Let (X & sup2; + Y & sup2;) (A & sup2; + B & sup2;) = (AX + by) & sup2;, and XY ≠ 0, ab ≠ 0, try vector method to prove X / a = Y / b

Let (X & sup2; + Y & sup2;) (A & sup2; + B & sup2;) = (AX + by) & sup2;, and XY ≠ 0, ab ≠ 0, try vector method to prove X / a = Y / b

The following (u.v) denotes u dot multiplied by V
= = = = = = = = =
Prove: let vector u = (x, y), vector V = (a, b)
Then u ^ 2 = x ^ 2 + y ^ 2,
v^2 = a^2 +b^2.
(u.v) = ax +by.
It is known that,
u^2 *v^2 =(u.v)^2.
That is | u | ^ 2 * | V | ^ 2 = | u | ^ 2 * | ^ 2 * (COS) ^ 2
Because XY ≠ 0, ab ≠ 0,
So u and V are all nonzero vectors
So cos = plus or minus 1
That is, vectors u and V are collinear
So there is a real number t such that
u =tv.
That is, x = TA,
y =tb.
So x / a = Y / b
= = = = = = = = =
Note the condition XY ≠ 0, ab ≠ 0
Otherwise, we can only get BX = ay