1 votes 1 votes Let $a,b,x,y$ be real numbers such that $a^{2}+b^{2}=25,x^{2}+y^{2}=169$, and $ax+by=65$. If $k=ay-bx$, then $k=0$ $0< k\leq \frac{5}{13}$ $k=\frac{5}{13}$ $k> \frac{5}{13}$ Quantitative Aptitude cat2019-2 quantitative-aptitude system-of-equations + – go_editor asked Mar 20, 2020 • retagged Mar 22, 2022 by Lakshman Bhaiya go_editor 13.9k points 711 views answer comment Share See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Given that, $a^{2}+b^{2}=25 \quad \longrightarrow (1)$ $x^{2}+y^{2}=169 \quad \longrightarrow (2) $ $ax+by = 65 \quad \longrightarrow (3)$ $ k=ay-bx \quad \longrightarrow (4)$ Multiply equation $(1)$ and equation $(2)$, we get $(a^{2}+b^{2}) (x^{2}+y^{2}) = 25 \ast 169$ $\Rightarrow a^{2}x^{2} + a^{2}y^{2} + b^{2}x^{2} + b^{2}y^{2} = 5^{2} \ast 13^{2}$ $ \Rightarrow a^{2}x^{2} + a^{2}y^{2} + b^{2}x^{2} + b^{2} y^{2}=(5 \ast 13)^{2}$ $ \Rightarrow a^{2}x^{2} +a^{2}y^{2} + b^{2}x^{2} + b^{2}y^{2} = (65)^{2}$ $ \Rightarrow a^{2}x^{2} + a^{2}y^{2} + b^{2}x^{2} + b^{2}y^{2} = (ax+by)^{2} \quad \left[\because \text {From equation (3)} \right]$ $ \Rightarrow a^{2}x^{2} + a^{2}y^{2} + b^{2}x^{2} + b^{2}y^{2} = a^{2}x^{2} + b^{2}y^{2} + 2abxy $ $ \Rightarrow a^{2}y^{2} + b^{2}x^{2} – 2abxy = 0$ $ \Rightarrow (ay – bx)^{2} = 0$ $ \Rightarrow ay – bx = 0 $ $ \Rightarrow \boxed{k=0} \quad \left[\because \text {From equation (4)}\right]$ Correct Answer : A Anjana5051 answered Jul 31, 2021 • edited Jul 31, 2021 by Anjana5051 Anjana5051 11.8k points comment Share See all 0 reply Please log in or register to add a comment.