Concepts! help with bijection and proof?

This is a terrible question but not impossible as

a*x_1 - b*y_1 = a*x_2 - b*y_2; x_1 = [a*x_2 - b*y_2 + b*y_1]/a
b*x_1 + a*y_1 = b*x_2 + a*y_2

let solve as x_1 = (b*x_2 + a*y_2 - a*y_1)/b

a*[(b*x_2 + a*y_2 - a*y_1)/b] - b*y_1 = a*x_2 - b*y_2

a*x_2 + (a^2/b)(y_2 - y_1) - b*y_1 = a*x_2 - b*y_2

(a^2/b)(y_2 - y_1) - b*y_1 = -b*y_2

(a^2/b)(y_2 - y_1) = b*y_1 - b*y_2

(a^2/b)(y_2 - y_1) = -b*(y_2 - y_1)

(a^2)*(y_2 - y_1) = -(b^2)*(y_2 - y_1)

(a^2)*(y_2 - y_1) + (b^2)*(y_2 - y_1) = 0

(y_2 - y_1)*[(a^2) + (b^2)] = 0

y_2 = y_1 therefore injective


for the last part

ax - by = c; bx + ay = d

solve for y of the first one as (a*x - c)/b = y

b*x + a*[(a*x - c)/b] = d

b*x + [(a^2)*x/b] - (a*c/b) = d

b*x + [(a^2)*x/b] = d + (a*c/b)

x*[b + ((a^2)/b)] = [d + (a*c/b)]

x = [d + (a*c/b)]/[b + ((a^2)/b)]

x = (b*d + a*c)/((a^2) + (b^2))

y = (a*x - c)/b -> (a*[(b*d + a*c)/((a^2) + (b^2))] - c)/b = (a*d - b*c)/[(a^2) + (b^2)]

f[(a*c + b*d)/((a^2) + (b^2)),(a*d - b*c)/((a^2) + (b^2))] as check for both equation which show that it is
true aka surjective

so which means both which means bijection

b) inverse is easy already solved as f^-1[(a*c + b*d)/((a^2) + (b^2)),(a*d - b*c)/((a^2) + (b^2))]

c) (a^2) + (b^2) = 1 is all the points on the unit circle...Show more