Question

The lines $x=\frac{1}{4}y+a$ and $y=\frac{1}{4}x+b$ intersect at the point $(1,2)$, then $a+b=$

• A. $0$
• B. $1$
• C. $3$
• D. $\frac{9}{4}$

Answer 1

Answer: (D)

$\frac{9}{4}$

$x=\frac{1}{4}y+a---\left(1\right)$

$y=\frac{1}{4}x+b---\left(2\right)$

since lines $1$ & $2$ intersect at point $(1,2)$

So,

$1=\frac{1}{4}\times 2+a\Rightarrow 1=\frac{1}{2}+a\Rightarrow a=\frac{1}{2}\&2=\frac{1}{4}\times 1+b\Rightarrow b=2-\frac{1}{4}=\frac{7}{4}$

Threrefore $a+b=\frac{1}{2}+\frac{7}{4}=\frac{9}{4}$