

To minimize this, take its derivative with respect to x, which is

(x-1)/sqrt[(x-1)^2 + 9] + (x-3)/sqrt[(3-x)^2 + 4]

Setting the derivative to zero gives

(x-1) sqrt[(3-x)^2 + 4] = (3-x) sqrt[(x-1)^2 + 9]

Square both sides;

(x^2 - 2x + 1) (13 - 6x + x^2) = (9 - 6x + x^2) (x^2 + 2x + 10)

When the polynomials are multiplied,
the x^4 term may be subtracted from both sides,
leaving a cubic equation,
which always has at least one real root.
If the cubic has no obvious factor,
you can find the real root by graphing,
then either trial and error or "Newton's method" if you've studied that.