ax+by-a+b=0 and bx-ay-a-b=0 by cross multiplication
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Answer:
Ax + by = a-b becomes
(ax + by)/(a - b) = 1
bx - ay = a + b becomes
(bx-ay)/(a+b) = 1
So now we can equate the two:
(ax + by)/(a-b) = (bx - ay)/(a+b)
So by cross multiplication:
(ax + by)(a + b) = (bx - ay)(a - b)
a^2x + abx + aby + b^2y = abx - b^2x - a^2y + aby
a^2x + b^2y = -b^2x - a^2y
a^2x + b^2x = -a^2y - b^2y
(a^2 + b^2)x = -(a^2 + b^2)y
x = -y
Substituting into the first equation:
-ay + by = a-b
-(a - b)y = a-b
y = -1
So x = 1
To check: a(1) + b(-1) = a-b, b(1) - a(-1) = a + b
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