Formula and proof of distance between two points

Formula and proof of distance between two points

Let the coordinates of point a be (x1, Y1)
The coordinates of point B are (X2, Y2)
AB=√（X1-X2）²+（Y1-Y2）²
Proof by Pythagorean theorem

How to work out the formula for the symmetric point of a point about the straight line y = KX + B when the coordinate of the point is known in the plane rectangular coordinate system
I want a formula, not a process

Given the coordinates of a point (x1, Y1)
If the symmetry point is set to (X2, Y2), then:
①[﹙y2-y1﹚/﹙x2-x1﹚]·k=-1
②﹙y1＋y2﹚/2=k[﹙x1＋x2﹚/2]＋b
The values of x 2 and y 2 can be obtained by solving the simultaneous equations

The coordinates of the symmetric point of P (2,5) with respect to the line x + y = 0 are ()
A. （5，2）B. （2，-5）C. （-5，-2）D. （-2，-5）

X + y = 0y = - XX = - y, so the symmetry point is (- 5, - 2), so choose C

Given the point P {a-2,3a-3}, and the distance from point P to the 2 coordinate axis is equal, the coordinates of point P are calculated

Because the distance from point P to axis 2 is equal
So | A-2 | = | 3a-3|
①a-2=3a-3
a=0.5
②a-2=-(3a-3)
a-2=-3a+3
a=1.25
So a = 0.5 or a = 1.25
So p (- 1.5, - 1.5) or P (- 0.75,0.75)