Solution of a compulsory double-line equation in senior high school mathematics

Let c (x3, y3) a (x 1, y 1) b (x2, y2) p (1, 0).

(x 1+X2)/2 = X3(y 1+Y2)/2 = Y3

And c is on Y=X/2, so Y3=X3 /2.

So (y1+y2)/2 = (x1+x2)/4 and: X 1=Y 1 X2= root number 3Y2.

Calculation: Y 1= (root number 3-2) Y2X 1 = (1- (root number 2 3)/3)X2.

And y-y0 = (y2-y1) (x-x0)/(x2-x1).

Let P( 1, 0) Y 1= (root number 3-2)Y2.

X 1=( 1-(2 root number 3)/3)X2X2= root number 3Y2 substitution.

Y-1= (root number 3 of 2)/2 x