difficulty-s-1: provide that locus is straight line

[sujoy]

How are you? We haven't talked for a long time indeed! How each of you teachers (Ted, Williams, Denis, Eliz, Gene, Soraban, pka, and all teachers whose name I have missed) doing healthwise?

Here is my question:

Prove that the locus of a point which is always equidistant from two given points is a straight line, using co-ordinate geometry.

The solution is given as:

Let [x₁, y₁] and [x₂, y₂] be be the co-ordinates of the two given points A & B , and let [x, y] be the co-ordinate of any point P which is equidistant from them. Then:

. . .PA = PB, so [PA]^2 = [PB]^2

. . .so [x - x₁]² + [y - y₁]² = [x- x₂]² + [y- y₂]²

. . .so 2x[x₂ - x₁] + 2y[y₂ - y₁] + [x₁² + y₁² - x₂² - y₂²] = 0

As this last represents a first degree equation, it is a straight line.

My difficulty:

q1) When squared parameters are involved, how can we say this is a straight line?

q2) How do we get that these squared parameters are to be taken as constants to appropriately explain the facts?

Regards,
Sujoy

 

[stapel]

q1) The squares are constants. As long as the variables (x and y) are linear, the constants can be anything they like. You'll still get a straight line when you graph.

q2) The exercise says that A and B are fixed. This makes their coordinates constant.

Eliz.

 

[sujoy]

Thank you