Algebra problem 1 by Dhaval Furia

Algebra Level pending

If a , b , x a, b, x and y y are real numbers, a 2 + b 2 = 25 , x 2 + y 2 = 169 , a x + b y = 65 , k = a y b x a^{2} + b^{2} = 25, x^{2} + y^{2} = 169, ax + by = 65, k = ay - bx , then _____

k > 5/13 k = 0 0 < k < 5/13 k = 5/13

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Dhaval Furia
May 14, 2020

(a^2 + b^2) * (x^2 + y^2) = 25 * 169

a^2 x^2 + a^2 y^2 + b^2 x^2 + b^2 y^2 = 5^2 * 13^2

(a^2 x^2 + b^2 y^2) + (a^2 y^2 + b^2 x^2) = 65^2

((ax+by)^2 - 2abxy) + ((ay-bx)^2 + 2abxy) = 65^2

65^2 - 2abxy + k^2 + 2abxy = 65^2

k^2 = 65^2 - 65^2 = 0

k = 0

Let a = 5 cos α , b = 5 sin α , x = 13 cos β , y = 13 sin β a=5\cos α, b=5\sin α,x=13\cos β,y=13\sin β . Then

a x + b y = 65 cos ( α β ) = 65 α = β ax+by=65\cos(α-β)=65\implies α=β .

k = 65 sin ( β α ) = 0 k=65\sin (β-α)=\boxed 0 .

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...